Question

Prove that $\sqrt{2}+\sqrt{3}$ is irrational

Answer 1

Let us assume that $\sqrt{2}+\sqrt{3}$ is a rational number
Then. there exist coprime integers $p$, $q$,$q\ne 0$ such that
$\sqrt{2}+\sqrt{3}=\frac{p}{q}$
$=>\frac{p}{q}-\sqrt{3}=\sqrt{2}$
Squaring on both sides, we get
$=>(\frac{p}{q}-\sqrt{3}{)}^2=(\sqrt{2}{)}^2$
$=>\frac{p^2}{q^2}-2\frac{p}{q}\sqrt{3}+(\sqrt{3}{)}^2=2$
$=>\frac{p^2}{q^2}-2\frac{p}{q}\sqrt{3}+3=2$
$=>\frac{p^2}{q^2}+1=2\frac{p}{q}\sqrt{3}$
$=>\frac{p^2+q^2}{q^2}\times \frac{q}{2p}=\sqrt{3}$
$=>\frac{p^2+q^2}{2pq}=\sqrt{3}$
Since, $p$,$q$ are integers, $\frac{p^2+q^2}{2pq}$ is a rational number.
$=>\sqrt{3}$ is a rational number.
This contradicts the fact that $\sqrt{3}$ is irrational.
Thus, our assumption is incorrect.
Therefore, $\sqrt{2}+\sqrt{3}$ is a irrational.