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 they 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.

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.

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.*

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.*

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.*

Therefore the product of *E* and *D* equals the product of *H* and *M.* Therefore *E* is to *H* as *M* is to *D.*

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.

Given three numbers *a*, *b*, etc., let *d* be their greatest common divisor. If *d* measures *a* *m* times, let *e* = *mu*, where *u* is the unit. Likewise, if *b* = *nd*, let *f* = *nb*, etc. (Euclid hasn’t defined division, but he’s essentially saying let *e* = *a*/*d*, *f* = *b*/*d*, etc.)
Then *a* = *kd*, *b* = *ke*, etc., where *d* = *ku*. Therefore, *a* : *b* : *c* = *e* : *f* : *g.*

The remainder of the proof shows that no smaller numbers are in the same ratio.

This proposition is unusual in that it discusses a ratio *a* : *b* : *c* of three (or more) numbers. It also has the proportion *a* : *b* : *c* = *e* : *f* : *g.* These multiterm ratios and proportions may have been left over from an earlier time. Euclid argues that the proportion holds because *e, f,* and *g* measure *a, b,* and *c,* respectively, the same number, *d,* times. By the definition of proportion, that observation directly implies

The desired proportion, *a* : *b* : *c* = *e* : *f* : *g* is an alternate form of that multiple proportion. See V.Def.13 for the definition of alternate ratios.