Show that the following numbers are irrational.
(i) $\frac{1}{\sqrt{2}}$
(ii) $7 \sqrt{5}$
(iii) $6+\sqrt{2}$
(iv) $3-\sqrt{5}$
(i) Let us assume that $\frac{1}{\sqrt{2}}$ is rational .Then, there exist positive co primes $a$ and $b$ such that
$\frac{1}{\sqrt{2}}=\frac{\mathrm{a}}{\mathrm{b}}$
$\frac{1}{\sqrt{2}}=\left(\frac{a}{b}\right)^{2}$
$\Rightarrow \frac{1}{2}=\frac{a^{2}}{b^{2}}$
$\Rightarrow b^{2}=2 a^{2}$
$\Rightarrow 2 \mid b^{2}\left(\because 2 \mid 2 a^{2}\right)$
$\Rightarrow 2 \mid b$
$\Rightarrow b=2 c$ for some positive integer $c$
$\Rightarrow 2 a^{2}=b^{2}$
$\Rightarrow 2 a^{2}=4 c^{2}(\because a=p c)$
$\Rightarrow a^{2}=2 c^{2}$
$\Rightarrow 2 \mid a^{2}\left(\because 2 \mid 2 c^{2}\right)$
$\Rightarrow 2 \mid a$
$\Rightarrow 2 \mid a$ and $2 \mid b$
This contradicts the fact that $a$ and $b$ are co primes.
Hence $\frac{1}{\sqrt{2}}$ is irrational
(ii) Let us assume that $7 \sqrt{5}$ is rational .Then, there exist positive co primes $a$ and $b$ such that
$7 \sqrt{5}=\frac{a}{b}$
$\sqrt{5}=\frac{a}{7 b}$
We know that $\sqrt{5}$ is an irrational number
Here we see that $\sqrt{5}$ is a rational number which is a contradiction
Hence $7 \sqrt{5}$ is irrational
(iii) Let us assume that $6+\sqrt{2}$ is rational. Then, there exist positive co primes $a$ and $b$ such that
$6+\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-6$
$\sqrt{2}=\frac{a-6 b}{b}$
Here we see that $\sqrt{2}$ is a rational number which is a contradiction as we know that $\sqrt{2}$ is an irrational number
Hence $6+\sqrt{2}$ is irrational
(iv) Let us assume that $3-\sqrt{5}$ is rational .Then, there exist positive co primes $a$ and $b$ such that
$3-\sqrt{5}=\frac{a}{b}$
$\sqrt{5}=3-\frac{a}{b}$
$\sqrt{5}=\frac{3 b-a}{b}$
Here we see that $\sqrt{5}$ is a rational number which is a contradiction as we know that $\sqrt{5}$ is an irrational number
Hence $3-\sqrt{5}$ is irrational
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