Proposition 7.33
Given as many numbers as we please, to find the least of those which have the same ratio with them.
Given as many numbers as we please, to find the least of those which have the same ratio with them.
Let A, B, C be the given numbers, as many as we please; thus it is required to find the least of
those which have the same ratio with A, B, C.
A, B, C are either prime to one another or not.
Now, if A, B, C are prime to one
another, they are the least of those which have the same ratio with them. [VII. 21]
But, if not, let D the greatest common measure of A, B, C be taken, [VII. 3]
respectively, so many units let there be in the numbers E, F, G respectively.
Therefore the numbers E, F, G measure the numbers A, B, C respectively according to the units in D. [VII. 16]
Therefore E, F, G measure A, B, C the same number of
times; therefore E, F, G are in the same ratio with A, B, C. [VII. Def. 20]
I say next that they are the least that are in that ratio.
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.