| If two magnitudes are equimultiples of two magnitudes, and any magnitudes subtracted from them are equimultiples of the same, then the remainders either equal the same or are equimultiples of them. | ||
| Let two magnitudes AB and CD be equimultiples of two magnitudes E and F, and let AG and CH subtracted from them be equimultiples of the same two E and F.
I say that the remainders GB and HD either equal E and F or are equimultiples of them. | ||
| First, let GB equal E.
I say that HD also equals F. Make CK equal to F. |
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| Since AG is the same multiple of E that CH is of F, while GB equals E, and KC equals F, therefore AB is the same multiple of E that KH is of F. | V.2 | |
| But, by hypothesis, AB is the same multiple of E that CD is of F, therefore KH is the same multiple of F that CD is of F.
Since then each of the magnitudes KH and CD is the same multiple of F, therefore KH equals CD. Subtract CH from each. Then the remainder KC equals the remainder HD. But F equals KC, therefore HD also equals F. Hence, if GB equals E, HD also equals F. Similarly we can prove that, even if GB is a multiple of E, HD is also the same multiple of F. | ||
| Therefore, if two magnitudes are equimultiples of two magnitudes, and any magnitudes subtracted from them are equimultiples of the same, then the remainders either equal the same or are equimultiples of them. | ||
| Q.E.D. | ||
Its proof depends on a distributivity, namely that multiplication by magnitudes distributes over subtraction of numbers: (m - n)a = ma - na. Euclid takes 4 as m and 3 as n. He has two cases since since he doesn't take 1 to be a number.
This proposition is not used in the rest of the Elements.