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Euclid: Elementa

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PROPOSITION 34. 
 
 
Given two numbers, to find the least number which they measure. 
 
 
Let A, B be the two given numbers;  thus it is required to find the least number which they measure. 
   
   
Now A, B are either prime to one another or not. First, let A, B be prime to one another, and let A by multiplying B make C;therefore also B by multiplying A has made C. [VII. 16] Therefore A, B measure C.  I say next that it is also the least number they measure. 
   
   
For, if not, A, B will measure some number which is less than C.  Let them measure D.  Then, as many times as A measures D, so many units let there be in E, and, as many times as B measures D, so many units let there be in F;  therefore A by multiplying E has made D, and B by multiplying F has made D; [VII. Def. 15]  therefore the product of A, E is equal to the product of B, F.  Therefore, as A is to B, so is F E. [VII. 19]  But A, B are prime, primes are also least, [VII. 21]  and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20]   therefore B measures E, as consequent consequent.  And, since A by multiplying B, E has made C, D, therefore, as B is to E, so is C to D. [VII. 17]  But B measures E;  therefore C also measures D, the greater the less: which is impossible.  Therefore A, B do not measure any number less than C;  therefore C is the least that is measured by A, B. 
                           
                           
Next, let A, B not be prime to one another,  and let F, E, the least numbers of those which have the same ratio with A, B, be taken; [VII. 33]  therefore the product of A, E is equal to the product of B, F. [VII. 19]  And let A by multiplying E make C;  therefore also B by multiplying F has made C;  therefore A, B measure C.  I say next that it is also the least number that they measure. 
             
             
For, if not, A, B will measure some number which is less than C.  Let them measure D.  And, as many times as A measures D, so many units let there be in G,  and, as many times as B measures D, so many units let there be in H.  Therefore A by multiplying G has made D,  and B by multiplying H has made D.  Therefore the product of A, G is equal to the product of B, H;  therefore, as A is to B, so is H to G. [VII. 19]  But, as A is to B, so is F to E.  Therefore also, as F is to E, so is H to G.  But F, E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20]  therefore E measures G.  And, since A by multiplying E, G has made C, D, therefore, as E is to G, so is C to D. [VII. 17]  But E measures G;  therefore C also measures D, the greater the less: which is impossible.  Therefore A, B will not measure any number which is less than C.  Therefore C is the least that is measured by A, B.  Q. E. D. 
                                   
                                   
 
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