Question:
Show that $3+\sqrt{2}$ is an irrational number.
Solution:
Let us assume that $3+\sqrt{2}$ is rational . Then, there exist positive co primes $a$ and $b$ such that
$3+\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-3$
$\sqrt{2}=\frac{a-3 b}{b}$
This implies,
$\sqrt{2}$ is a rational number which is a contradication.
Hence $3+\sqrt{2}$ is irrational