Prove that the following are irrationals :

Solution:

(i) Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational. That is we can find coprime integers a and $b(b \neq 0)$ such that,

$\frac{1}{\sqrt{2}}=\frac{\mathbf{P}}{q}$

Therefore, $q=\sqrt{\mathbf{z}_{\mathbf{p}}}$

Squaring on both sides, we get

$\mathrm{q}^{2}=2 \mathrm{p}^{2}$ ...(1)

Therefore, 2 divides $\mathrm{q}^{2}$

so, 2 divides q

so we can write $\mathrm{q}=2 \mathrm{r}$ for some integer $\mathrm{r}$

squaring both sides, we get

$\mathrm{g}^{2}=4 \mathrm{r}^{2}$ ...(2)

From (i) \& (ii), we get

$2 p^{2}=4 r^{2}$

Therefore, 2 divides $\mathrm{p}^{2}$

So, 2 divides p

So, p & q have atleast 2 as a common factor.

But this contradict the fact that $\mathrm{p} \& \mathrm{q}$ have no common factor other than 1 .

This contradict our assumption that $\frac{1}{\sqrt{2}}$ is rational. So, we condude that $\frac{1}{\sqrt{2}}$ is irrational.

(ii) Let us assume, to the contrary, that $7 \sqrt{5}$ is rational.

That is, we can find coprime integers a and $b(b \neq 0)$ such that $7 \sqrt{\mathbf{5}}=\frac{\mathbf{a}}{\mathbf{b}}$

Therefore, $\frac{\mathbf{a}}{\mathbf{7 b}}=\sqrt{\mathbf{5}}$

Since $a$ and $b$ are integers, we get $\frac{\mathbf{a}}{\mathbf{7 b}}$ is rational, and so $\frac{\mathbf{a}}{\mathbf{7 b}}=\sqrt{\mathbf{5}}$ is rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction has arisen because of

our incorrect assumption that $7 \sqrt{5}$ is rational.

So, we conclude that $7 \sqrt{5}$ is irrational.

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

That is, we can find coprime integers a and $b(b \neq 0)$ such that $6+\sqrt{\mathbf{2}}=\frac{\mathbf{a}}{\mathbf{b}}$

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

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

Since $\mathrm{a}$ and $\mathrm{b}$ are integers, we get $\frac{\mathbf{a}}{\mathbf{b}}-6$ is rational, and so $\frac{\mathbf{a}-\mathbf{6 b}}{\mathbf{b}}=\sqrt{\mathbf{2}}$ is rational.

But this contradicts the fact that $\sqrt{\mathbf{2}}$ is irrational. This contradiction has arisen because of our incorrect assumption that

$6+\sqrt{2}$ is rational.

So, we conclude that $6+\sqrt{2}$ is irrational.