Question

Prove that $3-sqrt5$ 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,
$\Rightarrow \surd 5=3-\frac{a}{b}$
$\Rightarrow \surd 5=\frac{3b-a}{b}$
This shows $\frac{3b-a}{b}$ is a rational number. But we know that $\surd 5$ is an irrational number.

So, it contradicts our assumption that $3-\surd 5$ is a rational number is incorrect.

Therefore, $3-\surd 5$ is an irrational number

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