Question

Question 3 (iii)
Prove that $6+\sqrt{2}$ is irrational.

Answer 1

Let's assume that $6+\sqrt{2}$ is a rational number.
So, we can write this number as;
$6+\sqrt{2}=\frac{a}{b}$
Here a and b are two co prime numbers and b $\ne$ 0.
$\Rightarrow$ $\sqrt{2}=\frac{a}{b}-6$
$\Rightarrow$ $\sqrt{2}=\frac{a-6b}{b}$ ----(i)
R.H.S of equation (i) is a rational number, but we know that $\sqrt{2}$ is an irrational number.
It is not possible.
That means our assumption is wrong.
Hence, $6+\sqrt{2}$ is an irrational number.