---
book: 10
number: 51
id: "X.51"
kind: "construction"
uses: ["[[book-10/proposition-6]]", "[[book-10/proposition-9]]", "[[book-5/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.51"
license: "CC-BY-SA-4.0"
---

# X.51

*To find the fourth binomial straight line*.

## Proof

Let two numbers *AC*, *CB* be set out such that *AB* neither has to *BC*, nor yet to *AC*, the ratio which a square number has to a square number.

Let a rational straight line *D* be set out, and let *EF* be commensurable in length with *D*; therefore *EF* is also rational.

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

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

Therefore *EF*, *FG* are rational straight lines commensurable in square only; so that *EG* is binomial.

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

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

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

But *AB* has not to *BC* the ratio which a square number has to a square number; therefore neither has the square on *EF* to the square on *H* the ratio which a square number has to a square number.

Therefore *EF* is incommensurable in length with *H*; [[book-10/proposition-9|X. 9]] therefore the square on *EF* is greater than the square on *GF* by the square on a straight line incommensurable with *EF*.

And *EF*, *FG* are rational straight lines commensurable in square only, and *EF* is commensurable in length with *D*.

Therefore *EG* is a fourth binomial straight line. Q. E. D.