---
book: 10
number: 28
id: "X.28"
kind: "construction"
uses: ["[[book-6/proposition-13]]", "[[book-6/proposition-12]]", "[[book-6/proposition-17]]", "[[book-10/proposition-21]]", "[[book-10/proposition-11]]", "[[book-10/proposition-23]]", "[[book-5/proposition-16]]", "[[book-6/proposition-16]]", "[[book-9/proposition-24]]", "[[book-9/proposition-26]]", "[[book-2/proposition-6]]", "[[book-9/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.28"
license: "CC-BY-SA-4.0"
---

# X.28

*To find medial straight lines commensurable in square only which contain a medial rectangle*.

## Proof

Let the rational straight lines *A*, *B*, *C* commensurable in square only be set out; let *D* be taken a mean proportional between *A*, *B*, [[book-6/proposition-13|VI. 13]] and let it be contrived that, as *B* is to *C*, so is *D* to *E*. [[book-6/proposition-12|VI. 12]]

Since *A*, *B* are rational straight lines commensurable in square only, therefore the rectangle *A*, *B*, that is, the square on *D* [[book-6/proposition-17|VI. 17]], is medial. [[book-10/proposition-21|X. 21]]

Therefore *D* is medial. [[book-10/proposition-21|X. 21]]

And since *B*, *C* are commensurable in square only, and, as *B* is to *C*, so is *D* to *E*, therefore *D*, *E* are also commensurable in square only. [[book-10/proposition-11|X. 11]]

But *D* is medial; therefore *E* is also medial. [[book-10/proposition-23|X. 23]], addition

Therefore *D*, *E* are medial straight lines commensurable in square only.

I say next that they also contain a medial rectangle.

For since, as *B* is to *C*, so is *D* to *E*, therefore, alternately, as *B* is to *D*, so is *C* to *E*. [[book-5/proposition-16|V. 16]]

But, as *B* is to *D*, so is *D* to *A*; therefore also, as *D* is to *A*, so is *C* to *E*; therefore the rectangle *A*, *C* is equal to the rectangle *D*, *E*. [[book-6/proposition-16|VI. 16]]

But the rectangle *A*, *C* is medial; [[book-10/proposition-21|X. 21]] therefore the rectangle *D*, *E* is also medial.

Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q. E. D.

LEMMA I. *To find two square numbers such that their sum is also square*.

Let two numbers *AB*, *BC* be set out, and let them be either both even or both odd.

Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [[book-9/proposition-24|IX. 24, 26]] therefore the remainder *AC* is even.

Let *AC* be bisected at *D*.

Let *AB*, *BC* also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers.

Now the product of *AB*, *BC* together with the square on *CD* is equal to the square on *BD*. [[book-2/proposition-6|II. 6]]

And the product of *AB*, *BC* is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [[book-9/proposition-1|IX. 1]]

Therefore two square numbers, the product of *AB*, *BC*, and the square on *CD*, have been found which, when added together, make the square on *BD*.

And it is manifest that two square numbers, the square on *BD* and the square on *CD*, have again been found such that their difference, the product of *AB*, *BC*, is a square, whenever *AB*, *BC* are similar plane numbers.

But when they are not similar plane numbers, two square numbers, the square on *BD* and the square on *DC*, have been found such that their difference, the product of *AB*, *BC*, is not square. Q. E. D.

LEMMA 2. *To find two square numbers such that their sum is not square*.

For let the product of *AB*, *BC*, as we said, be square, and *CA* even, and let *CA* be bisected by *D*.

It is then manifest that the square product of *AB*, *BC* together with the square on *CD* is equal to the square on *BD*. See [[book-10/proposition-28|Lemma 1]]

Let the unit *DE* be subtracted; therefore the product of *AB*, *BC* together with the square on *CE* is less than the square on *BD*.

I say then that the square product of *AB*, *BC* together with the square on *CE* will not be square.

For, if it is square, it is either equal to the square on *BE*, or less than the square on *BE*, but cannot any more be greater, lest the unit be divided.

First, if possible, let the product of *AB*, *BC* together with the square on *CE* be equal to the square on *BE*, and let *GA* be double of the unit *DE*.

Since then the whole *AC* is double of the whole *CD*, and in them *AG* is double of *DE*, therefore the remainder *GC* is also double of the remainder *EC*; therefore *GC* is bisected by *E*.

Therefore the product of *GB*, *BC* together with the square on *CE* is equal to the square on *BE*. [[book-2/proposition-6|II. 6]]

But the product of *AB*, *BC* together with the square on *CE* is also, by hypothesis, equal to the square on *BE*; therefore the product of *GB*, *BC* together with the square on *CE* is equal to the product of *AB*, *BC* together with the square on *CE*.

And, if the common square on *CE* be subtracted, it follows that *AB* is equal to *GB*: which is absurd.

Therefore the product of *AB*, *BC* together with the square on *CE* is not equal to the square on *BE*.

I say next that neither is it less than the square on *BE*.

For, if possible, let it be equal to the square on *BF*, and let *HA* be double of *DF*.

Now it will again follow that *HC* is double of *CF*; so that *CH* has also been bisected at *F*, and for this reason the product of *HB*, *BC* together with the square on *FC* is equal to the square on *BF*. [[book-2/proposition-6|II. 6]]

But, by hypothesis, the product of *AB*, *BC* together with the square on *CE* is also equal to the square on *BF*.

Thus the product of *HB*, *BC* together with the square on *CF* will also be equal to the product of *AB*, *BC* together with the square on *CE*: which is absurd.

Therefore the product of *AB*, *BC* together with the square on *CE* is not less than the square on *BE*.

And it was proved that neither is it equal to the square on *BE*.

Therefore the product of *AB*, *BC* together with the square on *CE* is not square. Q. E. D.