Irrational numbers and Pythagoras Theorem

Is it always true that if the right angled triangle with is also isosceles and having lengths that can be denoted in terms of a rational number, the length of its hypotenuse will always be an irrational number? Another way to look at it would be that the diagonal of a square is always irrational. Does this always hold true?

• Assuming that the two sides are of positive rational length, your question is whether $\sqrt{2x^2}$ is always irrational for any positive rational $x$. the answer is of course yes as it is equal to $x\sqrt{2}$ – JMoravitz Nov 6 '16 at 3:04

If the legs each have length $x$, then the hypotenuse has length $x\sqrt{2}$.

So if $x$ is rational, then the hypotenuse has irrational length. If $x$ is irrational, then the hypotenuse could have irrational or rational length.

For example:

If $x=5$, the hypotenuse has length $5\sqrt{2}$, which is irrational.

If $x=\sqrt{2}$, the hypotenuse has length $2$, which is rational.

If $x=\sqrt{3}$, the hypotenuse has length $\sqrt{6}$, which is irrational.

• I edited my question to include the length be denoted by a rational number. – naveen dankal Nov 6 '16 at 3:08
• @naveendankal I included both cases in this answer - In the case where both legs have rational length $x$, the hypotenuse must have irrational length because $x\sqrt{2}$ is irrational. – MightyTyGuy Nov 6 '16 at 3:31

If the adjacent sides of a right triangle are sqrt(2) then the hypotenuse will be 2 which is rational. However if the side lengths are rational then a$^2$+b$^2$=c$^2$ so 2a$^2$=c$^2$ and c = $\sqrt{2a}$ which is irrational since $\sqrt{2}$ is irrational and a is rational.

• I edited my question to include the length be denoted by a rational number. – naveen dankal Nov 6 '16 at 3:08