---
book: 8
number: 11
id: "VIII.11"
kind: "theorem"
uses: ["[[book-7/proposition-17]]", "[[book-7/proposition-18]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:8.prop.11"
license: "CC-BY-SA-4.0"
---

# VIII.11

*Between two square numbers there is one mean proportional number, and the square has to the square the ratio duplicate of that which the side has to the side.*

## Proof

Let *A*, *B* be square numbers, and let *C* be the side of *A*, and *D* of *B*; I say that between *A*, *B* there is one mean proportional number, and *A* has to *B* the ratio duplicate of that which *C* has to *D*.

For let *C* by multiplying *D* make *E*.

Now, since *A* is a square and *C* is its side, therefore *C* by multiplying itself has made *A*.

For the same reason also *D* by multiplying itself has made *B*.

Since then *C* by multiplying the numbers *C*, *D* has made *A*, *E* respectively, therefore, as *C* is to *D*, so is *A* to *E*. [[book-7/proposition-17|VII. 17]]

For the same reason also, as *C* is to *D*, so is *E* to *B*. [[book-7/proposition-18|VII. 18]]

Therefore also, as *A* is to *E*, so is *E* to *B*.

Therefore between *A*, *B* there is one mean proportional number.

I say next that *A* also has to *B* the ratio duplicate of that which *C* has to *D*.

For, since *A*, *E*, *B* are three numbers in proportion, therefore *A* has to *B* the ratio duplicate of that which *A* has to *E*. [[book-5/definitions#Definition 9|V. Def. 9]]

But, as *A* is to *E*, so is *C* to *D*.

Therefore *A* has to *B* the ratio duplicate of that which the side *C* has to *D*. Q. E. D.