Bibliotheca Polyglotta

BOOK X.

DEFINITIONS.

1. Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure.

2. Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure.

3. With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational.

4. And let the square on the assigned straight line be called rational and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them.

PROPOSITION 1.

Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the lesser magnitude set out.

Let AB, C be two unequal magnitudes of which AB is the greater: I say that, if from AB there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C.

For C if multiplied will sometime be greater than AB. [cf. v. Def. 4] Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE.

Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE. Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA. And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK. But DF is equal to C; therefore C is also greater than AK. Therefore AK is less than C.

Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C. Q. E. D. And the theorem can be similarly proved even if the parts subtracted be halves.

PROPOSITION 2.

If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable.

For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it; I say that the magnitudes AB, CD are incommensurable.

For, if they are commensurable, some magnitude will measure them. Let a magnitude measure them, if possible, and let it be E; let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E. Suppose this done, and let there be left AG less than E. Then, since E measures AB, while AB measures DF, therefore E will also measure FD. But it measures the whole CD also; therefore it will also measure the remainder CF. But CF measures BG; therefore E also measures BG. But it measures the whole AB also; therefore it will also measure the remainder AG, the greater the less: which is impossible. Therefore no magnitude will measure the magnitudes AB, CD; therefore the magnitudes AB, CD are incommensurable.

Therefore etc. [X. Def. 1]

PROPOSITION 3.

Given two commensurable magnitudes, to find their greatest common measure.

Let the two given commensurable magnitudes be AB, CD of which AB is the less; thus it is required to find the greatest common measure of AB, CD.

Now the magnitude AB either measures CD or it does not. If then it measures it—and it measures itself also—AB is a common measure of AB, CD. And it is manifest that it is also the greatest; for a greater magnitude than the magnitude AB will not measure AB.

Next, let AB not measure CD. Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable; [cf. X. 2] let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE.

Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB. But it measures itself also; therefore AF will also measure the whole AB. But AB measures DE; therefore AF will also measure ED. But it measures CE also; therefore it also measures the whole CD. Therefore AF is a common measure of AB, CD. I say next that it is also the greatest.

For, if not, there will be some magnitude greater than AF which will measure AB, CD. Let it be G. Since then G measures AB, while AB measures ED, therefore G will also measure ED. But it measures the whole CD also; therefore G will also measure the remainder CE. But CE measures FB; therefore G will also measure FB. But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible. Therefore no magnitude greater than AF will measure AB, CD; therefore AF is the greatest common measure of AB, CD.

Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found. Q. E. D.

PORISM.
From this it is manifest that, if a magnitude measure two magnitudes, it will also measure their greatest common measure.

PROPOSITION 4.

Given three commensurable magnitudes, to find their greatest common measure.

Let A, B, C be the three given commensurable magnitudes; thus it is required to find the greatest common measure of A, B, C.

Let the greatest common measure of the two magnitudes A, B be taken, and let it be D; [X. 3] then D either measures C, or does not measure it. First, let it measure it. Since then D measures C, while it also measures A, B, therefore D is a common measure of A, B, C. And it is manifest that it is also the greatest; for a greater magnitude than the magnitude D does not measure A, B.

Next, let D not measure C. I say first that C, D are commensurable.

For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A, B also; so that it will also measure the greatest common measure of A, B, namely D. [X. 3, Por.] But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable. Now let their greatest common measure be taken, and let it be E. [X. 3] Since then E measures D, while D measures A, B, therefore E will also measure A, B. But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C. I say next that it is also the greatest.

For, if possible, let there be some magnitude F greater than E, and let it measure A, B, C. Now, since F measures A, B, C, it will also measure A, B, and will measure the greatest common measure of A, B. [X. 3, Por.] But the greatest common measure of A, B is D; therefore F measures D. But it measures C also; therefore F measures C, D; therefore F will also measure the greatest common measure of C, D. [X. 3, Por.] But that is E; therefore F will measure E, the greater the less: which is impossible. Therefore no magnitude greater than the magnitude E will measure A, B, C; therefore E is the greatest common measure of A, B, C if D do not measure C, and, if it measure it, D is itself the greatest common measure.

Therefore the greatest common measure of the three given commensurable magnitudes has been found.

PORISM.
From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure.

Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended. Q. E. D.

PROPOSITION 5.

Commensurable magnitudes have to one another the ratio which a number has to a number.

Let A, B be commensurable magnitudes; I say that A has to B the ratio which a number has to a number.

For, since A, B are commensurable, some magnitude will measure them. Let it measure them, and let it be C. And, as many times as C measures A, so many units let there be in D; and, as many times as C measures B, so many units let there be in E.

Since then C measures A according to the units in D, while the unit also measures D according to the units in it, therefore the unit measures the number D the same number of times as the magnitude C measures A; therefore, as C is to A, so is the unit to D; [VII. Def. 20] therefore, inversely, as A is to C, so is D to the unit. [cf. V. 7, Por.] Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the unit to E. But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to B, so is the number D to E. [V. 22]

Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E. Q. E. D.

PROPOSITION 6.

If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable.

For let the two magnitudes A, B have to one another the ratio which the number D has to the number E; I say that the magnitudes A, B are commensurable.

For let A be divided into as many equal parts as there are units in D, and let C be equal to one of them; and let F be made up of as many magnitudes equal to C as there are units in E.

Since then there are in A as many magnitudes equal to C as there are units in D, whatever part the unit is of D, the same part is C of A also; therefore, as C is to A, so is the unit to D. [VII. Def. 20] But the unit measures the number D; therefore C also measures A. And since, as C is to A, so is the unit to D, therefore, inversely, as A is to C, so is the number D to the unit. [cf. V. 7, Por.] Again, since there are in F as many magnitudes equal to C as there are units in E, therefore, as C is to F, so is the unit to E. [VII. Def. 20] But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to F, so is D to E. [v. 22] But, as D is to E, so is A to B; therefore also, as A is to B, so is it to F also. [V. 11] Therefore A has the same ratio to each of the magnitudes B, F; therefore B is equal to F. [V. 9] But C measures F; therefore it measures B also. Further it measures A also; therefore C measures A, B. Therefore A is commensurable with B.

Therefore etc.

PORISM.
From this it is manifest that, if there be two numbers, as D, E, and a straight line, as A, it is possible to make a straight line [F] such that the given straight line is to it as the number D is to the number E.
And, if a mean proportional be also taken between A, F, as B, as A is to F, so will the square on A be to the square on B, that is, as the first is to the third, so is the figure on the first to that which is similar and similarly described on the second. [VI. 19, Por.]
But, as A is to F, so is the number D to the number E;
therefore it has been contrived that, as the number D is to the number E, so also is the figure on the straight line A to the figure on the straight line B.
Q. E. D.

PROPOSITION 7.

Incommensurable magnitudes have not to one another the ratio which a number has to a number.

Let A, B be incommensurable magnitudes; I say that A has not to B the ratio which a number has to a number.

For, if A has to B the ratio which a number has to a number, A will be commensurable with B. [X. 6] But it is not; therefore A has not to B the ratio which a number has to a number.

Therefore etc.

PROPOSITION 8.

If two magnitudes have not to one another the ratio which a number has to a number, the magnitudes will be incommensurable.

For let the two magnitudes A, B not have to one another the ratio which a number has to a number; I say that the magnitudes A, B are incommensurable.

For, if they are commensurable, A will have to B the ratio which a number has to a number. [X. 5] But it has not; therefore the magnitudes A, B are incommensurable.

Therefore etc.

PROPOSITION 9.

The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.

For let A, B be commensurable in length; I say that the square on A has to the square on B the ratio which a square number has to a square number.

For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [X. 5] Let it have to it the ratio which C has to D. Since then, as A is to B, so is C to D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, for similar figures are in the duplicate ratio of their corresponding sides; [VI. 20, Por.] and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, for between two square numbers there is one mean proportional number, and the square number has to the square number the ratio duplicate of that which the side has to the side; [VIII. 11] therefore also, as the square on A is to the square on B, so is the square on C to the square on D.

Next, as the square on A is to the square on B, so let the square on C be to the square on D; I say that A is commensurable in length with B.

For since, as the square on A is to the square on B, so is the square on C to the square on D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, therefore also, as A is to B, so is C to D. Therefore A has to B the ratio which the number C has to the number D; therefore A is commensurable in length with B. [X. 6]

Next, let A be incommensurable in length with B; I say that the square on A has not to the square on B the ratio which a square number has to a square number.

For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B. But it is not; therefore the square on A has not to the square on B the ratio which a square number has to a square number.

Again, let the square on A not have to the square on B the ratio which a square number has to a square number; I say that A is incommensurable in length with B.

For, if A is commensurable with B, the square on A will have to the square on B the ratio which a square number has to a square number. But it has not; therefore A is not commensurable in length with B.

Therefore etc.

PORISM.
And it is manifest from what has been proved that straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always commensurable in length also.

LEMMA. [It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number, [VIII. 26] and that, if two numbers have to one another the ratio which a square number has to a square number, they are similar plane numbers. [Converse of VIII. 26] And it is manifest from these propositions that numbers which are not similar plane numbers, that is, those which have not their sides proportional, have not to one another the ratio which a square number has to a square number. For, if they have, they will be similar plane numbers: which is contrary to the hypothesis. Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.]

[PROPOSITION 10.

To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line.

Let A be the assigned straight line; thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A.

Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers; and let it be contrived that, as B is to C, so is the square on A to the square on D – for we have learnt how to do this — [X. 6, Por.] therefore the square on A is commensurable with the square on D. [X. 6] And, since B has not to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number; therefore A is incommensurable in length with D. [X. 9] Let E be taken a mean proportional between A, D; therefore, as A is to D, so is the square on A to the square on E. [V. Def. 9] But A is incommensurable in length with D; therefore the square on A is also incommensurable with the square on E; [X. 11] therefore A is incommensurable in square with E.

Therefore two straight lines D, E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A.]

PROPOSITION 11.

If four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable with the fourth; and, if the first be incommensurable with the second, the third will also be incommensurable with the fourth.

Let A, B, C, D be four magnitudes in proportion, so that, as A is to B, so is C to D, and let A be commensurable with B; I say that C will also be commensurable with D.

For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X. 5] And, as A is to B, so is C to D; therefore C also has to D the ratio which a number has to a number; therefore C is commensurable with D. [X. 6]

Next, let A be incommensurable with B; I say that C will also be incommensurable with D.

For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7] And, as A is to B, so is C to D; therefore neither has C to D the ratio which a number has to a number; therefore C is incommensurable with D. [X. 8]

Therefore etc.

PROPOSITION 12.

Magnitudes commensurable with the same magnitude are commensurable with one another also.

For let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B.

For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [X. 5] Let it have the ratio which D has to E. Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. [X. 5] Let it have the ratio which F has to G. And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, let the numbers H, K, L be taken continuously in the given ratios; [cf. VIII. 4] so that, as D is to E, so is H to K, and, as F is to G, so is K to L.

Since, then, as A is to C, so is D to E, while, as D is to E, so is H to K, therefore also, as A is to C, so is H to K. [V. 11] Again, since, as C is to B, so is F to G, while, as F is to G, so is K to L, therefore also, as C is to B, so is K to L. [V. 11] But also, as A is to C, so is H to K; therefore, ex aequali, as A is to B, so is H to L. [V. 22] Therefore A has to B the ratio which a number has to a number; therefore A is commensurable with B. [X. 6]

Therefore etc. Q. E. D.

PROPOSITION 13.

If two magnitudes be commensurable, and the one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same.

Let A, B be two commensurable magnitudes, and let one of them, A, be incommensurable with any other magnitude C; I say that the remaining one, B, will also be incommensurable with C.

For, if B is commensurable with C, while A is also commensurable with B, A is also commensurable with C. [X. 12] But it is also incommensurable with it: which is impossible. Therefore B is not commensurable with C; therefore it is incommensurable with it.

Therefore etc.

LEMMA.
Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less.

Let AB, C be the given two unequal straight lines, and let AB be the greater of them; thus it is required to find by what square the square on AB is greater than the square on C.

Let the semicircle ADB be described on AB, and let AD be fitted into it equal to C; [IV. 1] let DB be joined. It is then manifest that the angle ADB is right, [III. 31] and that the square on AB is greater than the square on AD, that is, C, by the square on DB. [I. 47]

Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner.

Let AD, DB be the given two straight lines, and let it be required to find the straight line the square on which is equal to the sum of the squares on them. Let them be placed so as to contain a right angle, that formed by AD, DB; and let AB be joined. It is again manifest that the straight line the square on which is equal to the sum of the squares on AD, DB is AB. [I. 47] Q. E. D.

PROPOSITION 14.

If four straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line commensurable with the third. And, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line incommensurable with the third.

Let A, B, C, D be four straight lines in proportion, so that, as A is to B, so is C to D; and let the square on A be greater than the square on B by the square on E, and let the square on C be greater than the square on D by the square on F; I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F.

For since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. [VI. 22] But the squares on E, B are equal to the square on A, and the squares on D, F are equal to the square on C. Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D; therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; [V. 17] therefore also, as E is to B, so is F to D; [VI. 22] therefore, inversely, as B is to E, so is D to F. But, as A is to B, so also is C to D; therefore, ex aequali, as A is to E, so is C to F. [V. 22] Therefore, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. [X. 11]

Therefore etc.2

PROPOSITION 15.

If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.

For let the two commensurable magnitudes AB, BC be added together; I say that the whole AC is also commensurable with each of the magnitudes AB, BC.

For, since AB, BC are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC, it will also measure the whole AC. But it measures AB, BC also; therefore D measures AB, BC, AC; therefore AC is commensurable with each of the magnitudes AB, BC. [X. Def. 1]

Next, let AC be commensurable with AB; I say that AB, BC are also commensurable.

For, since AC, AB are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures CA, AB, it will also measure the remainder BC. But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. [X. Def. 1]

Therefore etc.

PROPOSITION 16.

If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.

For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC.

For, if CA, AB are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D. Since then D measures CA, AB, therefore it will also measure the remainder BC. But it measures AB also; therefore D measures AB, BC. Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [X. Def. 1] Similarly we can prove that AC, CB are also incommensurable. Therefore AC is incommensurable with each of the magnitudes AB, BC.

Next, let AC be incommensurable with one of the magnitudes AB, BC. First, let it be incommensurable with AB; I say that AB, BC are also incommensurable.

For, if they are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC. therefore it will also measure the whole AC. But it measures AB also; therefore D measures CA, AB. Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [X. Def. 1]

Therefore etc.

LEMMA.
If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application.

For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB.

This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectangle AC, CB.

Therefore etc.

PROPOSITION 17.

If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater. And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length.

Let A, BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma] and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

For let BC be bisected at the point E, and let EF be made equal to DE. Therefore the remainder DC is equal to BF. And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D, therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [II. 5] And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square on EC. But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE. And the square on BC is equal to four times the square on EC, for again BC is double of CE. Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A by the square on DF. It is to be proved that BC is also commensurable with DF. Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15] But CD is commensurable in length with CD, BF, for CD is equal to BF. [X. 6] Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC. It is to be proved that BD is commensurable in length with DC.

With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line commensurable with BC. Therefore BC is commensurable in length with FD, so that BC is also commensurable in length with the remainder, the sum of BF, DC. [X. 15] But the sum of BF, DC is commensurable with DC, [X. 6] so that BC is also commensurable in length with CD; [X. 12] and therefore, separando, BD is commensurable in length with DC. [X. 15]

Therefore etc.

PROPOSITION 18.

If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are incommensurable, the square on the greater will be greater than the square on the less by the square on a straight line incommensurable with the greater. And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable.

Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17] and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD. It is to be proved that BC is incommensurable in length with DF. Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16] But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16] And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure. Let this be the rectangle BD, DC. It is to be proved that BD is incommensurable in length with DC.

For, with the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC; therefore BC is incommensurable in length with FD, so that BC is also incommensurable with the remainder, the sum of BF, DC. [X. 16] But the sum of BF, DC is commensurable in length with DC; [X. 6] therefore BC is also incommensurable in length with DC, [X. 13] so that, separando, BD is also incommensurable in length with DC. [X. 16]

Therefore etc.

LEMMA.
[Since it has been proved that straight lines commensurable in length are always commensurable in square also, while those commensurable in square are not always commensurable in length also, but can of course be either commensurable or incommensurable in length, it is manifest that, if any straight line be commensurable in length with a given rational straight line, it is called rational and commensurable with the other not only in length but in square also, since straight lines commensurable in length are always commensurable in square also. But, if any straight line be commensurable in square with a given rational straight line, then, if it is also commensurable in length with it, it is called in this case also rational and commensurable with it both in length and in square; but, if again any straight line, being commensurable in square with a given rational straight line, be incommensurable in length with it, it is called in this case also rational but commensurable in square only.]

PROPOSITION 19.

The rectangle contained by rational straight lines commensurable in length is rational.

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in length; I say that AC is rational.

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4] And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC. And, as BD is to BC, so is DA to AC. [VI. 1] Therefore DA is commensurable with AC. [X. 11] But DA is rational; therefore AC is also rational. [X. Def. 4]

Therefore etc.

PROPOSITION 20.

If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.

For let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth; I say that BC is rational and commensurable in length with BA.

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4] But AC is also rational; therefore DA is commensurable with AC. And, as DA is to AC, so is DB to BC. [VI. 1] Therefore DB is also commensurable with BC; [X. 11] and DB is equal to BA; therefore AB is also commensurable with BC. But AB is rational; therefore BC is also rational and commensurable in length with AB.

Therefore etc.

PROPOSITION 21.

The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called medial.

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only; I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial.

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4] And, since AB is incommensurable in length with BC, for by hypothesis they are commensurable in square only, while AB is equal to BD, therefore DB is also incommensurable in length with BC. And, as DB is to BC, so is AD to AC; [VI. 1] therefore DA is incommensurable with AC. [X. 11] But DA is rational; therefore AC is irrational, so that the side of the square equal to AC is also irrational. [X. Def. 4] And let the latter be called medial. Q. E. D.

LEMMA.
If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.

Let FE, EG be two straight lines. I say that, as FE is to EG, so is the square on FE to the rectangle FE, EG.

For on FE let the square DF be described, and let GD be completed. Since then, as FE is to EG, so is FD to DG, [VI. 1] and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG, therefore, as FE is to EG, so is the square on FE to the rectangle FE, EG. Similarly also, as the rectangle GE, EF is to the square on EF, that is, as GD is to FD, so is GE to EF. Q. E. D.

PROPOSITION 22.

The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.

Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth; I say that CD is rational and incommensurable in length with CB.

For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only. [X. 21] Let the square on it be equal to GF. But the square on it is also equal to BD; therefore BD is equal to GF. But it is also equiangular with it; and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, proportionally, as BC is to EG, so is EF to CD. Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [VI. 22] But the square on CB is commensurable with the square on EG, for each of these straight lines is rational; therefore the square on EF is also commensurable with the square on CD. [X. 11] But the square on EF is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is rational. And, since EF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle FE, EG, [Lemma] therefore the square on EF is incommensurable with the rectangle FE, EG. [X. 11] But the square on CD is commensurable with the square on EF, for the straight lines are rational in square; and the rectangle DC, CB is commensurable with the rectangle FE, EG, for they are equal to the square on A; therefore the square on CD is also incommensurable with the rectangle DC, CB. [X. 13] But, as the square on CD is to the rectangle DC, CB, so is DC to CB; [Lemma] therefore DC is incommensurable in length with CB. [X. 11] Therefore CD is rational and incommensurable in length with CB. Q. E. D.

PROPOSITION 23.

A straight line commensurable with a medial straight line is medial.

Let A be medial, and let B be commensurable with A; I say that B is also medial.

For let a rational straight line CD be set out, and to CD let the rectangular area CE equal to the square on A be applied, producing ED as breadth; therefore ED is rational and incommensurable in length with CD. [X. 22] And let the rectangular area CF equal to the square on B be applied to CD, producing DF as breadth. Since then A is commensurable with B, the square on A is also commensurable with the square on B. But EC is equal to the square on A, and CF is equal to the square on B; therefore EC is commensurable with CF. And, as EC is to CF, so is ED to DF; [VI. 1] therefore ED is commensurable in length with DF. [X. 11] But ED is rational and incommensurable in length with DC; therefore DF is also rational [X. Def. 3] and incommensurable in length with DC. [X. 13] Therefore CD, DF are rational and commensurable in square only. But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable in square only is medial; [X. 21] therefore the side of the square equal to the rectangle CD, DF is medial. And B is the side of the square equal to the rectangle CD, DF; therefore B is medial.

PORISM.
From this it is manifest that an area commensurable with a medial area is medial.
[And in the same way as was explained in the case of rationals [Lemma following X. 18] it follows, as regards medials, that a straight line commensurable in length with a medial straight line is called medial and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also.
But, if any straight line be commensurable in square with a medial straight line, then, if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only.]

PROPOSITION 24.

The rectangle contained by medial straight lines commensurable in length is medial.

For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in length; I say that AC is medial.

For on AB let the square AD be described; therefore AD is medial. And, since AB is commensurable in length with BC, while AB is equal to BD, therefore DB is also commensurable in length with BC; so that DA is also commensurable with AC. [VI. 1, X. 11] But DA is medial; therefore AC is also medial. [X. 23, Por.] Q. E. D.

PROPOSITION 25.

The rectangle contained by medial straight lines commensurable in square only is either rational or medial.

For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in square only; I say that AC is either rational or medial.

For on AB, BC let the squares AD, BE be described; therefore each of the squares AD, BE is medial. Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth, to HM let there be applied the rectangular parallelogram MK equal to AC, producing HK as breadth, and further to KN let there be similarly applied NL equal to BE, producing KL as breadth; therefore FH, HK, KL are in a straight line. Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL, therefore each of the rectangles GH, NL is also medial. And they are applied to the rational straight line FG; therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22] And, since AD is commensurable with BE, therefore GH is also commensurable with NL. And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11] Therefore FH, KL are rational straight lines commensurable in length; therefore the rectangle FH, KL is rational. [X. 19] And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO. But, as DB is to BC, so is DA to AC, [VI. 1] and, as AB is to BO, so is AC to CO; [id.] therefore, as DA is to AC, so is AC to CO. But AD is equal to GH, AC to MK and CO to NL; therefore, as GH is to MK, so is MK to NL; therefore also, as FH is to HK, so is HK to KL; [VI. 1, V. 11] therefore the rectangle FH, KL is equal to the square on HK. [VI. 17] But the rectangle FH, KL is rational; therefore the square on HK is also rational. Therefore HK is rational. And, if it is commensurable in length with FG, HN is rational; [X. 19] but, if it is incommensurable in length with FG, KH, HM are rational straight lines commensurable in square only, and therefore HN is medial. [X. 21] Therefore HN is either rational or medial. But HN is equal to AC; therefore AC is either rational or medial.

Therefore etc.

PROPOSITION 26.

A medial area does not exceed a medial area by a rational area.

For, if possible, let the medial area AB exceed the medial area AC by the rational area DB, and let a rational straight line EF be set out; to EF let there be applied the rectangular parallelogram FH equal to AB, producing EH as breadth, and let the rectangle FG equal to AC be subtracted; therefore the remainder BD is equal to the remainder KH.

But DB is rational; therefore KH is also rational. Since, then, each of the rectangles AB, AC is medial, and AB is equal to FH, and AC to FG, therefore each of the rectangles FH, FG is also medial. And they are applied to the rational straight line EF; therefore each of the straight lines HE, EG is rational and incommensurable in length with EF. [X. 22] And, since [DB is rational and is equal to KH, therefore] KH is [also] rational; and it is applied to the rational straight line EF; therefore GH is rational and commensurable in length with EF. [X. 20] But EG is also rational, and is incommensurable in length with EF; therefore EG is incommensurable in length with GH. [X. 13] And, as EG is to GH, so is the square on EG to the rectangle EG, GH; therefore the square on EG is incommensurable with the rectangle EG, GH. [X. 11] But the squares on EG, GH are commensurable with the square on EG, for both are rational; and twice the rectangle EG, GH is commensurable with the rectangle EG, GH, for it is double of it; [X. 6] therefore the squares on EG, GH are incommensurable with twice the rectangle EG, GH; [X. 13] therefore also the sum of the squares on EG, GH and twice the rectangle EG, GH, that is, the square on EH [II. 4], is incommensurable with the squares on EG, GH. [X. 16] But the squares on EG, GH are rational; therefore the square on EH is irrational. [X. Def. 4] Therefore EH is irrational. But it is also rational: which is impossible.

Therefore etc. Q. E. D.

PROPOSITION 27.

To find medial straight lines commensurable in square only which contain a rational rectangle.

Let two rational straight lines A, B commensurable in square only be set out; let C be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as A is to B, so is C to D. [VI. 12]

Then, since A, B are rational and commensurable in square only, the rectangle A, B, that is, the square on C [VI.17], is medial. [X. 21] Therefore C is medial. [X. 21] And since, as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only. [X. 11] And C is medial; therefore D is also medial. [X. 23, addition]

Therefore C, D are medial and commensurable in square only. I say that they also contain a rational rectangle.

For since, as A is to B, so is C to D, therefore, alternately, as A is to C, so is B to D. [V. 16] But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D; therefore the rectangle C, D is equal to the square on B. But the square on B is rational; therefore the rectangle C, D is also rational.

Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle. Q. E. D.

PROPOSITION 28.

To find medial straight lines commensurable in square only which contain a medial rectangle.

Let the rational straight lines A, B, C commensurable in square only be set out; let D be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as B is to C, so is D to E. [VI. 12]

Since A, B are rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21] Therefore D is medial. [X. 21] And since B, C are commensurable in square only, and, as B is to C, so is D to E, therefore D, E are also commensurable in square only. [X. 11] But D is medial; therefore E is also medial. [X. 23, addition] Therefore D, E are medial straight lines commensurable in square only. I say next that they also contain a medial rectangle.

For since, as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E. [V. 16] But, as B is to D, so is D to A; therefore also, as D is to A, so is C to E; therefore the rectangle A, C is equal to the rectangle D, E. [VI. 16] But the rectangle A, C is medial; [X. 21] therefore the rectangle D, E is also medial.

Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q. E. D.

LEMMA I.
To find two square numbers such that their sum is also square. Let two numbers AB, BC be set out, and let them be either both even or both odd. Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [IX. 24, 26] therefore the remainder AC is even. Let AC be bisected at D. Let AB, BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers. Now the product of AB, BC together with the square on CD is equal to the square on BD. [II. 6] And the product of AB, BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [IX. 1] Therefore two square numbers, the product of AB, BC, and the square on CD, have been found which, when added together, make the square on BD.

And it is manifest that two square numbers, the square on BD and the square on CD, have again been found such that their difference, the product of AB, BC, is a square, whenever AB, BC are similar plane numbers. But when they are not similar plane numbers, two square numbers, the square on BD and the square on DC, have been found such that their difference, the product of AB, BC, is not square. Q. E. D.

LEMMA 2.
To find two square numbers such that their sum is not square.

For let the product of AB, BC, as we said, be square, and CA even, and let CA be bisected by D. It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See Lemma 1] Let the unit DE be subtracted; therefore the product of AB, BC together with the square on CE is less than the square on BD. I say then that the square product of AB, BC together with the square on CE will not be square.

For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the unit be divided. First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE. Since then the whole AC is double of the whole CD, and in them AG is double of DE, therefore the remainder GC is also double of the remainder EC; therefore GC is bisected by E. Therefore the product of GB, BC together with the square on CE is equal to the square on BE. [II. 6] But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE; therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE. And, if the common square on CE be subtracted, it follows that AB is equal to GB: which is absurd.

Therefore the product of AB, BC together with the square on CE is not equal to the square on BE. I say next that neither is it less than the square on BE. For, if possible, let it be equal to the square on BF, and let HA be double of DF. Now it will again follow that HC is double of CF; so that CH has also been bisected at F, and for this reason the product of HB, BC together with the square on FC is equal to the square on BF. [II. 6] But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BF. Thus the product of HB, BC together with the square on CF will also be equal to the product of AB, BC together with the square on CE: which is absurd. Therefore the product of AB, BC together with the square on CE is not less than the square on BE. And it was proved that neither is it equal to the square on BE. Therefore the product of AB, BC together with the square on CE is not square. Q. E. D.

PROPOSITION 29.

To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.] Let FB be joined.

Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF. [X. 6] But the square on AB is rational; [X. Def. 4] therefore the square on AF is also rational; [id.] therefore AF is also rational. And, since DC has not to CE the ratio which a square number has to a square number, neither has the square on BA to the square on AF the ratio which a square number has to a square number; therefore AB is incommensurable in length with AF. [X. 9] Therefore BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47] But CD has to DE the ratio which a square number has to a square number: therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF. [X. 9] And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB.

Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.

PROPOSITION 30.

To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater.

Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not square; [Lemma 2] let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, [X. 6, Por.] and let FB be joined.

Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47] But CD has not to DE the ratio which a square number has to a square number; therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF. [X. 9] And the square on AB is greater than the square on AF by the square on FB incommensurable with AB.

Therefore AB, AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB. Q. E. D.

PROPOSITION 31.

To find two medial straight lines commensurable in square only, containing a rational rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29] And let the square on C be equal to the rectangle A, B. Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial. [X. 21] Let the rectangle C, D be equal to the square on B. Now the square on B is rational; therefore the rectangle C, D is also rational. And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D. But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D. But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [X. 11] And C is medial; therefore D is also medial. [X. 23, addition] And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [X. 14]

Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C.

Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C, when the square on A is greater than the square on B by the square on a straight line incommensurable with A. [X. 30]

PROPOSITION 32.

To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on A is greater than the square on C by the square on a straight line commensurable with A, [X. 29] and let the square on D be equal to the rectangle A, B. Therefore the square on D is medial; therefore D is also medial. [X. 21] Let the rectangle D, E be equal to the rectangle B, C. Then since, as the rectangle A, B is to the rectangle B, C, so is A to C; while the square on D is equal to the rectangle A, B, and the rectangle D, E is equal to the rectangle B, C, therefore, as A is to C, so is the square on D to the rectangle D, E. But, as the square on D is to the rectangle D, E, so is D to E; therefore also, as A is to C, so is D to E. But A is commensurable with C in square only; therefore D is also commensurable with E in square only. [X. 11] But D is medial; therefore E is also medial. [X. 23, addition] And, since, as A is to C, so is D to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore also the square on D will be greater than the square on E by the square on a straight line commensurable with D.[X. 14] I say next that the rectangle D, E is also medial.

For, since the rectangle B, C is equal to the rectangle D, E, while the rectangle B, C is medial, [X. 21] therefore the rectangle D, E is also medial.

Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30]

LEMMA.
Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC.

And first that the rectangle CB, BD is equal to the square on BA.

For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8] And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [VI. 17]

For the same reason the rectangle BC, CD is also equal to the square on AC.

And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square on AD. [VI. 17] I say that the rectangle BC, AD is also equal to the rectangle BA, AC.

For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI. 4] Therefore the rectangle BC, AD is equal to the rectangle BA, AC. [VI. 16] Q. E. D.

PROPOSITION 33.

To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.

Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30] let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined.

Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB, while there has been applied to AB a parallelogram equal to the fourth part of the square on BC, that is, to the square on half of it, and deficient by a square figure, making the rectangle AE, EB, therefore AE is incommensurable with EB. [X. 18] And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE, while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable with the square on FB; therefore AF, FB are incommensurable in square. And, since AB is rational, therefore the square on AB is also rational; so that the sum of the squares on AF, FB is also rational. [I. 47] And since, again, the rectangle AE, EB is equal to the square on EF, and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD; therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF. But the rectangle AB, BC is medial; [X. 21] therefore the rectangle AB, EF is also medial. [X. 23, Por.] But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma] therefore the rectangle AF, FB is also medial. But it was also proved that the sum of the squares on these straight lines is rational.

Therefore two straight lines AF, FB incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial. Q. E. D.

PROPOSITION 34.

To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.

Let there be set out two medial straight lines AB, BC, commensurable in square only, such that the rectangle which they contain is rational, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 31, ad fin.] let the semicircle ADB be described on AB, let BC be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28] therefore AF is incommensurable in length with FB. [X. 18] Let FD be drawn from F at right angles to AB, and let AD, DB be joined.

Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [X. 11] But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to the square on DB; therefore the square on AD is also incommensurable with the square on DB. And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47] And, since BC is double of DF, therefore the rectangle AB, BC is also double of the rectangle AB, FD. But the rectangle AB, BC is rational; therefore the rectangle AB, FD is also rational. [X. 6] But the rectangle AB, FD is equal to the rectangle AD, DB; [Lemma] so that the rectangle AD, DB is also rational.

Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial, but the rectangle contained by them rational. Q. E. D.

PROPOSITION 35.

To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.

Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 32 , ad fin.] let the semicircle ADB be described on AB, and let the rest of the construction be as above.

Then, since AF is incommensurable in length with FB, [X. 18 ] AD is also incommensurable in square with DB. [X. 11 ] And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31 , I. 47 ] And, since the rectangle AF, FB is equal to the square on each of the straight lines BE, DF, therefore BE is equal to DF; therefore BC is double of FD, so that the rectangle AB, BC is also double of the rectangle AB, FD. But the rectangle AB, BC is medial; therefore the rectangle AB, FD is also medial. [X. 32, Por.] And it is equal to the rectangle AD, DB; [Lemma after X. 32 ] therefore the rectangle AD, DB is also medial. And, since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, [X. 13 ] so that the square on AB is also incommensurable with the rectangle AB, BE. [X. 11 ] But the squares on AD, DB are equal to the square on AB, [I. 47 ] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB.

Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Q. E. D.

PROPOSITION 36.

If two rational straight lines commensurable in square only be added together, the whole is irrational; and let it be called binomial.

For let two rational straight lines AB, BC commensurable in square only be added together; I say that the whole AC is irrational.

For, since AB is incommensurable in length with BC — for they are commensurable in square only — and, as AB is to BC, so is the rectangle AB, BC to the square on BC, therefore the rectangle AB, BC is incommensurable with the square on BC. [X. 11 ] But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [X. 6 ], and the squares on AB, BC are commensurable with the square on BC — for AB, BC are rational straight lines commensurable in square only — [X. 15 ] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13 ] And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [II. 4 ], is incommensurable with the sum of the squares on AB, BC. [X. 16 ] But the sum of the squares on AB, BC is rational; therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ] And let it be called binomial.

PROPOSITION 37.

If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational; and let it be called a first bimedial straight line.

For let two medial straight lines AB, BC commensurable in square only and containing a rational rectangle be added together; I say that the whole AC is irrational.

For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20] and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on AC [II. 4], is incommensurable with the rectangle AB, BC. [X. 16 ] But the rectangle AB, BC is rational, for, by hypothesis, AB, BC are straight lines containing a rational rectangle; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4 ] And let it be called a first bimedial straight line. Q. E. D.

PROPOSITION 38.

If two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a second bimedial straight line.

For let two medial straight lines AB, BC commensurable in square only and containing a medial rectangle be added together; I say that AC is irrational.

For let a rational straight line DE be set out, and let the parallelogram DF equal to the square on AC be applied to DE, producing DG as breadth. [I. 44 ] Then, since the square on AC is equal to the squares on AB, BC and twice the rectangle AB, BC, [II. 4 ] let EH, equal to the squares on AB, BC, be applied to DE; therefore the remainder HF is equal to twice the rectangle AB, BC. And, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial. But, by hypothesis, twice the rectangle AB, BC is also medial. And EH is equal to the squares on AB, BC, while FH is equal to twice the rectangle AB, BC; therefore each of the rectangle EH, HF is medial. And they are applied to the rational straight line DE; therefore each of the straight lines DH, HG is rational and incommensurable in length with DE. [X. 22 ] Since then AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11 ] But the sum of the squares on AB, BC is commensurable with the square on AB, [X. 15 ] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6 ] Therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC. [X. 13 ] But EH is equal to the squares on AB, BC, and HF is equal to twice the rectangle AB, BC. Therefore EH is incommensurable with HF, so that DH is also incommensurable in length with HG. [VI. 1 , X. 11 ] Therefore DH, HG are rational straight lines commensurable in square only; so that DG is irrational. [X. 36 ] But DE is rational; and the rectangle contained by an irrational and a rational straight line is irrational; [cf. X. 20 ] therefore the area DF is irrational, and the side of the square equal to it is irrational. [X. Def. 4 ] But AC is the side of the square equal to DF; therefore AC is irrational. And let it be called a second bimedial straight line. Q. E. D.

PROPOSITION 39.

If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational: and let it be called major.

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 33 ], be added together; I say that AC is irrational.

For, since the rectangle AB, BC is medial, twice the rectangle AB, BC is also medial. [X. 6 and 23, Por.] But the sum of the squares on AB, BC is rational; therefore twice the rectangle AB, BC is incommensurable with the sum of the squares on AB, BC, so that the squares on AB, BC together with twice the rectangle AB, BC that is, the square on AC, is also incommensurable with the sum of the squares on AB, BC; [X. 16 ] therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ] And let it be called major. Q. E. D.

PROPOSITION 40.

If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the side of a rational plus a medial area.

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34 ], be added together; I say that AC is irrational.

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC; so that the square on AC is also incommensurable with twice the rectangle AB, BC. [X. 16 ] But twice the rectangle AB, BC is rational; therefore the square on AC is irrational. Therefore AC is irrational. [X. Def. 4 ] And let it be called the side of a rational plus a medial area. Q. E. D.

PROPOSITION 41.

If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the side of the sum of two medial areas.

For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [X. 35 ] be added together; I say that AC is irrational.

Let a rational straight line DE be set out, and let there be applied to DE the rectangle DF equal to the squares on AB, BC, and the rectangle GH equal to twice the rectangle AB, BC; therefore the whole DH is equal to the square on AC. [II. 4 ] Now, since the sum of the squares on AB, BC is medial, and is equal to DF, therefore DF is also medial. And it is applied to the rational straight line DE; therefore DG is rational and incommensurable in length with DE. [X. 22 ] For the same reason GK is also rational and incommensurable in length with GF, that is, DE. And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, DF is incommensurable with GH; so that DG is also incommensurable with GK. [VI. 1 , X. 11 ] And they are rational; therefore DG, GK are rational straight lines commensurable in square only; therefore DK is irrational and what is called binomial. [X. 36 ] But DE is rational; therefore DH is irrational, and the side of the square which is equal to it is irrational. [X. Def. 4 ] But AC is the side of the square equal to HD; therefore AC is irrational. And let it be called the side of the sum of two medial areas. Q. E. D.

LEMMA.
And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question, we will now prove after premising the following lemma.

Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D, and let AC be supposed greater than DB; I say that the squares on AC, CB are greater than the squares on AD, DB.

For let AB be bisected at E. Then, since AC is greater than DB, let DC be subtracted from each; therefore the remainder AD is greater than the remainder CB. But AE is equal to EB; therefore DE is less than EC; therefore the points C, D are not equidistant from the point of bisection. And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5 ] and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.] therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE. And of these the square on DE is less than the square on EC; therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB, so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB. Therefore also the remainder, the sum of the squares on AC, CB, is greater than the sum of the squares on AD, DB. Q. E. D.

PROPOSITION 42.

A binomial straight line is divided into its terms at one point only.

Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36 ] I say that AB is not divided at another point into two rational straight lines commensurable in square only.

For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only. It is then manifest that AC is not the same with DB. For, if possible, let it be so. Then AD will also be the same as CB, and, as AC is to CB, so will BD be to DA; thus AB will be divided at D also in the same way as by the division at C: which is contrary to the hypothesis. Therefore AC is not the same with DB. For this reason also the points C, D are not equidistant from the point of bisection. Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [II. 4 ] But the squares on AC, CB differ from the squares on AD, DB by a rational area, for both are rational; therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [X. 21 ]: which is absurd, for a medial area does not exceed a medial by a rational area. [x. 26 ]

Therefore a binomial straight line is not divided at different points; therefore it is divided at one point only. Q. E. D.

PROPOSITION 43.

A first bimedial straight line is divided at one point only.

Let AB be a first bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a rational rectangle; I say that AB is not so divided at another point.

For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle. Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB, while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area — for both are rational — therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial: which is absurd. [x. 26 ]

Therefore a first bimedial straight line is not divided into its terms at different points; therefore it is so divided at one point only.

PROPOSITION 44.

A second bimedial straight line is divided at one point only.

Let AB be a second bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a medial rectangle; [X. 38 ] it is then manifest that C is not at the point of bisection, because the segments are not commensurable in length. I say that AB is not so divided at another point.

For, if possible, let it be divided at D also, so that AC is not the same with DB, but AC is supposed greater; it is then clear that the squares on AD, DB are also, as we proved above [Lemma], less than the squares on AC, CB; and suppose that AD, DB are medial straight lines commensurable in square only and containing a medial rectangle. Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB, and let EG equal to the squares on AC, CB be subtracted; therefore the remainder HK is equal to twice the rectangle AC, CB. [II. 4 ] Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma ]; therefore the remainder MK is also equal to twice the rectangle AD, DB. Now, since the squares on AC, CB are medial, therefore EG is medial. And it is applied to the rational straight line EF; therefore EH is rational and incommensurable in length with EF. [X. 22 ] For the same reason HN is also rational and incommensurable in length with EF. And, since AC, CB are medial straight lines commensurable in square only, therefore AC is incommensurable in length with CB. But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11 ] But the squares on AC, CB are commensurable with the square on AC; for AC, CB are commensurable in square. [x. 15 ] And twice the rectangle AC, CB is commensurable with the rectangle AC, CB. [X. 6 ] Therefore the squares on AC, CB are also incommensurable with twice the rectangle AC, CB. [X. 13 ] But EG is equal to the squares on AC, CB, and HK is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HK, so that EH is also incommensurable in length with HN. [VI. 1 , X. 11 ] And they are rational; therefore EH, HN are rational straight lines commensurable in square only. But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [X. 36 ] Therefore EN is a binomial straight line divided at H. In the same way EM, MN will also be proved to be rational straight lines commensurable in square only; and EN will be a binomial straight line divided at different points, H and M. And EH is not the same with MN. For the squares on AC, CB are greater than the squares on AD, DB. But the squares on AD, DB are greater than twice the rectangle AD, DB; therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB, that is, MK, so that EH is also greater than MN. Therefore EH is not the same with MN. Q. E. D.

PROPOSITION 45.

A major straight line is divided at one and the same point only.

Let AB be a major straight line divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB rational, but the rectangle AC, CB medial; [X. 39 ] I say that AB is not so divided at another point.

For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle contained by them medial. Then, since that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, while the squares on AC, CB exceed the squares on AD, DB by a rational area — for both are rational — therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial: which is impossible. [X. 26 ] Therefore a major straight line is not divided at different points; therefore it is only divided at one and the same point. Q. E. D.

PROPOSITION 46.

The side of a rational plus a medial area is divided at one point only.

Let AB be the side of a rational plus a medial area divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, but twice the rectangle AC, CB rational; [X. 40 ] I say that AB is not so divided at another point.

For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB medial, but twice the rectangle AD, DB rational. Since then that by which twice the rectangle AC, CB differs from twice the rectangle AD, DB is also that by which the squares on AD, DB differ from the squares on AC, CB, while twice the rectangle AC, CB exceeds twice the rectangle AD, DB by a rational area, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area, though they are medial: which is impossible. [X. 26 ] Therefore the side of a rational plus a medial area is not divided at different points; therefore it is divided at one point only. Q. E. D.

PROPOSITION 47.

The side of the sum of two medial areas is divided at one point only.

Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them; I say that AB is not divided at another point so as to fulfil the given conditions.

For, if possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC, CB, and the rectangle HK equal to twice the rectangle AC, CB; therefore the whole EK is equal to the square on AB. [II. 4 ] Again, let EL, equal to the squares on AD, DB, be applied to EF; therefore the remainder, twice the rectangle AD, DB, is equal to the remainder MK. And since, by hypothesis, the sum of the squares on AC, CB is medial, therefore EG is also medial. And it is applied to the rational straight line EF; therefore HE is rational and incommensurable in length with EF. [X. 22 ] For the same reason HN is also rational and incommensurable in length with EF. And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN, so that EH is also incommensurable with HN. [VI. 1 , X. 11 ] And they are rational; therefore EH, HN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at H. [X. 36 ] Similarly we can prove that it is also divided at M. And EH is not the same with MN; therefore a binomial has been divided at different points: which is absurd. [X. 42 ] Therefore a side of the sum of two medial areas is not divided at different points; therefore it is divided at one point only.

DEFINITIONS II.

1. Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line;

2. but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial;

3. and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial.

4. Again, if the square on the greater term be greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a fourth binomial;

5. if the lesser, a fifth binomial;

6. and if neither, a sixth binomial.

PROPOSITION 48.

To find the first binomial straight line.

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to CA the ratio which a square number has to a square number; [Lemma I after X. 28] let any rational straight line D be set out, and let EF be commensurable in length with D. Therefore EF is also rational. Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG. [X. 6, Por.] But AB has to AC the ratio which a number has to a number; therefore the square on EF also has to the square on FG the ratio which a number has to a number, so that the square on EF is commensurable with the square on FG. [X. 6] And EF is rational; therefore FG is also rational. And, since BA has not to AC the ratio which a square number has to a square number, neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9] Therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36] I say that it is also a first binomial straight line.

For since, as the number BA is to AC, so is the square on EF to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG. Let then the squares on FG, H be equal to the square on EF. Now since, as BA is to AC, so is the square on EF to the square on FG, therefore, convertendo, as AB is to BC, so is the square on EF to the square on H. [V. 19, Por.] But AB has to BC the ratio which a square number has to a square number; therefore the square on EF also has to the square on H the ratio which a square number has to a square number. Therefore EF is commensurable in length with H; [X. 9] therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF. And EF, FG are rational, and EF is commensurable in length with D.

Therefore EF is a first binomial straight line. Q. E. D.

PROPOSITION 49.

To find the second binomial straight line.

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number; let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is rational. Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG. [X. 6] Therefore FG is also rational. Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number. Therefore EF is incommensurable in length with FG; [X. 9] therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36] It is next to be proved that it is also a second binomial straight line.

For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE. Let the squares on EF, H be equal to the square on GF; therefore, convertendo, as AB is to BC, so is the square on FG to the square on H. [V. 19, Por.] But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on H the ratio which a square number has to a square number. Therefore FG is commensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG. And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out.

Therefore EG is a second binomial straight line. Q. E. D.

PROPOSITION 50.

To find the third binomial straight line.

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number. Let any other number D, not square, be set out also, and let it not have to either of the numbers BA, AC the ratio which a square number has to a square number. Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6] And E is rational; therefore FG is also rational. And, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9] Next let it be contrived that, as the number BA is to AC, so is the square on FG to the square on GH; [X. 6, Por.] therefore the square on FG is commensurable with the square on GH. [X. 6] But FG is rational; therefore GH is also rational. And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on HG the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36] I say next that it is also a third binomial straight line.

For since, as D is to AB, so is the square on E to the square on FG, and, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22] But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9] And since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH. Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.] But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number; therefore FG is commensurable in length with K. [X. 9] Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.

Therefore FH is a third binomial straight line. Q. E. D.

PROPOSITION 51.

To find the fourth binomial straight line.

Let two numbers AC, CB be set out such that AB neither has to BC, nor yet to AC, the ratio which a square number has to a square number. Let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is also rational. Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG; [X. 6] therefore FG is also rational. Now, since BA has not to AC the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9] Therefore EF, FG are rational straight lines commensurable in square only; so that EG is binomial. I say next that it is also a fourth binomial straight line.

For since, as BA is to AC, so is the square on EF to the square on FG, therefore the square on EF is greater than the square on FG. Let then the squares on FG, H be equal to the square on EF; therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19, Por.] But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on EF to the square on H the ratio which a square number has to a square number. Therefore EF is incommensurable in length with H; [X. 9] therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF. And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D.

Therefore EG is a fourth binomial straight line. Q. E. D.

PROPOSITION 52.

To find the fifth binomial straight line.

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; let any rational straight line D be set out, and let EF be commensurable with D; therefore EF is rational. Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [X. 6, Por.] But CA has not to AB the ratio which a square number has to a square number; therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number. Therefore EF, FG are rational straight lines commensurable in square only; [X. 9] therefore EG is binomial. [X. 36] I say next that it is also a fifth binomial straight line.

For since, as CA is to AB, so is the square on EF to the square on FG, inversely, as BA is to AC, so is the square on FG to the square on FE; therefore the square on GF is greater than the square on FE. Let then the squares on EF, H be equal to the square on GF; therefore, convertendo, as the number AB is to BC, so is the square on GF to the square on H. [V. 19, Por.] But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on FG to the square on H the ratio which a square number has to a square number. Therefore FG is incommensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG. And GF, FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out.

Therefore EG is a fifth binomial straight line. Q. E. D.

PROPOSITION 53.

To find the sixth binomial straight line.

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; and let there also be another number D which is not square and which has not to either of the numbers BA, AC the ratio which a square number has to a square number. Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6] And E is rational; therefore FG is also rational. Now, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9] Again, let it be contrived that, as BA is to AC, so is the square on FG to the square on GH. [X. 6, Por.] Therefore the square on FG is commensurable with the square on HG. [X. 6] Therefore the square on HG is rational; therefore HG is rational. And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36] It is next to be proved that it is also a sixth binomial straight line.

For since, as D is to AB, so is the square on E to the square on FG, and also, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22] But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9] But it was also proved incommensurable with FG; therefore each of the straight lines FG, GH is incommensurable in length with E. And, since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH. Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.] But AB has not to BC the ratio which a square number has to a square number; so that neither has the square on FG to the square on K the ratio which a square number has to a square number. Therefore FG is incommensurable in length with K; [X. 9] therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG. And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out.

Therefore FH is a sixth binomial straight line. Q. E. D.

LEMMA.
Let there be two squares AB, BC, and let them be placed so that DB is in a straight line with BE; therefore FB is also in a straight line with BG.

Let the parallelogram AC be completed; I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB.

For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG. But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34] therefore each of the straight lines AH, KC is also equal to each of the straight lines AK, HC. Therefore the parallelogram AC is equilateral. And it is also rectangular; therefore AC is a square.

And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC, [VI. 1] therefore also, as AB is to DG, so is DG to BC. [V. 11] Therefore DG is a mean proportional between AB, BC. I say next that DC is also a mean proportional between AC, CB.

For since, as AD is to DK, so is KG to GC — for they are equal respectively — and, componendo, as AK is to KD, so is KC to CG, [V. 18] while, as AK is to KD, so is AC to CD, and, as KC is to CG, so is DC to CB, [VI. 1] therefore also, as AC is to DC, so is DC to BC. [V. 11] Therefore DC is a mean proportional between AC, CB. Being what it was proposed to prove.

PROPOSITION 54.

If an area be contained by a rational straight line and the first binomial, the side of the area is the irrational straight line which is called binomial.

For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the “side” of the area AC is the irrational straight line which is called binomial.

For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term. It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1] Let ED be bisected at the point F. Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less, that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore AG is commensurable in length with EG. Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14] and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP. And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma] Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17] therefore also, as AH is to EL, so is EL to KG; [VI. 1] therefore EL is a mean proportional between AH, GK. But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ. But MR is also a mean proportional between the same SN, NQ; [Lemma] therefore EL is equal to MR, so that it is also equal to PO. But AH, GK are also equal to SN, NQ; therefore the whole AC is equal to the whole SQ, that is, to the square on MO; therefore MO is the “side” of AC. I say next that MO is binomial.

For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG, GE. [X. 15] But AE is also, by hypothesis, commensurable with AB; therefore AG, GE are also commensurable with AB. [X. 12] And AB is rational; therefore each of the straight lines AG, GE is also rational; therefore each of the rectangles AH, GK is rational, [X. 19] and AH is commensurable with GK. But AH is equal to SN, and GK to NQ; therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable. And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF, therefore AG is also incommensurable with EF, [X. 13] so that AH is also incommensurable with EL. [VI. 1, X. 11] But AH is equal to SN, and EL to MR; therefore SN is also incommensurable with MR. But, as SN is to MR, so is PN to NR; [VI. 1] therefore PN is incommensurable with NR. [X. 11] But PN is equal to MN, and NR to NO; therefore MN is incommensurable with NO. And the square on MN is commensurable with the square on NO, and each is rational; therefore MN, NO are rational straight lines commensurable in square only.

Therefore MO is binomial [X. 36] and the “side” of AC. Q. E. D. 5

PROPOSITION 55.

If an area be contained by a rational straight line and the second binomial, the side of the area is the irrational straight line which is called a first bimedial.

For let the area ABCD be contained by the rational straight line AB and the second binomial AD; I say that the “side” of the area AC is a first bimedial straight line.

For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term; therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. [X. Deff. II. 2] Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure; therefore AG is commensurable in length with GE. [X. 17] Through G, E, F let GH, EK, FL be drawn parallel to AB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP. Let the square SQ be completed. It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the “side” of the area AC. It is now to be proved that MO is a first bimedial straight line. Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [X. 13] And, since AG is commensurable with EG, AE is also commensurable with each of the straight lines AG, GE. [X. 15] But AE is incommensurable in length with AB; therefore AG, GE are also incommensurable with AB. [X. 13] Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only; so that each of the rectangles AH, GK is medial. [X. 21] Hence each of the squares SN, NQ is medial. Therefore MN, NO are also medial.

And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO. And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF; [X. 13] so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11] that is, MN is incommensurable in length with NO. But MN, NO were proved to be both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only. I say next that they also contain a rational rectangle.

For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, therefore EF is also commensurable with EK. [X. 12] And each of them is rational; therefore EL, that is, MR is rational, [X. 19] and MR is the rectangle MN, NO.

But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line. [X. 37]

Therefore MO is a first bimedial straight line. Q. E. D. 6

PROPOSITION 56.

If an area be contained by a rational straight line and the third binomial, the side of the area is the irrational straight line called a second bimedial.

For let the area ABCD be contained by the rational straight line AB and the third binomial AD divided into its terms at E, of which terms AE is the greater; I say that the “side” of the area AC is the irrational straight line called a second bimedial.

For let the same construction be made as before. Now, since AD is a third binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and neither of the terms AE, ED is commensurable in length with AB. [X. Deff. II. 3] Then, in manner similar to the foregoing, we shall prove that MO is the “side” of the area AC, and MN, NO are medial straight lines commensurable in square only; so that MO is bimedial. It is next to be proved that it is also a second bimedial straight line.

Since DE is incommensurable in length with AB, that is, with EK, and DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13] And they are rational; therefore FE, EK are rational straight lines commensurable in square only. Therefore EL, that is, MR, is medial. [X. 21] And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

Therefore MO is a second bimedial straight line. [X. 38] Q. E. D.

PROPOSITION 57.

If an area be contained by a rational straight line and the fourth binomial, the side of the area is the irrational straight line called major.

For let the area AC be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AE be the greater; I say that the “side” of the area AC is the irrational straight line called major.

For, since AD is a fourth binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. [X. Deff. II. 4] Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF; therefore AG is incommensurable in length with GE. [X. 18] Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before; it is then manifest that MO is the “side” of the area AC. It is next to be proved that MO is the irrational straight line called major.

Since AG is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ; [VI. 1, X. 11] therefore MN, NO are incommensurable in square. And, since AE is commensurable with AB, AK is rational; [X. 19] and it is equal to the squares on MN, NO; therefore the sum of the squares on MN, NO is also rational. And, since DE is incommensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13] Therefore EK, EF are rational straight lines commensurable in square only; therefore LE, that is, MR, is medial. [X. 21] And it is contained by MN, NO; therefore the rectangle MN, NO is medial. And the [sum] of the squares on MN, NO is rational, and MN, NO are incommensurable in square. But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole is irrational and is called major. [X. 39]

Therefore MO is the irrational straight line called major and is the “side” of the area AC. Q. E. D.

PROPOSITION 58.

If an area be contained by a rational straight line and the fifth binomial, the side of the area is the irrational straight line called the side of a rational plus a medial area.

For let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that AE is the greater term; I say that the “side” of the area AC is the irrational straight line called the side of a rational plus a medial area.

For let the same construction be made as before shown; it is then manifest that MO is the “side” of the area AC. It is then to be proved that MO is the side of a rational plus a medial area.

For, since AG is incommensurable with GE, [X. 18] therefore AH is also commensurable with HE, [VI. 1, X. 11] that is, the square on MN with the square on NO; therefore MN, NO are incommensurable in square. And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [X. Deff. II. 5] But AE is incommensurable with ED; therefore AB is also incommensurable in length with AE. [X. 13] Therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21] And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [X. 12] And EK is rational; therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. [X. 19] Therefore MN, NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational.

Therefore MO is the side of a rational plus a medial area [X. 40] and is the “side” of the area AC. Q. E. D.

PROPOSITION 59.

If an area be contained by a rational straight line and the sixth binomial, the side of the area is the irrational straight line called the side of the sum of two medial areas.

For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term; I say that the “side” of AC is the side of the sum of two medial areas.

Let the same construction be made as before shown. It is then manifest that MO is the “side” of AC, and that MN is incommensurable in square with NO. Now, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only; therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21] Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [X. 13] therefore FE, EK are rational straight lines commensurable in square only; therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. [X. 21] And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. 1, X. 11] But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO. And each of them is medial, and MN, NO are incommensurable in square.

Therefore MO is the side of the sum of two medial areas [X. 41], and is the “side” of AC. Q. E. D.

[LEMMA.
If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.

Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.

For let AB be bisected at D. Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle AC, CB is less than double of the square on AD. But the squares on AC, CB are double of the squares on AD, DC; [II. 9] therefore the squares on AC, CB are greater than twice the rectangle AC, CB. Q. E. D.]

PROPOSITION 60.

The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.

Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term; let a rational straight line DE be set out, and let DEFG equal to the square on AB be applied to DE producing DG as its breadth; I say that DG is a first binomial straight line.

For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC; therefore the remainder, twice the rectangle AC, CB, is equal to MF. Let MG be bisected at N, and let NO be drawn parallel [to ML or GF]. Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB. Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [X. 15] And it is equal to DL; therefore DL is rational. And it is applied to the rational straight line DE; therefore DM is rational and commensurable in length with DE. [X. 20] Again, since AC, CB are rational straight lines commensurable in square only, therefore twice the rectangle AC, CB, that is MF, is medial. [X. 21] And it is applied to the rational straight line ML; therefore MG is also rational and incommensurable in length with ML, that is, DE. [X. 22] But MD is also rational and is commensurable in length with DE; therefore DM is incommensurable in length with MG. [X. 13] And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] It is next to be proved that it is also a first binomial straight line.

Since the rectangle AC, CB is a mean proportional between the squares on AC, CB, [cf. Lemma after X. 53] therefore MO is also a mean proportional between DH, KL. Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [VI. 1] therefore the rectangle DK, KM is equal to the square on MN. [VI. 17] And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11] And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma] therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1] And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM. But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [X. 17] therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out.

Therefore DG is a first binomial straight line. [X. Deff. II. 1] Q. E. D.

PROPOSITION 61.

The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.

Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a second binominal straight line.

For let the same construction as before be made. Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37] so that the squares on AC, CB are also medial. [X. 21] Therefore DL is medial. [X. 15 and 23, Por.] And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. [X. 22] Again, since twice the rectangle AC, CB is rational, MF is also rational. And it is applied to the rational straight line ML; therefore MG is also rational and commensurable in length with ML, that is, DE; [X. 20] therefore DM is incommensurable in length with MG. [X. 13] And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] It is next to be proved that it is also a second binomial straight line.

For, since the squares on AC, CB are greater than twice the rectangle AC, CB, therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1] And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11] And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. [X. 17] And MG is commensurable is length with DE.

Therefore DG is a second binomial straight line. [X. Deff. II. 2]

PROPOSITION 62.

The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.

Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a third binomial straight line.

Let the same construction be made as before shown. Then, since AB is a second bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [X. 38] so that the sum of the squares on AC, CB is also medial. [X. 15 and 23 Por.] And it is equal to DL; therefore DL is also medial. And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE. [X. 22] For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. And, since AC is incommensurable in length with CB, and, as AC is to CB, so is the square on AC to the rectangle AC, CB, therefore the square on AC is also incommensurable with the rectangle AC, CB. [X. 11] Hence the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, [X. 12, 13] that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [VI. 1, X. 11] And they are rational; therefore DG is binomial. [X. 36] It is to be proved that it is also a third binomial straight line.

In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM. And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. And neither of the straight lines DM, MG is commensurable in length with DE.

Therefore DG is a third binomial straight line. [X. Deff. II. 3] Q. E. D.

PROPOSITION 63.

The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.

Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line.

Let the same construction be made as before shown. Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39] Since then the sum of the squares on AC, CB is rational, therefore DL is rational; therefore DM is also rational and commensurable in length with DE. [X. 20] Again, since twice the rectangle AC, CB, that is, MF, is medial, and it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with DE; [X. 22] therefore DM is also incommensurable in length with MG. [X. 13] Therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

It is to be proved that it is also a fourth binomial straight line.

In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN. Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. [VI. 1, X. 11] But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out.

Therefore DG is a fourth binomial straight line. [X. Deff. II. 4] Q. E. D.

PROPOSITION 64.

The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.

Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a fifth binomial straight line.

Let the same construction as before be made. Since then AB is the side of a rational plus a medial area, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40] Since then the sum of the squares on AC, CB is medial, therefore DL is medial, so that DM is rational and incommensurable in length with DE. [X. 22] Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. [X. 20] Therefore DM is incommensurable with MG; [X. 13] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] I say next that it is also a fifth binomial straight line.

For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18] And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE.

Therefore DG is a fifth binomial. Q. E. D.

PROPOSITION 65.

The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.

Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a sixth binomial straight line.

For let the same construction be made as before. Then, since AB is the side of the sum of two medial areas, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [X. 41] so that, in accordance with what was before proved, each of the rectangles DL, MF is medial. And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [X. 22] And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore DL is incommensurable with MF. Therefore DM is also incommensurable with MG; [VI. 1, X. 11] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] I say next that it is also a sixth binomial straight line.

Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM. And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out.

Therefore DG is a sixth binomial straight line. Q. E. D.

PROPOSITION 66.

A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.

Let AB be binomial, and let CD be commensurable in length with AB; I say that CD is binomial and the same in order with AB.

For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. [X. 36] Let it be contrived that, as AB is to CD, so is AE to CF; [VI. 12] therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19] But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with FD. [X. 11] And AE, EB are rational; therefore CF, FD are also rational. And, as AE is to CF, so is EB to FD. [V. 11] Therefore, alternately, as AE is to EB, so is CF to FD. [V. 16] But AE, EB are commensurable in square only; therefore CF, FD are also commensurable in square only. [X. 11] And they are rational; therefore CD is binomial. [X. 36] I say next that it is the same in order with AB.

For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it. If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14] And, if AE is commensurable with the rational straight line set out, CF will also be commensurable with it, [X. 12] and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. [X. Deff. II. 1] But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [X. 12] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. [X. Deff. II. 2] But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [X. 13] and each of the straight lines AB, CD is a third binomial. [X. Deff. II. 3] But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14] And, if AE is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. [X. Deff. II. 4] But, if EB is so commensurable, so is FD also, and each of the straight lines AB, CD will be a fifth binomial. [X. Deff. II. 5] But, if neither of the straight lines AE, EB is so commensurable, neither of the straight lines CF, FD is commensurable with the rational straight line set out, and each of the straight lines AB, CD will be a sixth binomial. [X. Deff. II. 6]

Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order. Q. E. D.

PROPOSITION 67.

A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.

Let AB be bimedial, and let CD be commensurable in length with AB; I say that CD is bimedial and the same in order with AB.

For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38] And let it be contrived that, as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19] But AB is commensurable in length with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11] But AE, EB are medial; therefore CF, FD are also medial. [X. 23] And since, as AE is to EB, so is CF to FD, [V. 11] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. [X. 11] But they were also proved medial; therefore CD is bimedial. I say next that it is also the same in order with AB.

For since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD; therefore, alternately, as the square on AE is to the square on CF, so is the rectangle AE, EB to the rectangle CF, FD. [V. 16] But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [X. 37] but if medial, medial, [X. 23, Por.] and each of the straight lines AB, CD is a second bimedial. [X. 38]

And for this reason CD will be the same in order with AB. Q. E. D.

PROPOSITION 68.

A straight line commensurable with a major straight line is itself also major.

Let AB be major, and let CD be commensurable with AB; I say that CD is major.

Let AB be divided at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39] Let the same construction be made as before. Then since, as AB is to CD, so is AE to CF, and EB to FD, therefore also, as AE is to CF, so is EB to FD. [V. 11] But AB is commensurable with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11] And since, as AE is to CF, so is EB to FD, alternately also, as AE is to EB, so is CF to FD; [V. 16] therefore also, componendo, as AB is to BE, so is CD to DF; [V. 18] therefore also, as the square on AB is to the square on BE, so is the square on CD to the square on DF. [VI. 20] Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF. Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD; therefore also, alternately, as the square on AB is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. [V. 16] But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD. And the squares on AE, EB together are rational; therefore the squares on CF, FD together are rational. Similarly also twice the rectangle AE, EB is commensurable with twice the rectangle CF, FD. And twice the rectangle AE, EB is medial; therefore twice the rectangle CF, FD is also medial. [X. 23, Por.] Therefore CF, FD are straight lines incommensurable in square which make, at the same time, the sum of the squares on them rational, but the rectangle contained by them medial; therefore the whole CD is the irrational straight line called major. [X. 39]

Therefore a straight line commensurable with the major straight line is major. Q. E. D.

PROPOSITION 69.

A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area.

Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB; it is to be proved that CD is also the side of a rational plus a medial area.

Let AB be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40] Let the same construction be made as before. We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational.

Therefore CD is the side of a rational plus a medial area. Q. E. D.

PROPOSITION 70.

A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.

Let AB be the side of the sum of two medial areas, and CD commensurable with AB; it is to be proved that CD is also the side of the sum of two medial areas.

For, since AB is the side of the sum of two medial areas, let it be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and furthermore the sum of the squares on AE, EB incommensurable with the rectangle AE, EB. [X. 41] Let the same construction be made as before. We can then prove similarly that CF, FD are also incommensurable in square, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, the rectangle CF, FD is medial, and moreover the sum of the squares on CF, FD is incommensurable with the rectangle CF, FD.

Therefore CD is the side of the sum of two medial areas. Q. E. D.

PROPOSITION 71.

If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.

Let AB be rational, and CD medial; I say that the “side” of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.

For AB is either greater or less than CD. First, let it be greater; let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB, producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth. Then, since AB is rational and is equal to EG, therefore EG is also rational. And it has been applied to EF, producing EH as breadth; therefore EH is rational and commensurable in length with EF. [X. 20] Again, since CD is medial and is equal to HI, therefore HI is also medial. And it is applied to the rational straight line EF, producing HK as breadth; therefore HK is rational and incommensurable in length with EF [X. 22] And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EG is also incommensurable with HI. But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is also incommensurable in length with HK. [X. 11] And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [X. 36] And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI; therefore EH is also greater than HK. The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable with itself. Now the greater straight line HE is commensurable in length with the rational straight line EF set out; therefore EK is a first binomial. [X. Deff. II. 1] But EF is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [X. 54] Therefore the “side” of EI is binomial; so that the “side” of AD is also binomial. Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH. Now the greater straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a fourth binomial. [X. Deff. II. 4] But EF is rational; and, if an area be contained by a rational straight line and the fourth binomial, the “side” of the area is the irrational straight line called major. [X. 57] Therefore the “side” of the area EI is major; so that the “side” of the area AD is also major.

Next, let AB be less than CD; therefore EG is also less than HI, so that EH is also less than HK. Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable in length with itself. Now the lesser straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a second binomial. [X. Deff. II. 2] But EF is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [X. 55] therefore the “side” of the area EI is a first bimedial, so that the “side” of AD is also a first bimedial. Next, let the square on HK be greater than the square on HE by the square on a straight line incommensurable with HK. Now the lesser straight line EH is commensurable with the rational straight line EF set out; therefore EK is a fifth binomial. [X. Deff. II. 5] But EF is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [X. 58] Therefore the “side” of the area EI is a side of a rational plus a medial area, so that the “side” of the area AD is also a side of a rational plus a medial area.

Therefore etc. Q. E. D.

PROPOSITION 72.

If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas.

For let two medial areas AB, CD incommensurable with one another be added together; I say that the “side” of the area AD is either a second bimedial or a side of the sum of two medial areas.

For AB is either greater or less than CD. First, if it so chance, let AB be greater than CD. Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth. Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial. And they are applied to the rational straight line FE, producing EH, HK as breadth; therefore each of the straight lines EH, HK is rational and incommensurable in length with EF. [X. 22] And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, therefore EG is also incommensurable with HI. But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is incommensurable in length with HK. [X. 11] Therefore EH, HK are rational straight lines commensurable in square only; therefore EK is binomial. [X. 36] But the square on EH is greater than the square on HK either by the square on a straight line commensurable with EH or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable in length with itself. Now neither of the straight lines EH, HK is commensurable in length with the rational straight line EF set out; therefore EK is a third binomial. [X. Deff. II. 3] But EF is rational; and, if an area be contained by a rational straight line and the third binomial, the “side” of the area is a second bimedial; [X. 56] therefore the “side” of EI, that is, of AD, is a second bimedial. Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH. Now each of the straight lines EH, HK is incommensurable in length with EF; therefore EK is a sixth binomial. [X. Deff. II. 6] But, if an area be contained by a rational straight line and the sixth binomial, the “side” of the area is the side of the sum of two medial areas; [X. 59] so that the “side” of the area AD is also the side of the sum of two medial areas.

Therefore etc. Q. E. D.

PROPOSITION 73.

If from a rational straight line there be subtracted a rational straight line commensurable with the whole in square only, the remainder is irrational; and let it be called an apotome.

For from the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subtracted; I say that the remainder AC is the irrational straight line called apotome.

For, since AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11] But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6] And, inasmuch as the squares on AB, BC are equal to twice the rectangle AB, BC together with the square on CA, [II. 7] therefore the squares on AB, BC are also incommensurable with the remainder, the square on AC. [X. 13, 16] But the squares on AB, BC are rational; therefore AC is irrational. [X. Def. 4] And let it be called an apotome. Q. E. D.

PROPOSITION 74.

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a first apotome of a medial straight line.

For from the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in square only and with AB makes the rectangle AB, BC rational; I say that the remainder AC is irrational; and let it be called a first apotome of a medial straight line.

For, since AB, BC are medial, the squares on AB, BC are also medial. But twice the rectangle AB, BC is rational; therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore twice the rectangle AB, BC is also incommensurable with the remainder, the square on AC, [Cf. II. 7] since, if the whole is incommensurable with one of the magnitudes, the original magnitudes will also be incommensurable. [X. 16] But twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4] And let it be called a first apotome of a medial straight line.

PROPOSITION 75

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a second apotome of a medial straight line.

For from the medial straight line AB let there be subtracted the medial straight line CB which is commensurable with the whole AB in square only and such that the rectangle AB, BC, which it contains with the whole AB, is medial; [X. 28] I say that the remainder AC is irrational; and let it be called a second apotome of a medial straight line.

For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth; therefore the remainder FE is equal to the square on AC. [II. 7] Now, since the squares on AB, BC are medial and commensurable, therefore DE is also medial. [X. 15 and 23, Por.] And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22] Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23, Por.] And it is equal to DH; therefore DH is also medial. And it has been applied to the rational straight line DI, producing DF as breadth; therefore DF is rational and incommensurable in length with DI. [X. 22] And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC; therefore the square on AB is also incommensurable with the rectangle AB, BC. [X. 11] But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC; [X. 6] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13] But DE is equal to the squares on AB, BC, and DH to twice the rectangle AB, BC; therefore DE is incommensurable with DH. But, as DE is to DH, so is GD to DF; [VI. 1] therefore GD is incommensurable with DF. [X. 11] And both are rational; therefore GD, DF are rational straight lines commensurable in square only; therefore FG is an apotome. [X. 73] But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from X. 20] and its ’side’ is irrational. And AC is the ’side’ of FE; therefore AC is irrational. And let it be called a second apotome of a medial straight line. Q. E. D.

PROPOSITION 76

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the squares on them added together rational, but the rectangle contained by them medial, the remainder is irrational; and let it be called minor.

For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with the whole and fulfils the given conditions. [X. 33] I say that the remainder AC is the irrational straight line called minor.

For, since the sum of the squares on AB, BC is rational, while twice the rectangle AB, BC is medial, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; and, convertendo, the squares on AB, BC are incommensurable with the remainder, the square on AC. [II. 7, X. 16] But the squares on AB, BC are rational; therefore the square on AC is irrational; therefore AC is irrational. And let it be called minor.

PROPOSITION 77

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial, but twice the rectangle contained by them rational, the remainder is irrational: and let it be called that which produces with a rational area a medial whole.

For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [X. 34] I say that the remainder AC is the irrational straight line aforesaid.

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore the remainder also, the square on AC, is incommensurable with twice the rectangle AB, BC. [II. 7, X. 16] And twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational. And let it be called that which produces with a rational area a medial whole. Q. E. D.

PROPOSITION 78

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them, the remainder is irrational; and let it be called that which produces with a medial area a medial whole.

For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions; [X. 35] I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole.

For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted. Therefore the remainder FE is equal to the square on AC, [II. 7] so that AC is the “side” of FE. Now, since the sum of the squares on AB, BC is medial and is equal to DE, therefore DE is medial. And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22] Again, since twice the rectangle AB, BC is medial and is equal to DH, therefore DH is medial. And it is applied to the rational straight line DI, producing DF as breadth; therefore DF is also rational and incommensurable in length with DI. [X. 22] And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, therefore DE is also incommensurable with DH. But, as DE is to DH, so also is DG to DF; [VI. 1] therefore DG is incommensurable with DF. [X. 11] And both are rational; therefore GD, DF are rational straight lines commensurable in square only. Therefore FG is an apotome. [X. 73] And FH is rational; but the rectangle contained by a rational straight line and an apotome is irrational, [deduction from X. 20] and its “side” is irrational. And AC is the “side” of FE; therefore AC is irrational. And let it be called that which produces with a medial area a medial whole. Q. E. D.

PROPOSITION 79

To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only.

Let AB be an apotome, and BC an annex to it; therefore AC, CB are rational straight lines commensurable in square only. [X. 73] I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

For, if possible, let BD be so annexed; therefore AD, DB are also rational straight lines commensurable in square only. [X. 73] Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB. But the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational; therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial [X. 21], and a medial area does not exceed a medial by a rational area. [X. 26] Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

Therefore only one rational straight line can be annexed to an apotome which is commensurable with the whole in square only. Q. E. D.

PROPOSITION 80.

To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle.

For let AB be a first apotome of a medial straight line, and let BC be an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is rational; [X. 74] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a rational area.

For, if possible, let DB also be so annexed; therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational. [X. 74] Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for they exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB. But twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational. Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area. which is impossible, for both are medial [X. 15 and 23, Por.], and a medial area does not exceed a medial by a rational area. [X. 26]

Therefore etc. Q. E. D.

PROPOSITION 81.

To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

Let AB be a second apotome of a medial straight line and BC an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is medial. [X. 75] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

For, if possible, let BD also be so annexed; therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is medial. [X. 75] Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be subtracted, producing HM as breadth; therefore the remainder EL is equal to the square on AB, [II. 7] so that AB is the “side” of EL. Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth. But EL is also equal to the square on AB; therefore the remainder HI is equal to twice the rectangle AD, DB. [II. 7] Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial. And they are equal to EG; therefore EG is also medial. [X. 15 and 23, Por.] And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22] Again, since the rectangle AC, CB is medial, twice the rectangle AC, CB is also medial. [X. 23, Por.] And it is equal to HG; therefore HG is also medial. And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is also rational and incommensurable in length with EF. [X. 22] And, since AC, CB are commensurable in square only, therefore AC is incommensurable in length with CB. But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11] But the squares on AC, CB are commensurable with the square on AC, while twice the rectangle AC, CB is commensurable with the rectangle AC, CB; [X. 6] therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB. [X. 13] And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HG. But, as EG is to HG, so is EM to HM; [VI. 1] therefore EM is incommensurable in length with MH. [X. 11] And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73] Similarly we can prove that HN is also an annex to it; therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only: which is impossible. [X. 79]

Therefore etc. Q. E. D.

PROPOSITION 82.

To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial.

Let AB be the minor straight line, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76] I say that no other straight line can be annexed to AB fulfilling the same conditions.

For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 76] Now, since the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial. [X. 26]

Therefore to a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes the squares on them added together rational, but twice the rectangle contained by them medial. Q. E. D.

PROPOSITION 83.

To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.

Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which fulfil the given conditions. [X. 77] I say that no other straight line can be annexed to AB which fulfils the same conditions.

For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the given conditions. [X. 77] Since then, as in the preceding cases, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area: which is impossible, for both are medial. [X. 26]

Therefore no other straight line can be annexed to AB which is incommensurable in square with the whole and which with the whole fulfils the aforesaid conditions; therefore only one straight line can be so annexed. Q. E. D.

PROPOSITION 84.

To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.

Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it; therefore AC, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 78] I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions.

For, if possible, let BD be so annexed, so that AD, DB are also straight lines incommensurable in square which make the squares on AD, DB added together medial, twice the rectangle AD, DB medial, and also the squares on AD, DB incommensurable with twice the rectangle AD, DB. [X. 78] Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be applied to EF, producing HM as breadth; therefore the remainder, the square on AB [II. 7], is equal to EL; therefore AB is the “side” of EL. Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth. But the square on AB is also equal to EL; therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI. Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial. And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22] Again, since twice the rectangle AC, CB is medial and is equal to HG, therefore HG is also medial. And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is rational and incommensurable in length with EF. [X. 22] And, since the squares on AC, CB are incommensurable with twice the rectangle AC, CB, EG is also incommensurable with HG; therefore EM is also incommensurable in length with MH. [VI. 1, X. 11] And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73] Similarly we can prove that EH is again an apotome and HN an annex to it. Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only: which was proved impossible. [X. 79] Therefore no other straight line can be so annexed to AB.

Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the squares on them incommensurable with twice the rectangle contained by them. Q. E. D.

DEFINITIONS III.

1. Given a rational straight line and an apotome, if the square on the whole be greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole be commensurable in length with the rational straight line set out, let the apotome be called a first apotome.

2. But if the annex be commensurable in length with the rational straight line set out, and the square on the whole be greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome.

3. But if neither be commensurable in length with the rational straight line set out, and the square on the whole be greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome.

4. Again, if the square on the whole be greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole be commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome;

5. if the annex be so commensurable, a fifth;

6. and, if neither, a sixth.

PROPOSITION 85.

To find the first apotome.

Let a rational straight line A be set out, and let BG be commensurable in length with A; therefore BG is also rational. Let two square numbers DE, EF be set out, and let their difference FD not be square; therefore neither has ED to DF the ratio which a square number has to a square number. Let it be contrived that, as ED is to DF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6] But the square on BG is rational; therefore the square on GC is also rational; therefore GC is also rational. And, since ED has not to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a first apotome.

For let the square on H be that by which the square on BG is greater than the square on GC. Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H. But DE has to EF the ratio which a square number has to a square number, for each is square; therefore the square on GB also has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a first apotome. [X. Deff. III. 1]

Therefore the first apotome BC has been found. (Being) that which it was required to find.

PROPOSITION 86.

To find the second apotome.

Let a rational straight line A be set out, and GC commensurable in length with A; therefore GC is rational. Let two square numbers DE, EF be set out, and let their difference DF not be square. Now let it be contrived that, as FD is to DE, so is the square on CG to the square on GB. [X. 6, Por.] Therefore the square on CG is commensurable with the square on GB. [X. 6] But the square on CG is rational; therefore the square on GB is also rational; therefore BG is rational. And, since the square on GC has not to the square on GB the ratio which a square number has to a square number, CG is incommensurable in length with GB. [X. 9] And both are rational; therefore CG, GB are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a second apotome.

For let the square on H be that by which the square on BG is greater than the square on GC. Since then, as the square on BG is to the square on GC, so is the number ED to the number DF, therefore, convertendo, as the square on BG is to the square on H, so is DE to EF. [V. 19, Por.] And each of the numbers DE, EF is square; therefore the square on BG has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And CG, the annex, is commensurable with the rational straight line A set out. Therefore BC is a second apotome. [X. Deff. III. 2]

Therefore the second apotome BC has been found. Q. E. D.

PROPOSITION 87.

To find the third apotome.

Let a rational straight line A be set out, let three numbers E, BC, CD be set out which have not to one another the ratio which a square number has to a square number, but let CB have to BD the ratio which a square number has to a square number. Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.] Since then, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6] But the square on A is rational; therefore the square on FG is also rational; therefore FG is rational. And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9] Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6] But the square on FG is rational; therefore the square on GH is also rational; therefore GH is rational. And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73] I say next that it is also a third apotome.

For since, as E is to BC, so is the square on A to the square on FG and, as BC is to CD, so is the square on FG to the square on HG, therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22] But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH. [X. 9] Therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out. Now let the square on K be that by which the square on FG is greater than the square on GH. Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as BC is to BD, so is the square on FG to the square on K. [V. 19, Por.] But BC has to BD the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number. Therefore FG is commensurable in length with K, [X. 9] and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. And neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out; therefore FH is a third apotome. [X. Deff. III. 3]

Therefore the third apotome FH has been found. Q. E. D.

PROPOSITION 88.

To find the fourth apotome.

Let a rational straight line A be set out, and BG commensurable in length with it; therefore BG is also rational. Let two numbers DF, FE be set out such that the whole DE has not to either of the numbers DF, EF the ratio which a square number has to a square number. Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6] But the square on BG is rational; therefore the square on GC is also rational; therefore GC is rational. Now, since DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

Now let the square on H be that by which the square on BG is greater than the square on GC. Since then, as DE is to EF, so is the square on BG to the square on GC, therefore also, convertendo, as ED is to DF, so is the square on GB to the square on H. [v. 19, Por.] But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on GB to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line incommensurable with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a fourth apotome. [X. Deff. III. 4]

Therefore the fourth apotome has been found. Q. E. D.

PROPOSITION 89.

To find the fifth apotome.

Let a rational straight line A be set out, and let CG be commensurable in length with A; therefore CG is rational. Let two numbers DF, FE be set out such that DE again has not to either of the numbers DF, FE the ratio which a square number has to a square number; and let it be contrived that, as FE is to ED, so is the square on CG to the square on GB. Therefore the square on GB is also rational; [X. 6] therefore BG is also rational. Now since, as DE is to EF, so is the square on BG to the square on GC, while DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a fifth apotome.

For let the square on H be that by which the square on BG is greater than the square on GC. Since then, as the square on BG is to the square on GC, so is DE to EF, therefore, convertendo, as ED is to DF, so is the square on BG to the square on H. [V. 19, Por.] But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on BG to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on GB is greater than the square on GC by the square on a straight line incommensurable in length with GB. And the annex CG is commensurable in length with the rational straight line A set out; therefore BC is a fifth apotome. [X. Deff. III. 5]

Therefore the fifth apotome BC has been found. Q. E. D.

PROPOSITION 90.

To find the sixth apotome.

Let a rational straight line A be set out, and three numbers E, BC, CD not having to one another the ratio which a square number has to a square number; and further let CB also not have to BD the ratio which a square number has to a square number. Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]

Now since, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6] But the square on A is rational; therefore the square on FG is also rational; therefore FG is also rational. And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9] Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6] But the square on FG is rational; therefore the square on GH is also rational; therefore GH is also rational. And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73] I say next that it is also a sixth apotome.

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH, therefore, ex aequali, as E is to CD, so is the square on A to the square on GH. [v. 22] But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH; [X. 9] therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A. Now let the square on K be that by which the square on FG is greater than the square on GH. Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as CB is to BD, so is the square on FG to the square on K. [v. 19, Por.] But CB has not to BD the ratio which a square number has to a square number; therefore neither has the square on FG to the square on K the ratio which a square number has to a square number; therefore FG is incommensurable in length with K. [X. 9] And the square on FG is greater than the square on GH by the square on K; therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG. And neither of the straight lines FG, GH is commensurable with the rational straight line A set out. Therefore FH is a sixth apotome. [X. Deff. III. 6]

Therefore the sixth apotome FH has been found. Q. E. D.

PROPOSITION 91.

If an area be contained by a rational straight line and a first apotome, the side of the area is an apotome.

For let the area AB be contained by the rational straight line AC and the first apotome AD; I say that the “side” of the area AB is an apotome.

For, since AD is a first apotome, let DG be its annex; therefore AG, GD are rational straight lines commensurable in square only. [X. 73] And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1] if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable with FG. And through the points E, F, G let EH, FI, GK be drawn parallel to AC.

Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12] And AC is rational; therefore each of the straight lines AF, FG is also rational, so that each of the rectangles AI, FK is also rational. [X. 19] Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15] But DG is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21]

Now let the square LM be made equal to AI, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Since then the rectangle contained by AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17] But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to KF; [VI. 1] therefore EK is a mean proportional between AI, KF. [V. 11] But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after X. 53] and AI is equal to the square LM, and KF to NO; therefore MN is also equal to EK. But EK is equal to DH, and MN to LO; therefore DK is equal to the gnomon UVW and NO. But AK is also equal to the squares LM, NO; therefore the remainder AB is equal to ST. But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN is the “side” of AB. I say next that LN is an apotome.

For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO, therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational; therefore each of the straight lines LP, PN is also rational. Again, since DH is medial and is equal to LO, therefore LO is also medial. Since then LO is medial, while NO is rational, therefore LO is incommensurable with NO. But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP is incommensurable in length with PN. [X. 11] And both are rational; therefore LP, PN are rational straight lines commensurable in square only; therefore LN is an apotome. [X. 73] And it is the “side” of the area AB; therefore the “side” of the area AB is an apotome.

Therefore etc.

PROPOSITION 92.

If an area be contained by a rational straight line and a second apotome, the side of the area is a first apotome of a medial straight line.

For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the “side” of the area AB is a first apotome of a medial straight line.

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, [X. 73] and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2] Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable in length with FG. Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial. [X. 21] Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15] But DG is commensurable in length with AC. Therefore each of the rectangles DH, EK is rational. [X. 19]

Let then the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM, namely the angle LPM; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore LP, PN are also medial straight lines commensurable in square only. And, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG, [VI. 17] while, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11] But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore MN is also equal to EK. But DH is equal to EK, and LO equal to MN; therefore the whole DK is equal to the gnomon UVW and NO. Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO, therefore the remainder AB is equal to TS. But TS is the square on LN; therefore the square on LN is equal to the area AB; therefore LN is the “side” of the area AB. I say that LN is a first apotome of a medial straight line.

For, since EK is rational and is equal to LO, therefore LO, that is, the rectangle LP, PN, is rational. But NO was proved medial; therefore LO is incommensurable with NO. But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP, PN are incommensurable in length. [X. 11] Therefore LP, PN are medial straight lines commensurable in square only which contain a rational rectangle; therefore LN is a first apotome of a medial straight line. [X. 74] And it is the “side” of the area AB.

Therefore the “side” of the area AB is a first apotome of a medial straight line. Q. E. D.

PROPOSITION 93.

If an area be contained by a rational straight line and a third apotome, the side of the area is a second apotome of a medial straight line.

For let the area AB be contained by the rational straight line AC and the third apotome AD; I say that the “side” of the area AB is a second apotome of a medial straight line.

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, and neither of the straight lines AG, GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. [X. Deff. III. 3] Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into commensurable parts. [X. 17] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG. Let EH, FI, GK be drawn through the points E, F, G parallel to AC. Therefore AF, FG are commensurable; therefore AI is also commensurable with FK. [VI. 1, X. 11] And, since AF, FG are commensurable in length, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is rational and incommensurable in length with AC; so that AF, FG are so also. [X. 13] Therefore each of the rectangles AI, FK is medial. [X. 21] Again, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15] But GD is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21] And, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD. But AG is commensurable in length with AF, and DG with EG; therefore AF is incommensurable in length with EG. [X. 13] But, as AF is to EG, so is AI to EK; [VI. 1] therefore AI is incommensurable with EK. [X. 11]

Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM; therefore LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Now, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17] But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore also, as AI is to EK, so is EK to FK; [V. 11] therefore EK is a mean proportional between AI, FK. But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN. But MN is equal to LO, and EK equal to DH; therefore the whole DK is also equal to the gnomon UVW and NO. But AK is also equal to LM, NO; therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the “side” of the area AB. I say that LN is a second apotome of a medial straight line.

For, since AI, FK were proved medial, and are equal to the squares on LP, PN, therefore each of the squares on LP, PN is also medial; therefore each of the straight lines LP, PN is medial. And, since AI is commensurable with FK, [VI. 1, X. 11] therefore the square on LP is also commensurable with the square on PN. Again, since AI was proved incommensurable with EK, therefore LM is also incommensurable with MN, that is, the square on LP with the rectangle LP, PN; so that LP is also incommensurable in length with PN; [VI. 1, X. 11] therefore LP, PN are medial straight lines commensurable in square only. I say next that they also contain a medial rectangle.

For, since EK was proved medial, and is equal to the rectangle LP, PN, therefore the rectangle LP, PN is also medial, so that LP, PN are medial straight lines commensurable in square only which contain a medial rectangle. Therefore LN is a second apotome of a medial straight line; [X. 75] and it is the “side” of the area AB.

Therefore the “side” of the area AB is a second apotome of a medial straight line. Q. E. D.

PROPOSITION 94.

If an area be contained by a rational straight line and a fourth apotome, the side of the area is minor.

For let the area AB be contained by the rational straight line AC and the fourth apotome AD; I say that the “side” of the area AB is minor.

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [X. Deff. III. 4] Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG. Let EH, FI, GK be drawn through E, F, G parallel to AC, BD. Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19] Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. [X. 21] Again, since AF is incommensurable in length with FG, therefore AI is also incommensurable with FK. [VI. 1, X. 11] Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and about the same angle, the angle LPM. Therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Since then the rectangle AF, FG is equal to the square on EG, therefore, proportionally, as AF is to EG, so is EG to FG. [VI. 17] But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11] But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN. But DH is equal to EK, and LO is equal to MN; therefore the whole DK is equal to the gnomon UVW and NO. Since, then, the whole AK is equal to the squares LM, NO, and, in these, DK is equal to the gnomon UVW and the square NO, therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the “side” of the area AB. I say that LN is the irrational straight line called minor.

For, since AK is rational and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is rational. Again, since DK is medial, and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial. And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN. Therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. Therefore LN is the irrational straight line called minor; [X. 76] and it is the “side” of the area AB.

Therefore the “side” of the area AB is minor. Q. E. D.

PROPOSITION 95.

If an area be contained by a rational straight line and a fifth apotome, the side of the area is a straight line which produces with a rational area a medial whole.

For let the area AB be contained by the rational straight line AC and the fifth apotome AD; I say that the “side” of the area AB is a straight line which produces with a rational area a medial whole.

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5] Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at the point E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG. Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21] Again, since DG is rational and commensurable in length with AC, DK is rational. [X. 19]

Now let the square LM be constructed equal to AI, and let the square NO equal to FK and about the same angle, the angle LPM, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Similarly then we can prove that LN is the “side” of the area AB. I say that LN is the straight line which produces with a rational area a medial whole.

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial. Again, since DK is rational and is equal to twice the rectangle LP, PN, the latter is itself also rational. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; [X. 77] and it is the “side” of the area AB.

Therefore the “side” of the area AB is a straight line which produces with a rational area a medial whole. Q. E. D.

PROPOSITION 96.

If an area be contained by a rational straight line and a sixth apotome, the side of the area is a straight line which produces with a medial area a medial whole.

For let the area AB be contained by the rational straight line AC and the sixth apotome AD; I say that the “side” of the area AB is a straight line which produces with a medial area a medial whole.

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, neither of them is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6] Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG. But, as AF is to FG, so is AI to FK, [VI. 1] therefore AI is incommensurable with FK. [X. 11] And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21] Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21] Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD. But, as AG is to GD, so is AK to KD; [VI. 1] therefore AK is incommensurable with KD. [X. 11]

Now let the square LM be constructed equal to AI, and let NO equal to FK, and about the same angle, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Then in manner similar to the above we can prove that LN is the “side” of the area AB. I say that LN is a straight line which produces with a medial area a medial whole.

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial. Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial. And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them. Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78] and it is the “side” of the area AB.

Therefore the “side” of the area is a straight line which produces with a medial area a medial whole. Q. E. D.

PROPOSITION 97.

The square on an apotome applied to a rational straight line produces as breadth a first apotome.

Let AB be an apotome, and CD rational, and to CD let there be applied CE equal to the square on AB and producing CF as breadth; I say that CF is a first apotome.

For let BG be the annex to AB; therefore AG, GB are rational straight lines commensurable in square only. [X. 73] To CD let there be applied CH equal to the square on AG, and KL equal to the square on BG. Therefore the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB; therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] Let FM be bisected at the point N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, LN is equal to the rectangle AG, GB. Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB, therefore DM is rational. And it has been applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and commensurable in length with CD. [X. 20] Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial. And it is applied to the rational straight line CD, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial, therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB; therefore DM is incommensurable with FL. But, as DM is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a first apotome.

For, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on BG, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM. And, since the square on AG is commensurable with the square on GB, CH is also commensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is commensurable with KM. [X. 11] Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17] And CM is commensurable in length with the rational straight line CD set out; therefore CF is a first apotome. [X. Deff. III. 1]

Therefore etc. Q. E. D.

PROPOSITION 98.

The square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome.

Let AB be a first apotome of a medial straight line and CD a rational straight line, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a second apotome.

For let BG be the annex to AB; therefore AG, GB are medial straight lines commensurable in square only which contain a rational rectangle. [X. 74] To CD let there be applied CH equal to the square on AG, producing CK as breadth, and KL equal to the square on GB, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.] And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22] Now, since CL is equal to the squares on AG, GB, and, in these, the square on AB is equal to CE, therefore the remainder, twice the rectangle AG, GB, is equal to FL. [II. 7] But twice the rectangle AG, GB is rational; therefore FL is rational. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is also rational and commensurable in length with CD. [X. 20] Now, since the sum of the squares on AG, GB, that is, CL, is medial, while twice the rectangle AG, GB, that is, FL, is rational, therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a second apotome.

For let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. Now, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the rectangle AG, GB to NL, and the square on BG to KL, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1] therefore, as CK is to NM, so is NM, so is KM; [V. 11] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM. Since the CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to the greater, CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17] And the annex FM is commensurable in length with the rational straight line CD set out; therefore CF is a second apotome. [X. Deff. III. 2]

Therefore etc. Q. E. D.

PROPOSITION 99.

The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.

Let AB be a second apotome of a medial straight line, and CD rational, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a third apotome.

For let BG be the annex to AB; therefore AG, GB are medial straight lines commensurable in square only which contain a medial rectangle. [X. 75] Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.] And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22] Now, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder LF is equal to twice the rectangle AG, GB. [II. 7] Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. But the rectangle AG, GB is medial; therefore FL is also medial. And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is also rational and incommensurable in length with CD. [X. 22] And, since AG, GB are commensurable in square only, therefore AG is incommensurable in length with GB; therefore the square on AG is also incommensurable with the rectangle AG, GB. [VI. 1, X. 11] But the squares on AG, GB are commensurable with the square on AG, and twice the rectangle AG, GB with the rectangle AG, GB; therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. [X. 13] But CL is equal to the squares on AG, GB, and FL is equal to twice the rectangle AG, GB; therefore CL is also incommensurable with FL. But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a third apotome.

For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. [VI. 1, X. 11] And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM. Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. [X. 17] And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out; therefore CF is a third apotome. [X. Deff. III. 3]

Therefore etc. Q. E. D.

PROPOSITION 100.

The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome.

Let AB be a minor and CD a rational straight line, and to the rational straight line CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fourth apotome.

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on AG, GB rational, but twice the rectangle AG, GB medial. [X. 76] To CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB. And the sum of the squares on AG, GB is rational; therefore CL is also rational. And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is also rational and commensurable in length with CD. [X. 20] And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] Let then FM be bisected at the point N, and let NO be drawn through N parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. And, since twice the rectangle AG, GB is medial and is equal to FL, therefore FL is also medial. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the sum of the squares on AG, GB is rational, while twice the rectangle AG, GB is medial, the squares on AG, GB are incommensurable with twice the rectangle AG, GB. But CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB; therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say that it is also a fourth apotome.

For, since AG, GB are incommensurable in square, therefore the square on AG is also incommensurable with the square on GB. And CH is equal to the square on AG, and KL equal to the square on GB; therefore CH is incommensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11] And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the square on GB to KL, and the rectangle AG, GB to NL, therefore NL is a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to the square on MN [VI. 17], that is, to the fourth part of the square on FM. Since then CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to CM and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18] And the whole CM is commensurable in length with the rational straight line CD set out; therefore CF is a fourth apotome. [X. Deff. III. 4]

Therefore etc. Q. E. D.

PROPOSITION 101.

The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.

Let AB be the straight line which produces with a rational area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fifth apotome.

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [X. 77] To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB; therefore the whole CL is equal to the squares on AG, GB. But the sum of the squares on AG, GB together is medial; therefore CL is medial. And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable with CD. [X. 22] And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] Let then FM be bisected at N, and through N let NO be drawn parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB: And, since twice the rectangle AG, GB is rational and equal to FL, therefore FL is rational. And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is rational and commensurable in length with CD. [X. 20] Now, since CL is medial, and FL rational, therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a fifth apotome.

For we can prove similarly that the rectangle CK, KM is equal to the square on NM, that is, to the fourth part of the square on FM. And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11] Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18] And the annex FM is commensurable with the rational straight line CD set out; therefore CF is a fifth apotome. [X. Deff. III. 5] Q. E. D.

PROPOSITION 102.

The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome.

Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a sixth apotome.

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle AG, GB medial, and the squares on AG, GB incommensurable with twice the rectangle AG, GB. [X. 78] Now to CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22] Since now CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] And twice the rectangle AG, GB is medial; therefore FL is also medial. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the squares on AG, GB are incommensurable with twice the rectangle AG, GB, and CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB, therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11] And both are rational. Therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a sixth apotome.

For, since FL is equal to twice the rectangle AG, GB, let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB. But CH is equal to the square on AG, and KL is equal to the square on GB; therefore CH is incommensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable with KM. [X. 11] And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18] And neither of them is commensurable with the rational straight line CD set out; therefore CF is a sixth apotome. [X. Deff. III. 6] Q. E. D.

PROPOSITION 103.

A straight line commensurable in length with an apotome is an apotome and the same in order.

Let AB be an apotome, and let CD be commensurable in length with AB; I say that CD is also an apotome and the same in order with AB.

For, since AB is an apotome, let BE be the annex to it; therefore AE, EB are rational straight lines commensurable in square only. [X. 73] Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD; [VI. 12] therefore also, as one is to one, so are all to all; [V. 12] therefore also, as the whole AE is to the whole CF, so is AB to CD. But AB is commensurable in length with CD. Therefore AE is also commensurable with CF, and BE with DF. [X. 11] And AE, EB are rational straight lines commensurable in square only; therefore CF, FD are also rational straight lines commensurable in square only. [X. 13]

Now since, as AE is to CF, so is BE to DF, alternately therefore, as AE is to EB, so is CF to FD. [V. 16] And the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it. If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14] And, if AE is commensurable in length with the rational straight line set out, CF is so also, [X. 12] if BE, then DF also, [id.] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13] But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14] And, if AE is commensurable in length with the rational straight line set out, CF is so also, if BE, then DF also, [X. 12] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]

Therefore CD is an apotome and the same in order with AB. Q. E. D.

PROPOSITION 104.

A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order.

Let AB be an apotome of a medial straight line, and let CD be commensurable in length with AB; I say that CD is also an apotome of a medial straight line and the same in order with AB.

For, since AB is an apotome of a medial straight line, let EB be the annex to it. Therefore AE, EB are medial straight lines commensurable in square only. [X. 74, 75] Let it be contrived that, as AB is to CD, so is BE to DF; [VI. 12] therefore AE is also commensurable with CF, and BE with DF. [V. 12, X. 11] But AE, EB are medial straight lines commensurable in square only; therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13] therefore CD is an apotome of a medial straight line. [X. 74, 75] I say next that it is also the same in order with AB.

Since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD. But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. [V. 16, X. 11] Therefore, if the rectangle AE, EB is rational, the rectangle CF, FD will also be rational, [X. Def. 4] and if the rectangle AE, EB is medial, the rectangle CF, FD is also medial. [X. 23, Por.]

Therefore CD is an apotome of a medial straight line and the same in order with AB. [X. 74, 75] Q. E. D.

PROPOSITION 105.

A straight line commensurable with a minor straight line is minor.

Let AB be a minor straight line, and CD commensurable with AB; I say that CD is also minor.

Let the same construction be made as before; then, since AE, EB are incommensurable in square, [X. 76] therefore CF, FD are also incommensurable in square. [X. 13] Now since, as AE is to EB, so is CF to FD, [V. 12, V. 16] therefore also, as the square on AE is to the square on EB, so is the square on CF to the square on FD. [VI. 22] Therefore, componendo, as the squares on AE, EB are to the square on EB, so are the squares on CF, FD to the square on FD. [V. 18] But the square on BE is commensurable with the square on DF; therefore the sum of the squares on AE, EB is also commensurable with the sum of the squares on CF, FD. [V. 16, X. 11] But the sum of the squares on AE, EB is rational; [X. 76] therefore the sum of the squares on CF, FD is also rational. [X. Def. 4] Again, since, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD, while the square on AE is commensurable with the square on CF, therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. But the rectangle AE, EB is medial; [X. 76] therefore the rectangle CF, FD is also medial; [X. 23, Por.] therefore CF, FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial.

Therefore CD is minor. [X. 76] Q. E. D.

PROPOSITION 106.

A straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole.

Let AB be a straight line which produces with a rational area a medial whole, and CD commensurable with AB; I say that CD is also a straight line which produces with a rational area a medial whole.

For let BE be the annex to AB; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on AE, EB medial, but the rectangle contained by them rational. [X. 77] Let the same construction be made. Then we can prove, in manner similar to the foregoing, that CF, FD are in the same ratio as AE, EB, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that CF, FD are also straight lines incommensurable in square which make the sum of the squares on CF, FD medial, but the rectangle contained by them rational.

Therefore CD is a straight line which produces with a rational area a medial whole. [X. 77] Q. E. D.

PROPOSITION 107.

A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole.

Let AB be a straight line which produces with a medial area a medial whole, and let CD be commensurable with AB; I say that CD is also a straight line which produces with a medial area a medial whole.

For let BE be the annex to AB, and let the same construction be made; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them. [X. 78] Now, as was proved, AE, EB are commensurable with CF, FD, the sum of the squares on AE, EB with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; therefore CF, FD are also straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them.

Therefore CD is a straight line which produces with a medial area a medial whole. [X. 78]

PROPOSITION 108.

If from a rational area a medial area be subtracted, the side of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line.

For from the rational area BC let the medial area BD be subtracted; I say that the “side” of the remainder EC becomes one of two irrational straight lines, either an apotome or a minor straight line.

For let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to BC, and let GK equal to DB be subtracted; therefore the remainder EC is equal to LH. Since then BC is rational, and BD medial, while BC is equal to GH, and BD to GK, therefore GH is rational, and GK medial. And they are applied to the rational straight line FG; therefore FH is rational and commensurable in length with FG, [X. 20] while FK is rational and incommensurable in length with FG; [X. 22] therefore FH is incommensurable in length with FK. [X. 13] Therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome [X. 73], and KF the annex to it. Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable.

First, let the square on it be greater by the square on a straight line commensurable with it. Now the whole HF is commensurable in length with the rational straight line FG set out; therefore KH is a first apotome. [X. Deff. III. 1] But the “side” of the rectangle contained by a rational straight line and a first apotome is an apotome. [X. 91] Therefore the “side” of LH, that is, of EC, is an apotome.

But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the whole FH is commensurable in length with the rational straight line FG set out, KH is a fourth apotome. [X. Deff. III. 4] But the “side” of the rectangle contained by a rational straight line and a fourth apotome is minor. [X. 94] Q. E. D.

PROPOSITION 109.

If from a medial area a rational area be subtracted, there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

For from the medial area BC let the rational area BD be subtracted. I say that the “side” of the remainder EC becomes one of two irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

For let a rational straight line FG be set out, and let the areas be similarly applied. It follows then that FH is rational and incommensurable in length with FG, while KF is rational and commensurable in length with FG; therefore FH, FK are rational straight lines commensurable in square only; [X. 13] therefore KH is an apotome, and FK the annex to it. [X. 73] Now the square on HF is greater than the square on FK either by the square on a straight line commensurable with HF or by the square on a straight line incommensurable with it.

If then the square on HF is greater than the square on FK by the square on a straight line commensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a second apotome. [X. Deff. III. 2] But FG is rational; so that the “side” of LH, that is, of EC, is a first apotome of a medial straight line. [X. 92]

But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a fifth apotome; [X. Deff. III. 5] so that the “side” of EC is a straight line which produces with a rational area a medial whole. [X. 95]

PROPOSITION 110.

If from a medial area there be subtracted a medial area incommensurable with the whole, the two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

For, as in the foregoing figures, let there be subtracted from the medial area BC the medial area BD incommensurable with the whole; I say that the “side” of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

For, since each of the rectangles BC, BD is medial, and BC is incommensurable with BD, it follows that each of the straight lines FH, FK will be rational and incommensurable in length with FG. [X. 22] And, since BC is incommensurable with BD, that is, GH with GK, HF is also incommensurable with FK; [VI. 1, X. 11] therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome. [X. 73]

If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH, FK is commensurable in length with the rational straight line FG set out, KH is a third apotome. [X. Deff. III. 3] But KL is rational, and the rectangle contained by a rational straight line and a third apotome is irrational, and the “side” of it is irrational, and is called a second apotome of a medial straight line; [X. 93] so that the “side” of LH, that is, of EC, is a second apotome of a medial straight line.

But, if the square on FH is greater than the square on FK by the square on a straight line incommensurable with FH, while neither of the straight lines HF, FK is commensurable in length with FG, KH is a sixth apotome. [X. Deff. III. 6] But the “side” of the rectangle contained by a rational straight line and a sixth apotome is a straight line which produces with a medial area a medial whole. [X. 96] Therefore the “side” of LH, that is, of EC, is a straight line which produces with a medial area a medial whole. Q. E. D.

PROPOSITION 111.

The apotome is not the same with the binomial straight line.

Let AB be an apotome; I say that AB is not the same with the binomial straight line.

For, if possible, let it be so; let a rational straight line DC be set out, and to CD let there be applied the rectangle CE equal to the square on AB and producing DE as breadth. Then, since AB is an apotome, DE is a first apotome. [X. 97] Let EF be the annex to it; therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1] Again, since AB is binomial, therefore DE is a first binomial straight line. [X. 60] Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1] Therefore DF is also commensurable in length with DG; [X. 12] therefore the remainder GF is also commensurable in length with DF. [X. 15] But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF. [X. 13] Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [X. 73] But it is also rational: which is impossible.

Therefore the apotome is not the same with the binomial straight line. Q. E. D.

PROPOSITION 112.

The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.

Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A; I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC.

For again let the rectangle BD, G be equal to the square on A. Since then the rectangle BC, EF is equal to the rectangle BD, G, therefore, as CB is to BD, so is G to EF. [VI. 16] But CB is greater than BD; therefore G is also greater than EF. [V. 16, V. 14] Let EH be equal to G; therefore, as CB is to BD, so is HE to EF; therefore, separando, as CD is to BD, so is HF to FE. [V. 17] Let it be contrived that, as HF is to FE, so is FK to KE; therefore also the whole HK is to the whole KF as FK is to KE; for, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. [V. 12] But, as FK is to KE, so is CD to DB; [V. 11] therefore also, as HK is to KF, so is CD to DB. [id.] But the square on CD is commensurable with the square on DB; [X. 36] therefore the square on HK is also commensurable with the square on KF. [VI. 22, X. 11] And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9] Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK. [X. 15] Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational. And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD. Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only, therefore FK, KE are also commensurable in square only. [X. 11] But KE is rational; therefore FK is also rational. Therefore FK, KE are rational straight lines commensurable in square only; therefore EF is an apotome. [X. 73]

Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it.

If then the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. [X. 14] And, if CD is commensurable in length with the rational straight line set out, so also is FK; [X. 11, 12] if BD is so commensurable, so also is KE; [X. 12] but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so.

But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK. [X. 14] And, if CD is commensurable with the rational straight line set out, so also is FK; if BD is so commensurable, so also is KE; but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so; so that FE is an apotome, the terms of which FK, KE are commensurable with the terms CD, DB of the binomial straight line and in the same ratio, and it has the same order as BC. Q. E. D.

PROPOSITION 113.

The square on a rational straight line, if applied to an apotome, produces as, breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome.

Let A be a rational straight line and BD an apotome, and let the rectangle BD, KH be equal to the square on A, so that the square on the rational straight line A when applied to the apotome BD produces KH as breadth; I say that KH is a binomial straight line the terms of which are commensurable with the terms of BD and in the same ratio; and further KH has the same order as BD.

For let DC be the annex to BD; therefore BC, CD are rational straight lines commensurable in square only. [X. 73] Let the rectangle BC, G be also equal to the square on A. But the square on A is rational; therefore the rectangle BC, G is also rational. And it has been applied to the rational straight line BC; therefore G is rational and commensurable in length with BC. [X. 20] Since now the rectangle BC, G is equal to the rectangle BD, KH, therefore, proportionally, as CB is to BD, so is KH to G. [VI. 16] But BC is greater than BD; therefore KH is also greater than G. [V. 16, V. 14] Let KE be made equal to G; therefore KE is commensurable in length with BC. And since, as CB is to BD, so is HK to KE, therefore, convertendo, as BC is to CD, so is KH to HE. [V. 19, Por.] Let it be contrived that, as KH is to HE, so is HF to FE; therefore also the remainder KF is to FH as KH is to HE, that is, as BC is to CD. [V. 19] But BC, CD are commensurable in square only; therefore KF, FH are also commensurable in square only. [X. 11] And since, as KH is to HE, so is KF to FH, while, as KH is to HE, so is HF to FE, therefore also, as KF is to FH, so is HF to FE, [V. 11] so that also, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9] therefore also, as KF is to FE, so is the square on KF to the square on FH. But the square on KF is commensurable with the square on FH, for KF, FH are commensurable in square; therefore KF is also commensurable in length with FE, [X. 11] so that KF is also commensurable in length with KE. [X. 15] But KE is rational and commensurable in length with BC; therefore KF is also rational and commensurable in length with BC. [X. 12] And, since, as BC is to CD, so is KF to FH, alternately, as BC is to KF, so is DC to FH. [V. 16] But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11] But BC, CD are rational straight lines commensurable in square only; therefore KF, FH are also rational straight lines [X. Def. 3] commensurable in square only; therefore KH is binomial. [X. 36]

If now the square on BC is greater than the square on CD by the square on a straight line commensurable with BC, the square on KF will also be greater than the square on FH by the square on a straight line commensurable with KF. [X 14] And, if BC is commensurable in length with the rational straight line set out, so also is KF; if CD is commensurable in length with the rational straight line set out, so also is FH, but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH.

But, if the square on BC is greater than the square on CD by the square on a straight line incommensurable with BC, the square on KF is also greater than the square on FH by the square on a straight line incommensurable with KF. [X. 14] And, if BC is commensurable with the rational straight line set out, so also is KF; if CD is so commensurable, in length with the rational straight line set out, so also is FH; but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH.

Therefore KH is a binomial straight line, the terms of which KF, FH are commensurable with the terms BC, CD of the apotome and in the same ratio, and further KH has the same order as BD. Q. E. D.

PROPOSITION 114.

If an area be contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, the side of the area is rational.

For let an area, the rectangle AB, CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter; let the terms CE, ED of the binomial straight line be commensurable with the terms AF, FB of the apotome and in the same ratio; and let the “side” of the rectangle AB, CD be G; I say that G is rational.

For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth. Therefore KL is an apotome. Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112] But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML. Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19] But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL. [X. 11] And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1] therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11] But the rectangle CD, KL is equal to the square on H; therefore the rectangle CD, AB is commensurable with the square on H. But the square on G is equal to the rectangle CD, AB; therefore the square on G is commensurable with the square on H. But the square on H is rational; therefore the square on G is also rational; therefore G is rational. And it is the “side” of the rectangle CD, AB.

Therefore etc.

PORISM.
And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines. Q. E. D.

PROPOSITION 115.

From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.

Let A be a medial straight line; I say that from A there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.

Let a rational straight line B be set out, and let the square on C be equal to the rectangle B, A; therefore C is irrational; [X. Def. 4] for that which is contained by an irrational and a rational straight line is irrational. [deduction from X. 20] And it is not the same with any of the preceding; for the square on none of the preceding, if applied to a rational straight line produces as breadth a medial straight line. Again, let the square on D be equal to the rectangle B, C; therefore the square on D is irrational. [deduction from X. 20] Therefore D is irrational; [X. Def. 4] and it is not the same with any of the preceding, for the square on none of the preceding, if applied to a rational straight line, produces C as breadth. Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding. Q. E. D.

Permanent link

http://www2.hf.uio.no/common/apps/permlink/permlink.php?app=polyglotta&context=ctext&uid=72583de8-3ed4-11e1-aecb-00215aecadea

Choose languages

Choose images, etc.

Choose languages

Choose display