Question

By using the method of contradiction, prove that the sum of an irrational number and a rational number is irrational.

Answer 1

$Let\sqrt{a}$ be irrational and b be rational.

Then, we have to prove that $(\sqrt{a}+b)$ is irrational.

If possible, let $(\sqrt{a}+b)$be rational.Then,

$(\sqrt{a}+b)$is rational, b is rational

$\Rightarrow \left[\right(\sqrt{a}+b\left)\right]$ is rational

$\therefore$difference of rationals is rational.

$\Rightarrow \sqrt{a}$ is rational.

This contradicts the fact that $\sqrt{a}$ is irrational.

Since the contradiction arises by assuming that $\sqrt{a}+b)$ is rational. hence $(\sqrt{a}+b)$ is irrational.