Question

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

Answer 1

Assume that $\sqrt{2}+\sqrt{3}$ is a rational number. Then,

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

Where $p$ and $q$ are co-prime numbers and $q\ne 0$.

Squaring both sides,

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

$5+2\sqrt{3}=\frac{p^2}{q^2}$

$2\sqrt{3}=\frac{p^2}{q^2}-5$

$2\sqrt{3}=\frac{p^2-5q^2}{q^2}$

$\sqrt{3}=\frac{p^2-5q^2}{2q^2}$

It can be observed that $\frac{p^2-5q^2}{2q^2}$ is a rational number but $\sqrt{3}$ is an irrational number which cannot be possible.

Therefore, the assumption is wrong.

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