Question

Prove that the following are irrationals:
(i) $\frac{1}{\sqrt{2}}$

(ii) $7\sqrt{5}$

(iii) $6+\sqrt{2}$

Answer 1

(1) $\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
Let $a=\left(\frac{1}{2}\right)\sqrt{2}$ be a rational number.
$\Rightarrow 2a=\sqrt{2}$
2a is a rational number since product of two rational number is a rational number .
Which will imply that $\sqrt{2}$ is a rational number.But it is a contradiction since $\sqrt{2}$ is an irrational number
Therefore 2a is irrational or a is irrational.
Therefore $\frac{1}{\sqrt{2}}$ is irrational .Hence proved.
(2) $7\sqrt{5}$
Let $a=7\sqrt{5}bearationalnumber$
$\Rightarrow \frac{a}{7}=\sqrt{5}$
Now , $\frac{a}{7}$ is a rational number since quotient of two rational number is a rational number.
The above will imply that $\sqrt{5}$ is a rational number. But $\sqrt{5}$ is an irrational number.
This contradicts our assumption.Therefore we can conclude that $7\sqrt{5}$ is an irrational number and hence the result.
(3) $6+\sqrt{2}$
If possible let $a=6+\sqrt{2}$ be a rational number.
Squaring both side
$a^2={(6+\sqrt{2})}^2$
$a^2=38+12\sqrt{2}$
$\sqrt{2}=\frac{a^2-38}{12}$ --(1)
Since a is a rational number the expression $\frac{a^2-38}{12}$ is also rational number.
$\Rightarrow \sqrt{2}$ is rational number.
This is a contradiction .Hence $6+\sqrt{2}$ is irrational
Hence proved.