# Both 2^(0.5) is irrational. [duplicate]

Isn't 2^(0.5) rational?

Method for proving: Contradiction. So show not P:2^(0.5) is rational.

## marked as duplicate by Matthew Conroy, Chase Ryan Taylor, Community♦Feb 8 '18 at 7:08

Let $7^{\frac 12} = a$ and $2^{\frac 12} = b$.
If you're allowed to assume (or you can show, and it's quite easy) that both $a$ and $b$ are irrational, then the problem becomes very easy.
Assume to the contrary that $a-b = X$ is rational.
Then $a+b = Y$ has to be irrational because $X+Y = 2a$ which is irrational (as assumed or previously shown).
Now consider $XY = (a+b)(a-b) = a^2 - b^2 = 7-2 = 5$.
This is rational. But the product of a rational ($X$, by assumption) and an irrational ($Y$, as deduced) number cannot be rational. We have arrived at a contradiction. Hence $X$ is irrational.