---
book: 5
number: 8
id: "V.8"
kind: "theorem"
uses: ["[[book-5/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.8"
license: "CC-BY-SA-4.0"
---

# V.8

*Of unequal magnitudes*, *the greater has to the same a greater ratio than the less has*; *and the same has to the less a greater ratio than it has to the greater*.

## Proof

Let *AB*, *C* be unequal magnitudes, and let *AB* be greater; let *D* be another, chance, magnitude; I say that *AB* has to *D* a greater ratio than *C* has to *D*, and *D* has to *C* a greater ratio than it has to *AB*.

For, since *AB* is greater than *C*, let *BE* be made equal to *C*; then the less of the magnitudes *AE*, *EB*, if multiplied, will sometime be greater than *D*. [[book-5/definitions#Definition 4|V. Def. 4]]

[*Case* I.]

First, let *AE* be less than *EB*; let *AE* be multiplied, and let *FG* be a multiple of it which is greater than *D*; then, whatever multiple *FG* is of *AE*, let *GH* be made the same multiple of *EB* and *K* of *C*; and let *L* be taken double of *D*, *M* triple of it, and successive multiples increasing by one, until what is taken is a multiple of *D* and the first that is greater than *K*. Let it be taken, and let it be *N* which is quadruple of *D* and the first multiple of it that is greather than *K*.

Then, since *K* is less than *N* first, therefore *K* is not less than *M*.

And, since *FG* is the same multiple of *AE* that *GH* is of *EB*, therefore *FG* is the same multiple of *AE* that *FH* is of *AB*. [[book-5/proposition-1|V. 1]]

But *FG* is the same multiple of *AE* that *K* is of *C*; therefore *FH* is the same multiple of *AB* that *K* is of *C*; therefore *FH*, *K* are equimultiples of *AB*, *C*.

Again, since *GH* is the same multiple of *EB* that *K* is of *C*, and *EB* is equal to *C*, therefore *GH* is equal to *K*.

But *K* is not less than *M*; therefore neither is *GH* less than *M*.

And *FG* is greater than *D*; therefore the whole *FH* is greater than *D*, *M* together.

But *D*, *M* together are equal to *N*, inasmuch as *M* is triple of *D*, and *M*, *D* together are quadruple of *D*, while *N* is also quadruple of *D*; whence *M*, *D* together are equal to *N*.

But *FH* is greater than *M*, *D*; therefore *FH* is in excess of *N*, while *K* is not in excess of *N*.

And *FH*, *K* are equimultiples of *AB*, *C*, while *N* is another, chance, multiple of *D*; therefore *AB* has to *D* a greater ratio than *C* has to *D*. [[book-5/definitions#Definition 7|V. Def. 7]]

I say next, that *D* also has to *C* a greater ratio than *D* has to *AB*.

For, with the same construction, we can prove similarly that *N* is in excess of *K*, while *N* is not in excess of *FH*.

And *N* is a multiple of *D*, while *FH*, *K* are other, chance, equimultiples of *AB*, *C*; therefore *D* has to *C* a greater ratio than *D* has to *AB*. [[book-5/definitions#Definition 7|V. Def. 7]]

[*Case* 2.]

Again, let *AE* be greater than *EB*.

Then the less, *EB*, if multiplied, will sometime be greater than *D*. [[book-5/definitions#Definition 4|V. Def. 4]]

Let it be multiplied, and let *GH* be a multiple of *EB* and greater than *D*; and, whatever multiple *GH* is of *EB*, let *FG* be made the same multiple of *AE*, and *K* of *C*.

Then we can prove similarly that *FH*, *K* are equimultiples of *AB*, *C*; and, similarly, let *N* be taken a multiple of *D* but the first that is greater than *FG*, so that *FG* is again not less than *M*.

But *GH* is greater than *D*; therefore the whole *FH* is in excess of *D*, *M*, that is, of *N*.

Now *K* is not in excess of *N*, inasmuch as *FG* also, which is greater than *GH*, that is, than *K*, is not in excess of *N*.

And in the same manner, by following the above argument, we complete the demonstration.

Therefore etc. Q. E. D.