| Given as many numbers as we please, to find the least of those which have the same ratio with them. | ||
| Let A, B, and C be the given numbers, as many as we please.
It is required to find the least of those which have the same ratio with A, B, and C. Either A, B, and C are relatively prime or thy are not. | ||
| Now, if A, B, and C are relatively prime, then they are the least of those which have the same ratio with them. | VII.21 | |
| But, if not, take D the greatest common measure of A, B, and C. Let there be as many units in the numbers E, F, and G as the times that D measures the numbers A, B, and C respectively. | VII.3 | |
| Therefore the numbers E, F, and G measure the numbers A, B, and C respectively according to the units in D. Therefore E, F, and G measure A, B, and C the same number of times. Therefore E, F, and G are in the same ratio with A, B, and C. | VII.16 | |
| I say next that they are the least that are in that ratio.
If E, F, and G are not the least of those which have the same ratio with A, B, and C, then there are numbers less than E, F, and G in the same ratio with A, B, and C. Let them be H, K, and L. Therefore H measures A the same number of times that the numbers K and L measure the numbers B and C respectively. Let there be as many units in M as the times that H measures A. Then the numbers K and L also measure the numbers B and 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. For the same reason M also measures the numbers B and C according to the units in the numbers K and L respectively. Therefore M measures A, B, and C. | VII.16 | |
| Now, since H measures A according to the units in M, therefore H multiplied by M makes A. For the same reason also E multiplied by D makes A. | VII.Def.15 | |
| Therefore the product of E and D equals the product of H and M. Therefore E is to H as M is to D. | VII.19 | |
| But E is greater than H, therefore M is also greater than D. And it measures A, B, and C, which is impossible, for by hypothesis, D is the greatest common measure of A, B, and C.
Therefore there cannot be any numbers less than E, F, and G which are in the same ratio with A, B, and C. Therefore E, F, and G are the least of those which have the same ratio with A, B, and C. | ||
| Q.E.D. | ||
Book VII Introduction - Proposition VII.32 - Proposition VII.34.