Question

Prove that $3+\surd 5$ is irrational

Answer 1

Let us assume that $3+\surd 5$ is a rational number.

So, it can be written in the form $\frac{a}{b}$
$3+\surd 5=\frac{a}{b}$

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

Solving $3+\surd 5=\frac{a}{b}$ we get,
$\Rightarrow \surd 5=\frac{a}{b}-3$
$\Rightarrow \surd 5=\frac{a-3b}{b}$
$\Rightarrow \surd 5=\frac{a-3b}{b}$

This shows $\frac{a-3b}{b}$ is a rational number. But we know that $\surd 5$ is an irrational number.

So, it contradicts our assumption. Our assumption of $3+\surd 5$ is a rational number is incorrect.

$3+\surd 5$ is an irrational number

Hence, it is proved that $3+\surd 5$ is irrational.