Question

Prove that $6+\surd 2$ is an irrational number.

Answer 1

Let us assume that $6+\surd 2$ is a rational number.

So it can be written in the form $\frac{a}{b}$

$6+\surd 2=\frac{a}{b}$

Here $a$ and $b$ are coprime numbers and $b\ne 0$

$6+\surd 2=\frac{a}{b}$

By solving the equation we get,

$\Rightarrow \surd 2=\frac{a}{b}–6$

$\Rightarrow \surd 2=\frac{a-6b}{b}$

This shows $\frac{a-6b}{b}$ is a rational number.

But we know that $\surd 2$ is an irrational number, it contradicts our assumption.

Our assumption $6+\surd 2$ is a rational number is incorrect.

Therefore, $6+\surd 2$ is an irrational number

Hence, it is proved that $6+\surd 2$ is an irrational number.