A medial area does not exceed a medial area by a rational area.

If possible, let the medial area *AB* exceed the medial area *AC* by the rational area *DB*.

Set out a rational straight line *EF*. Apply to *EF* the rectangular parallelogram *FH* equal to *AB* producing *EH* as breadth. Subtract the rectangle *FG* equal to *AC*. Then the remainder *BD* equals the remainder *KH*.

But *DB* is rational, therefore *KH* is also rational.

Since each of the rectangles *AB* and *AC* is medial, and *AB* equals *FH* while *AC* equals *FG*, therefore each of the rectangles *FH* and *FG* is also medial.

They are applied to the rational straight line *EF*, therefore each of the straight lines *HE* and *EG* is rational and incommensurable in length with *EF*.

Since *DB* is rational and equals *KH*, therefore *KH* is rational. And it is applied to the rational straight line *EF*, therefore *GH* is rational and commensurable in length with *EF*.

But *EG* is also rational, and is incommensurable in length with *EF*, therefore *EG* is incommensurable in length with *GH*.

And *EG* is to *GH* as the square on *EG* is to the rectangle *EG* by *GH*, therefore the square on *EG* is incommensurable with the rectangle *EG* by *GH*.

But the squares on *EG* and *GH* are commensurable with the square on *EG*, for both are rational, and twice the rectangle *EG* by *GH* is commensurable with the rectangle *EG* by *GH*, for it is double it, therefore the sum of the squares on *EG* and *GH* is incommensurable with twice the rectangle *EG* by *GH*.

Therefore the sum of the squares on *EG* and *GH* plus twice the rectangle *EG* by *GH*, that is, the square on *EH* is incommensurable with the sum of the squares on *EG* and *GH*.

But the squares on *EG* and *GH* are rational, therefore the square on *EH* is irrational.

Therefore *EH* is irrational. But it is also rational, which is impossible.

Therefore, *a medial area does not exceed a medial area by a rational area.*

Q.E.D.

It’s hard to follow this proof. An equivalent proof in modern terms is easier to comprehend. Let √*a* and √*b* be the areas of the larger and smaller rectangles, where both are irrational numbers but *a* and *b* are both rational numbers.

Suppose that √*a* – √*b* is a rational number. Then its square *a* – 2√(*ab*) + *b* is also rational. Then √(*ab*) is rational, as is √(*a*/*b*). Then √*a* – √*b* = √*b*(√(*a*/*b*) – 1) is the product of an irrational number and a rational number making it irrational, a contradiction. (For the modern proof, we assume *a* is not equal to *b*, so that √(*a*/*b*) – 1 does not equal 0, and then we can conclude that the product is irrational.)

This is an interesting proof in that in order to show the difference is irrational, we assumed it was rational and used that to prove it was irrational.

The same modern proof works for sums, of course.

This proposition is used in several others in Book X starting with X.42.