# Proving <msqrt> 2 </msqrt>

Proving $\sqrt{2}+\sqrt{3}$ is irrational
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kovilovop2
Note that $\sqrt{2}=\frac{1}{2}\left(\sqrt{2}+\sqrt{3}-\frac{1}{\sqrt{2}+\sqrt{3}}\right)\notin \mathbb{Q}$, hence $\sqrt{2}+\sqrt{3}\notin \mathbb{Q}$
or,
Suppose $\sqrt{2}+\sqrt{3}$ is rational, then so is $\left(\sqrt{2}+\sqrt{3}{\right)}^{2}=5+2\sqrt{6}$. Hence, $\sqrt{6}$ is rational which is of course not true.
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babajijwerz
Assume that $\sqrt{2}+\sqrt{3}=$$\frac{p}{q}$$p,q\in \mathbb{Z}$. Then $\sqrt{3}=$ $\frac{p}{q}$
Squaring gives $\sqrt{2}=$ which is a contradiction since $\sqrt{2}\notin \mathbb{Q}$