prove that 1/2- root 5 is irrational by contradiction method

Answer :

We have : $\frac{1}{2 - \sqrt{5}}$ , So
$\Rightarrow \frac{1}{2 - \sqrt{5}}\times \frac{2 + \sqrt{5}}{2 + \sqrt{5}} = \frac{2 + \sqrt{5}}{2^2 - {\left(\sqrt{5}\right)}^2} = \frac{2 + \sqrt{5}}{4 -5} = \mathbf{-}\mathbf{ }\left(\mathbf{2}\mathbf{ }\mathbf{+}\mathbf{ }\sqrt{\mathbf{5}}\right)$
Let  2 + $\sqrt{5}$  is a rational number .
Then we we can represent it in form of $\frac{p}{q}$ , where p and q are co-prime integers, So
 2 + $\sqrt{5}$    =  $\frac{p}{q}$

$\frac{p}{q}$  - ​$\sqrt{5}$     =  2

now squaring both side ,we get

( $\frac{p}{q}$  - ​$\sqrt{5}$     )²=  (  2 )²

$\frac{p^2}{q^2} + 5 - 2\sqrt{5}\frac{p}{q}$  = 4

$\frac{p^2}{q^2}$ + 5 - 4 = 2$\sqrt{5}$  $\frac{p}{q}$

$\frac{p^2}{q^2}$ +  1= 2$\sqrt{5}$  $\frac{p}{q}$

$\left(\frac{p^2 + q^2}{q^2}\right) \times \frac{q}{2p}$ = $\sqrt{5}$  

$\frac{p^2 + q^2}{2pq}$  = $\sqrt{5}$  

Hence 

$\sqrt{5}$    is a rational number  .            $$\begin{gathered} \left(\because p ,q are integers\right) \\ \left( \therefore \frac{p^2 + q^2}{2pq } is a rational number\right) \end{gathered}$$

But we know that $\sqrt{5}$   is a irrational number , So that contradict fact that $\sqrt{5}$   is a irrational number .
So,
Our assumption is incorrect ,

Hence 2 + $\sqrt{5}$   is a irrational number .   So , $\frac{1}{2 - \sqrt{5}}$ also a irrational number  .                                (  Hence proved )