For, if E, F, G are not the least of those which have the same ratio with A, B, C,
there will be numbers less than E, F, G which are in the same ratio with A, B, C.
Let them be H, K, L;
therefore H measures A the same number of times that the numbers K, L measure the numbers B, C respectively.
Now, as many times as H measures A, so many units let there be in M;
therefore the numbers K, L also measure the numbers B, C respectively according to the units in M.
And, since H measures A according to the units in M,
therefore M also measures A according to the units in H. [VII. 16]
For the same reason M also measures the numbers B, C according to the units in the numbers K, L respectively;
Therefore M measures A, B, C.
Now, since H measures A according to the units in M,
therefore H by multiplying M has made A. [VII. Def. 15]
For the same reason also E by multiplying D has made A.
Therefore the product of E, D is equal to the product of H, M.
Therefore, as E is to H, so is M to D. [VII. 19]
But E is greater than H;
therefore M is also greater than D.
And it measures A, B, C: which is impossible,
for by hypothesis D is the greatest common measure of A, B, C.
Therefore there cannot be any numbers less than E, F, G which are in the same ratio with A, B, C.
Therefore E, F, G are the least of those which have the same ratio with A, B, C.
Q. E. D.
Complete text
Book I