Book 7
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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]
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I say next that they are the least that are in that ratio.
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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.
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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.
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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.
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And, since H measures A according to the units in M, therefore M also measures A according to the units in H. [VII. 16]
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For the same reason M also measures the numbers B, C according to the units in the numbers K, L respectively;
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Therefore M measures A, B, C.
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Now, since H measures A according to the units in M, therefore H by multiplying M has made A. [VII. Def. 15]
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For the same reason also E by multiplying D has made A.
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Therefore the product of E, D is equal to the product of H, M.
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Therefore, as E is to H, so is M to D. [VII. 19]
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But E is greater than H; therefore M is also greater than D.
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And it measures A, B, C: which is impossible, for by hypothesis D is the greatest common measure of A, B, C.
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Therefore there cannot be any numbers less than E, F, G which are in the same ratio with A, B, C.
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Therefore E, F, G are the least of those which have the same ratio with A, B, C. Q. E. D. literally (as usual) each of the numbers E, F, G measures each of the numbers A, B, C.
PROPOSITION 34.
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Given two numbers, to find the least number which they measure.
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Let A, B be the two given numbers; thus it is required to find the least number which they measure.
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Now A, B are either prime to one another or not.
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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]
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Therefore A, B measure C
