---
book: 10
number: 53
id: "X.53"
kind: "construction"
uses: ["[[book-10/proposition-6]]", "[[book-10/proposition-9]]", "[[book-10/proposition-36]]", "[[book-5/proposition-22]]", "[[book-5/proposition-19]]", "[[book-1/proposition-34]]", "[[book-6/proposition-1]]", "[[book-5/proposition-11]]", "[[book-5/proposition-18]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.53"
license: "CC-BY-SA-4.0"
---

# X.53

*To find the sixth binomial straight line.*

## Proof

Let two numbers *AC*, *CB* be set out such that *AB* has not to either of them the ratio which a square number has to a square number; and let there also be another number *D* which is not square and which has not 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.

Now, 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]]

Again, let it be contrived that, as *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 *HG*. [[book-10/proposition-6|X. 6]]

Therefore the square on *HG* is rational; therefore *HG* is 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 *GH* 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]]

It is next to be proved that it is also a sixth binomial straight line.

For since, as *D* is to *AB*, so is the square on *E* to the square on *FG*, and also, 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]]

But it was also proved incommensurable with *FG*; therefore each of the straight lines *FG*, *GH* is incommensurable in length with *E*.

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 not to *BC* the ratio which a square number has to a square number; so that neither has the square on *FG* to the square on *K* the ratio which a square number has to a square number.

Therefore *FG* is incommensurable 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 incommensurable with *FG*.

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

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

Lemma. Let there be two squares *AB*, *BC*, and let them be placed so that *DB* is in a straight line with *BE*; therefore *FB* is also in a straight line with *BG*.

Let the parallelogram *AC* be completed; I say that *AC* is a square, that *DG* is a mean proportional between *AB*, *BC*, and further that *DC* is a mean proportional between *AC*, *CB*.

For, since *DB* is equal to *BF*, and *BE* to *BG*, therefore the whole *DE* is equal to the whole *FG*.

But *DE* is equal to each of the straight lines *AH*, *KC*, and *FG* is equal to each of the straight lines *AK*, *HC*; [[book-1/proposition-34|I. 34]] therefore each of the straight lines *AH*, *KC* is also equal to each of the straight lines *AK*, *HC*.

Therefore the parallelogram *AC* is equilateral.

And it is also rectangular; therefore *AC* is a square.

And since, as *FB* is to *BG*, so is *DB* to *BE*, while, as *FB* is to *BG*, so is *AB* to *DG*, and, as *DB* is to *BE*, so is *DG* to *BC*, [[book-6/proposition-1|VI. 1]] therefore also, as *AB* is to *DG*, so is *DG* to *BC*. [[book-5/proposition-11|V. 11]]

Therefore *DG* is a mean proportional between *AB*, *BC*.

I say next that *DC* is also a mean proportional between *AC*, *CB*.

For since, as *AD* is to *DK*, so is *KG* to *GC*— for they are equal respectively— and, componendo, as *AK* is to *KD*, so is *KC* to *CG*, [[book-5/proposition-18|V. 18]] while, as *AK* is to *KD*, so is *AC* to *CD*, and, as *KC* is to *CG*, so is *DC* to *CB*, [[book-6/proposition-1|VI. 1]] therefore also, as *AC* is to *DC*, so is *DC* to *BC*. [[book-5/proposition-11|V. 11]]

Therefore *DC* is a mean proportional between *AC*, *CB*. Being what it was proposed to prove.