---
book: 10
number: 48
id: "X.48"
kind: "construction"
uses: ["[[book-10/proposition-28]]", "[[book-10/proposition-6]]", "[[book-10/proposition-9]]", "[[book-10/proposition-36]]", "[[book-5/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.48"
license: "CC-BY-SA-4.0"
---

# X.48

*To find the first 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 *CA* the ratio which a square number has to a square number; [[book-10/proposition-28|Lemma I after X. 28]] let any 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.]]

But *AB* has to *AC* the ratio which a number has to a number; therefore the square on *EF* also has to the square on *FG* the ratio which a number has to a number, so that the square on *EF* is commensurable with the square on *FG*. [[book-10/proposition-6|X. 6]]

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

And, since *BA* has not to *AC* the ratio which a square number has to a square number. neither, therefore, 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; therefore *EG* is binomial. [[book-10/proposition-36|X. 36]]

I say that it is also a first binomial straight line.

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

Let then the squares on *FG*, *H* be equal to the square on *EF*.

Now since, as *BA* is to *AC*, so is the square on *EF* to the square on *FG*, therefore, *convertendo*, as *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 to *BC* the ratio which a square number has to a square number; therefore the square on *EF* also has to the square on *H* the ratio which a square number has to a square number.

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

And *EF*, *FG* are rational, and *EF* is commensurable in length with *D*.

Therefore *EF* is a first binomial straight line. Q. E. D.