Prove that the following are irrationals.
(i) $\frac{1}{\sqrt{2}}$
(ii) $7\sqrt{5}$

Given: 

The given numbers are:

(i) $\frac{1}{\sqrt{2}}$

(ii) $7\sqrt{5}$

To prove: Here we have to prove that, the given numbers are irrational.

Solution:

(i)  $\mathbf{\frac{1}{\sqrt{2}}}$

Let us assume that  $\frac{1}{\sqrt{2}}$ is a rational number.

So, $\frac{1}{\sqrt{2}}$ can be written in the form $\frac{a}{b}$, where a and b are co prime and b is not equal to 0.

Hence,   

$\frac{1}{\sqrt{2}}\ =\ \frac{a}{b}$

$\frac{b}{a}\ =\ \sqrt{2}$

Here, $\frac{b}{a}$ is rational but $\sqrt{2}$ is irrational.

Rational can't be equal to Irrational.

This contradicts our assumption, that $\frac{1}{\sqrt{2}}$ is  a rational number. 

Therefore, $\frac{1}{\sqrt{2}}$ is irrational number.

( ii) $\mathbf{7\sqrt{5}}$

Let us assume that $7\sqrt{5}$ is a rational number.

Hence, $7\sqrt{5}$ can be written in the form of $\frac{a}{b}$, where a and b are co prime and b is not equal to 0.

$7\sqrt{5}\ =\ \frac{a}{b}$

$\sqrt{5}\ =\ \frac{a}{7b}$  ​

Here, $\sqrt{5}$ is irrational but, $\frac{a}{7b}$​ is rational.

Rational can't be equal to Irrational. 

This contradicts our assumption, that the number $7\sqrt{5}$ is a rational number.

Therefore, $7\sqrt{5}$ is irrational number.

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Updated on: 10-Oct-2022

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