# Prove that the following are irrationals.(i) $\frac{1}{\sqrt{2}}$(ii) $7 \sqrt{5}$(iii) $6+\sqrt{2}$.

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To prove:

Here we have to prove that, the given numbers are irrationals.

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.

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

Let us assume, to the contrary, that $6\ +\ \sqrt{2}$ is rational.

So, we can find integers a and b ($≠$ 0) such that  $6\ +\ \sqrt{2}\ =\ \frac{a}{b}$.

Where a and b are co-prime.

Now,

$6\ +\ \sqrt{2}\ =\ \frac{a}{b}$

$\sqrt{2}\ =\ \frac{a}{b}\ -\ 6$

$\sqrt{2}\ =\ \frac{a\ -\ 6b}{b}$

Here, $\frac{a\ -\ 6b}{b}$ is a rational number but $\sqrt{2}$ is irrational number.

But, Irrational number  $≠$  Rational number.

This contradiction has arisen because of our incorrect assumption that $6\ +\ \sqrt{2}$ is rational.

So, this proves that $6\ +\ \sqrt{2}$ is an irrational number.

Updated on 10-Oct-2022 13:19:30