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

# V.1

*If there be any number of magnitudes whatever which are*, *respectively*, *equimultiples of any magnitudes equal in multitude*, *then*, *whatever multiple one of the magnitudes is of one*, *that multiple also will all be of all*.

## Proof

Let any number of magnitudes whatever *AB*, *CD* be respectively equimultiples of any magnitudes *E*, *F* equal in multitude; I say that, whatever multiple *AB* is of *E*, that multiple will *AB*, *CD* also be of *E*, *F*.

For, since *AB* is the same multiple of *E* that *CD* is of *F*, as many magnitudes as there are in *AB* equal to *E*, so many also are there in *CD* equal to *F*.

Let *AB* be divided into the magnitudes *AG*, *GB* equal to *E*, and *CD* into *CH*, *HD* equal to *F*; then the multitude of the magnitudes *AG*, *GB* will be equal to the multitude of the magnitudes *CH*, *HD*.

Now, since *AG* is equal to *E*, and *CH* to *F*, therefore *AG* is equal to *E*, and *AG*, *CH* to *E*, *F*.

For the same reason

*GB* is equal to *E*, and *GB*, *HD* to *E*, *F*; therefore, as many magnitudes as there are in *AB* equal to *E*, so many also are there in *AB*, *CD* equal to *E*, *F*; therefore, whatever multiple *AB* is of *E*, that multiple will *AB*, *CD* also be of *E*, *F*.

Therefore etc. Q. E. D.