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My apologies if this question has been asked before, but a quick search gave no results. This is not homework, but I would just like a hint please. The question asks

Show that $\{1, \sqrt{2}, \sqrt{3}\}$ is linearly independent over $\mathbb{Q}$.

To begin, I consider some linear relation $a + b \sqrt{2} + c \sqrt{3} = 0$, where $a, b, c \in \mathbb{Q}$. There are several cases to consider.

If $c = 0$, then it must be the case that $a = b = 0$, because $\sqrt{2}$ is not rational. Similarly, if $b = 0$, then $a = c = 0$ because $\sqrt{3}$ is not rational. If $a = 0$, then we must have that $b = c = 0$ because $\frac{\sqrt{2}}{\sqrt{3}}$ is not rational (because $\sqrt{6}$ is not rational).

My issue is drawing a contradiction when $a$, $b$, and $c$ are all nonzero. It is not always the case that the sum of two irrational numbers is irrational (e.g. $1 - \sqrt{2}$ and $\sqrt{2}$). Hints or suggestions would be greatly appreciated!

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    $\begingroup$ If you're interested, here is a bit more challenging problem along the same lines. Let $S= \{n\in \mathbf{N} : $no square greater than 1 divides $n\}$ Then, for any $p_1,\dots,p_k \in S, \{1, \sqrt{p_1}, \dots, \sqrt{p_k}\}$ is linearly independent over $\mathbf{Q}$ $\endgroup$
    – doppz
    Nov 9, 2013 at 14:07

1 Answer 1

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Hint:

$$a+b\sqrt2+c\sqrt3=0\;,\;\;a,b,c\in\Bbb Q\implies 2b^2+3c^2-a^2=-2bc\sqrt6$$

so if $\,bc\neq 0\;$ we get that $\;\sqrt6\in\Bbb Q\;$ , contradiction. So either

$$a+c\sqrt3=0\;\;or\;\;a+b\sqrt2=0$$

and then if $\;b\neq0\;$ or $\;c\neq0\;$ we get a straightforward contradiction again, so $\;b=c=0\;$ and etc. (there still lacks half a line to end the proof.)

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