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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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2 Answers 2

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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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    $\begingroup$ This gets about 80% of the way... but what if, e.g., $x=2$ and $y=8$? $\endgroup$ Jul 28, 2014 at 1:04
  • $\begingroup$ Should get the whole way now :) $\endgroup$
    – Mathmo123
    Jul 28, 2014 at 1:19
  • $\begingroup$ With the tiny caveat that $x$ itself can't be a rational square, of course, but this looks good! $\endgroup$ Jul 28, 2014 at 1:25
  • $\begingroup$ That was implicit in the question I believe - otherwise of course the theorem is false. $\endgroup$
    – Mathmo123
    Jul 28, 2014 at 1:27
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    $\begingroup$ @just1question Well, you know that $1+\frac nx$ is rational, and $\sqrt{x}$ is irrational, so... $\endgroup$ Jul 28, 2014 at 7:31
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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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