Question

Prove that $4-5\sqrt{3}$ is irrational

Answer 1

Let us assume that $4-5\sqrt{3}$ is a rational number.

So, it can be written in the form $\frac{a}{b}$
$4-5\surd 3=\frac{a}{b}$

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

Solving $4-5\surd 3=\frac{a}{b}$ we get,
$\Rightarrow 5\surd 3=4-\frac{a}{b}$
$\Rightarrow 5\surd 3=\frac{4b-a}{b}$

$\Rightarrow \surd 3=\frac{4b-a}{5b}$
This shows $\frac{4b-a}{5b}$ is a rational number. But we know that $\surd 3$ is an irrational number.

So, it contradicts our assumption. Our assumption of $4-5\sqrt{3}$ is a rational number is incorrect.

$4-5\sqrt{3}$ is an irrational number

Hence, it is proved that $4-5\sqrt{3}$ is irrational.