<s>[INST] prove that $2\sqrt5 +\sqrt{11}$ is irrational how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational?
I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields
$\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.
 [/INST] If $2\sqrt{5}+ \sqrt{11}=r$ is rational then  $4*5 + 4*\sqrt{55} + 11 = r^2$ is rational.  Then $\sqrt{55} = \frac {r^2 - 31}4$ is rational.
NOW do the proof by contradiction
Let $\sqrt{55} = \frac ab$ so $55b^2 = a^2$ so $5|a$ and so $25|a^2$ so $5|b^2$ so $5|b$ so $a$ and $b$ aren't in lowest terms.... yadda, yadda, yadda....
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