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

linear-algebra radicals irrational-numbers rational-numbers

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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