---
book: 7
number: 8
id: "VII.8"
kind: "theorem"
uses: ["[[book-7/proposition-7]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:7.prop.8"
license: "CC-BY-SA-4.0"
---

# VII.8

*If a number be the same parts of a number that a number subtracted is of a number subtracted, the remainder will also be the same parts of the remainder that the whole is of the whole.*

## Proof

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

For let *GH* be made equal to *AB*.

Therefore, whatever parts *GH* is of *CD*, the same parts also is *AE* of *CF*.

Let *GH* be divided into the parts of *CD*, namely *GK*, *KH*, and *AE* into the parts of *CF*, namely *AL*, *LE*; thus the multitude of *GK*, *KH* will be equal to the multitude of *AL*, *LE*.

Now since, whatever part *GK* is of *CD*, the same part also is *AL* of *CF*, while. *CD* is greater than *CF*, therefore *GK* is also greater than *AL*.

Let *GM* be made equal to *AL*.

Therefore, whatever part *GK* is of *CD*, the same part also is *GM* of *CF*; therefore also the remainder *MK* is the same part of the remainder *FD* that the whole *GK* is of the whole *CD*. [[book-7/proposition-7|VII. 7]]

Again, since, whatever part *KH* is of *CD*, the same part also is *EL* of *CF*, while *CD* is greater than *CF*, therefore *HK* is also greater than *EL*.

Let *KN* be made equal to *EL*.

Therefore, whatever part *KH* is of *CD*, the same part also is *KN* of *CF*; therefore also the remainder *NH* is the same part of the remainder *FD* that the whole *KH* is of the whole *CD*. [[book-7/proposition-7|VII. 7]]

But the remainder *MK* was also proved to be the same part of the remainder *FD* that the whole *GK* is of the whole *CD*; therefore also the sum of *MK*, *NH* is the same parts of *DF* that the whole *HG* is of the whole *CD*.

But the sum of *MK*, *NH* is equal to *EB*, and *HG* is equal to *BA*; therefore the remainder *EB* is the same parts of the remainder *FD* that the whole *AB* is of the whole *CD*. Q. E. D.