Question

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

Answer 1

Let us assume $\sqrt{3}+\sqrt{2}$ be a rational number

$\Rightarrow \sqrt{3}+\sqrt{2}=\frac{p}{q}$, where $p,q\in z,q\ne 0$

$\Rightarrow \sqrt{3}=\frac{p}{q}-\sqrt{2}$

By squaring on both sodes, $(\sqrt{3}{)}^2={(\frac{p}{q}-\sqrt{2})}^2$

$3=\frac{p^2}{q^2}-2.\sqrt{2}.\frac{p}{q}+2$

$2\sqrt{2}.\frac{p}{q}=\frac{p^2}{q^2}+2-3$

$\Rightarrow 2\sqrt{2}.\frac{p}{q}=\frac{p^2}{q^2}-1$

$2\left(\sqrt{2}\right)\frac{p}{q}=\frac{p^2-q^2}{q^2}$

$\sqrt{2}=\left(\frac{p^2-q^2}{q^2}\right)\left(\frac{q}{2p}\right)$

$\sqrt{2}=\frac{p^2-q^2}{2pq}$

$\Rightarrow \sqrt{2}$ is a rational number $\because \frac{p^2-q^2}{2pq}$ is rational.

But $\sqrt{2}$ is not a rational number. This leads us to a contradiction.

$\therefore$ our assumption that $\sqrt{3}+\sqrt{2}$, is a ab be rational number is wrong

$\Rightarrow \sqrt{3}+\sqrt{2}$ is an irrational number. .