---
book: 5
number: 5
id: "V.5"
kind: "theorem"
uses: ["[[book-5/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.5"
license: "CC-BY-SA-4.0"
---

# V.5

*If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted*, *the remainder will also be the same multiple of the remainder that the whole is of the whole*.

## Proof

For let the magnitude *AB* be the same multiple of the magnitude *CD* that the part *AE* subtracted is of the part *CF* subtracted; I say that the remainder *EB* is also the same multiple of the remainder *FD* that the whole *AB* is of the whole *CD*.

For, whatever multiple *AE* is of *CF*, let *EB* be made that multiple of *CG*.

Then, since *AE* is the same multiple of *CF* that *EB* is of *GC*, therefore *AE* is the same multiple of *CF* that *AB* is of *GF*. [[book-5/proposition-1|V. 1]]

But, by the assumption, *AE* is the same multiple of *CF* that *AB* is of *CD*.

Therefore *AB* is the same multiple of each of the magnitudes *GF*, *CD*; therefore *GF* is equal to *CD*.

Let *CF* be subtracted from each; therefore the remainder *GC* is equal to the remainder *FD*.

And, since *AE* is the same multiple of *CF* that *EB* is of *GC*, and *GC* is equal to *DF*, therefore *AE* is the same multiple of *CF* that *EB* is of *FD*.

But, by hypothesis,

*AE* is the same multiple of *CF* that *AB* is of *CD*; therefore *EB* is the same multiple of *FD* that *AB* is of *CD*.

That is, the remainder *EB* will be the same multiple of the remainder *FD* that the whole *AB* is of the whole *CD*.

Therefore etc. Q. E. D.