# Proposition 25

The rectangle contained by medial straight lines commensurable in square only is either rational or medial.

Let the rectangle AC be contained by the medial straight lines AB and BC which are commensurable in square only.

I say that AC is either rational or medial.

Describe the squares AD and BE on AB and BC. Then each of the squares AD and BE is medial.

Set out a rational straight line FG. Apply the rectangular parallelogram GH to FG equal to AD, producing FH as breadth, apply the rectangular parallelogram MK to HM equal to AC, producing HK as breadth, and further apply similarly NL to KN equal to BE, producing KL as breadth. Then FH, HK, and KL are in a straight line.

Since each of the squares AD and BE is medial, and AD equals GH while BE equals NL, therefore each of the rectangles GH and NL is also medial. X.22

And they are applied to the rational straight line FG, therefore each of the straight lines FH and KL is rational and incommensurable in length with FG.

And, since AD is commensurable with BE, therefore GH is commensurable with NL. And GH is to NL as FH is to KL, therefore FH is commensurable in length with KL.

X.19

Therefore FH and KL are rational straight lines commensurable in length, therefore the rectangle FH by KL is rational.

And, since DB equals BA while OB equals BC, therefore DB is to BC as AB is to BO.

VI.1

But DB is to BC as DA is to AC, and AB is to BO as AC is to CO, therefore DA is to AC as AC is to CO.

But AD equals GH, AC equals MK, and CO equals NL, therefore GH is to MK as MK is to NL. Therefore FH is to HK as HK is to KL. Therefore the rectangle FH by KL equals the square on HK.

But the rectangle FH by KL is rational, therefore the square on HK is also rational.

Therefore HK is rational. And, if it is commensurable in length with FG, then HN is rational, but, if it is incommensurable in length with FG, then KH and HM are rational straight lines commensurable in square only, and therefore HN is medial.

Therefore HN is either rational or medial. But HN equals AC, therefore AC is either rational or medial.

Therefore, the rectangle contained by medial straight lines commensurable in square only is either rational or medial.

Q.E.D.

## Guide

To say a rectangle with sides a and b is rational says that ab is a rational number. To say that it’s medial says that (ab)2 is a rational number but ab is an irrational number. Therefore, to say it is either rational or medial says only that (ab)2 is a rational number.

In this proposition we are given that a and b are medial, so both a4 and b4 are rational numbers, and we are given that a2/b2 is a rational number. Therefore, (ab)2, which equals b4(a2/b2) is also a rational number.

This proposition is not used in the rest of the Elements.