---
book: 5
number: 6
id: "V.6"
kind: "theorem"
uses: ["[[book-5/proposition-2]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.6"
license: "CC-BY-SA-4.0"
---

# V.6

*If two magnitudes be equimultiples of two magnitudes*, *and any magnitudes subtracted from them be equimultiples of the same*, *the remainders also are either equal to the same or equimultiples of them*.

## Proof

For let two magnitudes *AB*, *CD* be equimultiples of two magnitudes *E*, *F*, and let *AG*, *CH* subtracted from them be equimultiples of the same two *E*, *F*; I say that the remainders also, *GB*, *HD*, are either equal to *E*, *F* or equimultiples of them.

For, first, let *GB* be equal to *E*; I say that *HD* is also equal to *F*.

For let *CK* be made equal to *F*.

Since *AG* is the same multiple of *E* that *CH* is of *F*, while *GB* is equal to *E* and *KC* to *F*, therefore *AB* is the same multiple of *E* that *KH* is of *F*. [[book-5/proposition-2|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*, *CD* is the same multiple of *F*, therefore *KH* is equal to *CD*.

Let *CH* be subtracted from each; therefore the remainder *KC* is equal to the remainder *HD*.

But *F* is equal to *KC*; therefore *HD* is also equal to *F*.

Hence, if *GB* is equal to *E*, *HD* is also equal to *F*.

Similarly we can prove that, even if *GB* be a multiple of *E*, *HD* is also the same multiple of *F*.

Therefore etc. Q. E. D.