---
book: 10
number: 71
id: "X.71"
kind: "theorem"
uses: ["[[book-10/proposition-20]]", "[[book-10/proposition-22]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-36]]", "[[book-10/proposition-54]]", "[[book-10/proposition-57]]", "[[book-10/proposition-55]]", "[[book-10/proposition-58]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.71"
license: "CC-BY-SA-4.0"
---

# X.71

*If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area*.

## Proof

Let. *AB* be rational, and *CD* medial; I say that the side of the area *AD* is a binomial or a first bimedial or a major or a side of a rational plus a medial area.

For *AB* is either greater or less than *CD*.

First, let it be greater; let a rational straight line *EF* be set out, let there be applied to *EF* the rectangle *EG* equal to *AB*, producing *EH* as breadth, and let *HI*, equal to *DC*, be applied to *EF*, producing *HK* as breadth.

Then, since *AB* is rational and is equal to *EG*, therefore *EG* is also rational.

And it has been applied to *EF*, producing *EH* as breadth; therefore *EH* is rational and commensurable in length with *EF*. [[book-10/proposition-20|X. 20]]

Again, since *CD* is medial and is equal to *HI*, therefore *HI* is also medial.

And it is applied to the rational straight line *EF*, producing *HK* as breadth; therefore *HK* is rational and incommensurable in length with *EF* [[book-10/proposition-22|X. 22]]

And, since *CD* is medial, while *AB* is rational, therefore *AB* is incommensurable with *CD*, so that *EG* is also incommensurable with *HI*.

But, as *EG* is to *HI*, so is *EH* to *HK*; [[book-6/proposition-1|VI. 1]] therefore *EH* is also incommensurable in length with *HK*. [[book-10/proposition-11|X. 11]]

And both are rational; therefore *EH*, *HK* are rational straight lines commensurable in square only; therefore *EK* is a binomial straight line, divided at *H*. [[book-10/proposition-36|X. 36]]

And, since *AB* is greater than *CD*, while *AB* is equal to *EG* and *CD* to *HI*, therefore *EG* is also greater than *HI*; therefore *EH* is also greater than *HK*.

The square, then, on *EH* is greater than the square on *HK* either by the square on a straight line commensurable in length with *EH* or by the square on a straight line incommensurable with it.

First, let the square on it be greater by the square on a straight line commensurable with itself.

Now the greater straight line *HE* is commensurable in length with the rational straight line *EF* set out; therefore *EK* is a first binomial. [[book-10/definitions#Definition 1 (part 2)|X. Deff. II. 1]]

But *EF* is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [[book-10/proposition-54|X. 54]]

Therefore the side of *EI* is binomial; so that the side of *AD* is also binomial.

Next, let the square on *EH* be greater than the square on *HK* by the square on a straight line incommensurable with *EH*.

Now the greater straight line *EH* is commensurable in length with the rational straight line *EF* set out; therefore *EK* is a fourth binomial. [[book-10/definitions#Definition 4 (part 2)|X. Deff. II. 4]]

But *EF* is rational; and, if an area be contained by a rational straight line and the fourth binomial, the side of the area is the irrational straight line called major. [[book-10/proposition-57|X. 57]]

Therefore the side of the area *EI* is major; so that the side of the area *AD* is also major.

Next, let *AB* be less than *CD*; therefore *EG* is also less than *HI*, so that *EH* is also less than *HK*.

Now the square on *HK* is greater than the square on *EH* either by the square on a straight line commensurable with *HK* or by the square on a straight line incommensurable with it.

First, let the square on it be greater by the square on a straight line commensurable in length with itself.

Now the lesser straight line *EH* is commensurable in length with the rational straight line *EF* set out; therefore *EK* is a second binomial. [[book-10/definitions#Definition 2 (part 2)|X. Deff. II. 2]]

But *EF* is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [[book-10/proposition-55|X. 55]] therefore the side of the area *EI* is a first bimedial, so that the side of *AD* is also a first bimedial.

Next, let the square on *HK* be greater than the square on *HE* by the square on a straight line incommensurable with *HK*.

Now the lesser straight line *EH* is commensurable with the rational straight line *EF* set out; therefore *EK* is a fifth binomial. [[book-10/definitions#Definition 5 (part 2)|X. Deff. II. 5]]

But *EF* is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [[book-10/proposition-58|X. 58]]

Therefore the side of the area *EI* is a side of a rational plus a medial area, so that the side of the area *AD* is also a side of a rational plus a medial area.

Therefore etc. Q. E. D.