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Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn’t seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don’t think is possible, but how to show it?

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Squaring we get, $m=a^2+n-2a\sqrt n\implies \sqrt n=\frac{a^2+n-m}{2a}$ which is rational

Assume $m$ is a non-square integer. Then $\sqrt{m}$ is irrational, and if $x=\sqrt{m}+\sqrt{n}$, then

$$(x-\sqrt{m})^2=x^2-2x\sqrt{m}+m=n$$

Or

$$\frac{x^2+m-n}{2x}=\sqrt{m}$$

If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.

Same argument as here and here. It should be put in the FAQ ðŸ™‚

It’s easy to show that if the result is rational, it has to be natural.

**Hint:**

$$m=(\lfloor\sqrt{m}\rfloor+\epsilon)^{2}=\lfloor\sqrt m \rfloor ^2+2\epsilon \lfloor\sqrt m \rfloor + \epsilon^2$$

$$n=(\lceil \sqrt n \rceil-\epsilon)^2=\lceil \sqrt n \rceil^2-2\epsilon\lceil \sqrt n \rceil+\epsilon^2$$

Subtract these two from each other and show that $\epsilon$ has to be rational.

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