---
book: 5
number: 13
id: "V.13"
kind: "theorem"
uses: []
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.13"
license: "CC-BY-SA-4.0"
---

# V.13

*If a first magnitude have to a second the same ratio as a third to a fourth*, *and the third have to the fourth a greater ratio than a fifth has to a sixth*, *the first will also have to the second a greater ratio than the fifth to the sixth*.

## Proof

For let a first magnitude *A* have to a second *B* the same ratio as a third *C* has to a fourth *D*, and let the third *C* have to the fourth *D* a greater ratio than a fifth *E* has to a sixth *F*; I say that the first *A* will also have to the second *B* a greater ratio than the fifth *E* to the sixth *F*.

For, since there are some equimultiples of *C*, *E*, and of *D*, *F* other, chance, equimultiples, such that the multiple of *C* is in excess of the multiple of *D*, while the multiple of *E* is not in excess of the multiple of *F*, [[book-5/definitions#Definition 7|V. Def. 7]] let them be taken, and let *G*, *H* be equimultiples of *C*, *E*, and *K*, *L* other, chance, equimultiples of *D*, *F*, so that *G* is in excess of *K*, but *H* is not in excess of *L*; and, whatever multiple *G* is of *C*, let *M* be also that multiple of *A*, and, whatever multiple *K* is of *D*, let *N* be also that multiple of *B*.

Now, since, as *A* is to *B*, so is *C* to *D*, and of *A*, *C* equimultiples *M*, *G* have been taken, and of *B*, *D* other, chance, equimultiples *N*, *K*, therefore, if *M* is in excess of *N*, *G* is also in excess of *K*, if equal, equal, and if less, less. [[book-5/definitions#Definition 5|V. Def. 5]]

But *G* is in excess of *K*; therefore *M* is also in excess of *N*.

But *H* is not in excess of *L*; and *M*, *H* are equimultiples of *A*, *E*, and *N*, *L* other, chance, equimultiples of *B*, *F*; therefore *A* has to *B* a greater ratio than *E* has to *F*. [[book-5/definitions#Definition 7|V. Def. 7]]

Therefore etc. Q. E. D.