Solution

Let $\sqrt{5}$ is a rational number.

Therefore, we can find two integers $a, b(b \neq 0)$ such that $b$

$ \sqrt{5}=\dfrac{a}{b} $

Let $a$ and $b$ have a common factor other than 1 . Then we can divide them by the common factor, and assume that $a$ and bare co-prime.

$a=\sqrt{5} b$

$a^{2}=5 b^{2}$

Therefore, $a^{2}$ is divisible by 5 and it can be said that $a$ is divisible by 5 .

Let $a=5 k$, where $k$ is an integer

$(5 k)^{2}=5 b^{2}$ This means that $b^{2}$ is divisible by 5 and hence, $b$ is divisible by 5 .

$b^{2}=5 k^{2} \quad$ This implies that $a$ and $b$ have 5 as a common factor.

And this is a contradiction to the fact that $a$ and $b$ are co-prime.

Hence, $\sqrt{5}$ cannot be expressed as $\dfrac{p}{q}$ or it can be said that $\sqrt{5}$ is irrational.

Solution

Let $3+2 \sqrt{5}$ is rational.

Therefore, we can find two integers $a, b(b \neq 0)$ such that

$3+2 \sqrt{5}=\dfrac{a}{b}$

$2 \sqrt{5}=\dfrac{a}{b}-3$

$\sqrt{5}=\dfrac{1}{2}(\dfrac{a}{b}-3)$

Since $a$ and $b$ are integers, $\dfrac{1}{2}(\dfrac{a}{b}-3)$ will also be rational and therefore, $\sqrt{5}$ is rational.

This contradicts the fact that $\sqrt{5}$ is irrational. Hence, our assumption that $3+2 \sqrt{5}$ is rational is false.

Therefore, $3+2 \sqrt{5}$ is irrational.

Solution

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

Let $\dfrac{1}{\sqrt{2}}$ is rational.

Therefore, we can find two integers $a, b(b \neq 0)$ such that

$ \begin{aligned} & \dfrac{1}{\sqrt{2}}=\dfrac{a}{b} \\ & \sqrt{2}=\dfrac{b}{a} \end{aligned} $

$\dfrac{b}{a}$ is rational as $a$ and $b$ are integers.

Therefore, $\sqrt{2}$ is rational which contradicts to the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is false and $\dfrac{1}{\sqrt{2}}$ is irrational.

(ii) $7 \sqrt{5}$

Let $7 \sqrt{5}$ is rational.

Therefore, we can find two integers $a, b(b \neq 0)$ such that

$7 \sqrt{5}=\dfrac{a}{b}$ for some integers $a$ and $b$

$\therefore \sqrt{5}=\dfrac{a}{7 b}$

$\dfrac{a}{7 b}$ is rational as $a$ and $b$ are integers.

Therefore, $\sqrt{5}$ should be rational.

This contradicts the fact that $\sqrt{5}$ is irrational. Therefore, our assumption that $7 \sqrt{5}$ is rational is false. Hence, $7 \sqrt{5}$ is irrational.

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

Let $6+\sqrt{2}$ be rational.

Therefore, we can find two integers $a, b(b \neq 0)$ such that

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

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

Since $a$ and $b$ are integers, $\dfrac{a}{b}-6$ is also rational and hence, $\sqrt{2}$ should be rational. This contradicts the fact that $\sqrt{2}$ is irrational. Therefore, our assumption is false and hence,