---
book: 10
number: 50
id: "X.50"
kind: "construction"
uses: ["[[book-10/proposition-6]]", "[[book-10/proposition-9]]", "[[book-10/proposition-36]]", "[[book-5/proposition-22]]", "[[book-5/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.50"
license: "CC-BY-SA-4.0"
---

# X.50

*To find the third binomial straight line*.

## Proof

Let two numbers *AC*, *CB* be set out such that the sum of them *AB* has to *BC* the ratio which a square number has to a square number, but has not to *AC* the ratio which a square number has to a square number.

Let any other number *D*, not square, be set out also, and let it not have to either of the numbers *BA*. *AC* the ratio which a square number has to a square number.

Let any rational straight line *E* be set out, and let it be contrived that, as *D* is to *AB*, so is the square on *E* to the square on *FG*; [[book-10/proposition-6|X. 6, Por.]] therefore the square on *E* is commensurable with the square on *FG*. [[book-10/proposition-6|X. 6]]

And *E* is rational; therefore *FG* is also rational.

And, since *D* has not to *AB* the ratio which a square number has to a square number, neither has the square on *E* to the square on *FG* the ratio which a square number has to a square number; therefore *E* is incommensurable in length with *FG*. [[book-10/proposition-9|X. 9]]

Next let it be contrived that, as the number *BA* is to *AC*, so is the square on *FG* to the square on *GH*; [[book-10/proposition-6|X. 6, Por.]] therefore the square on *FG* is commensurable with the square on *GH*. [[book-10/proposition-6|X. 6]]

But *FG* is rational; therefore *GH* is also rational.

And, since *BA* has not to *AC* the ratio which a square number has to a square number, neither has the square on *FG* to the square on *HG* the ratio which a square number has to a square number; therefore *FG* is incommensurable in length with *GH*. [[book-10/proposition-9|X. 9]]

Therefore *FG*, *GH* are rational straight lines commensurable in square only; therefore *FH* is binomial. [[book-10/proposition-36|X. 36]]

I say next that it is also a third binomial straight line.

For since, as *D* is to *AB*, so is the square on *E* to the square on *FG*, and, as *BA* is to *AC*, so is the square on *FG* to the square on *GH*, therefore, ex aequali, as *D* is to *AC*, so is the square on *E* to the square on *GH*. [[book-5/proposition-22|V. 22]]

But *D* has not to *AC* the ratio which a square number has to a square number; therefore neither has the square on *E* to the square on *GH* the ratio which a square number has to a square number; therefore *E* is incommensurable in length with *GH*. [[book-10/proposition-9|X. 9]]

And since, as *BA* is to *AC*, so is the square on *FG* to the square on *GH*, therefore the square on *FG* is greater than the square on *GH*.

Let then the squares on *GH*, *K* be equal to the square on *FG*; therefore, convertendo, as *AB* is to *BC*, so is the square on *FG* to the square on *K*. [[book-5/proposition-19|V. 19, Por.]]

But *AB* has to *BC* the ratio which a square number has to a square number; therefore the square on *FG* also has to the square on *K* the ratio which a square number has to a square number; therefore *FG* is commensurable in length with *K*. [[book-10/proposition-9|X. 9]]

Therefore the square on *FG* is greater than the square on *GH* by the square on a straight line commensurable with *FG*.

And *FG*, *GH* are rational straight lines commensurable in square only, and neither of them is commensurable in length with *E*.

Therefore *FH* is a third binomial straight line. Q. E. D.