# The sum of two irrational square roots

This is very similar to this question, but I was wondering if there was a simpler proof. In particular, a proof that would prove that $\sqrt{x}+\sqrt{y}$ is an irrational number if both $\sqrt{x}$ and $\sqrt{y}$ are irrational.

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Hint: If $xy$ is not a perfect square, show that $(\sqrt x + \sqrt y )^2 = x + y + 2\sqrt{ xy}$ is irrational, and use the fact that and if $u^2$ is irrational then so is $u$.

If $xy = n^2$ then show that $\sqrt x + \sqrt y = \sqrt x\left(1+ \frac n {x}\right)$is irrational.

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This gets about 80% of the way... but what if, e.g., $x=2$ and $y=8$? – Steven Stadnicki Jul 28 '14 at 1:04
Should get the whole way now :) – Mathmo123 Jul 28 '14 at 1:19
With the tiny caveat that $x$ itself can't be a rational square, of course, but this looks good! – Steven Stadnicki Jul 28 '14 at 1:25
That was implicit in the question I believe - otherwise of course the theorem is false. – Mathmo123 Jul 28 '14 at 1:27
@just1question Well, you know that $1+\frac nx$ is rational, and $\sqrt{x}$ is irrational, so... – Steven Stadnicki Jul 28 '14 at 7:31

If $x$ and $y$ are integers then so is $x - y$ and further if $\sqrt x + \sqrt y$ is rational then $\sqrt x - \sqrt y = \frac{x - y}{\sqrt{x} + \sqrt{y}}$ is also rational. Then by sums and differences so are both $\sqrt{x}$ and $\sqrt{y}$. Therefore if we assume that at least one of $\sqrt{x}$ and $\sqrt{y}$ are irrational than so is their sum.

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