Question

Prove $3-2\surd 5$ is irrational.

Answer 1

We have to prove that $3-2\sqrt{5}$ is irrational.

Let us assume the opposite.

$3-2\sqrt{5}$ is rational.

Hence $3-2\sqrt{5}$ can be written in the form $\frac{a}{b}$ where $a$ and $b$ are co-prime and $b\ne 0$

Hence $3-2\sqrt{5}=\frac{a}{b}$

$\Rightarrow 2\sqrt{5}=3-\frac{a}{b}=\frac{3b-a}{b}$

$\Rightarrow \sqrt{5}=\frac{3b-a}{2b}$

where $\sqrt{5}$ is irrational and $\frac{3b-a}{2b}$ is rational.

Here, $\frac{3b-a}{2b}$ is rational.

and $\sqrt{5}$ is irrational.

Since rational$\ne$ irrational

This is a contradiction

$\therefore$ our assumption is incorrect.

Hence, $3-2\sqrt{5}$ is irrational.

Hence proved.

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