Question 1

Prove that $\sqrt{5}$ is irrational.

Accepted Answer

Let $\sqrt{5}$ is a rational number.
Therefore, we can find two integers $a$, $b(b eq 0)$ such that $\sqrt{5}=\frac{a}{b}$ 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 $b$ are co-prime.
$$
\begin{gathered}
a=\sqrt{5} b \
\Rightarrow a^{2}=5 b^{2}
\end{gathered}
$$

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
$$
\begin{aligned}
& (5 k)^{2}=5 b^{2} \
& \Rightarrow 5 k^{2}=b^{2}
\end{aligned}
$$

This means that $b^{2}$ is divisible by 5 and hence, $b$ is divisible by 5 .
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 $\frac{p}{q}$ or it can be said that $\sqrt{5}$ is irrational.

Question 2

Prove that $3+2 \sqrt{5}$ is irrational.

Accepted Answer

Let $3+2 \sqrt{5}$ is rational.
Therefore, we can find two co-prime integers $a, b(b eq 0)$ such that
$$
\begin{aligned}
& 3+2 \sqrt{5}=\frac{a}{b} \
\Rightarrow & 2 \sqrt{5}=\frac{a}{b}-3 \
\Rightarrow & \sqrt{5}=\frac{1}{2}\left(\frac{a}{b}-3 ight)
\end{aligned}
$$

Since $a$ and $b$ are integers, $\frac{1}{2}\left(\frac{a}{b}-3 ight)$ 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.

Question 3

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

Answer 3:
(i) $\frac{1}{\sqrt{2}}$

Let $\frac{1}{\sqrt{2}}$ is rational.
Therefore, we can find two co-prime integers $a, b(b eq 0)$ such that
$$
\frac{1}{\sqrt{2}}=\frac{a}{b}
$$

Or
$$
\sqrt{2}=\frac{b}{a}
$$
$\frac{b}{a}$ 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 $\frac{1}{\sqrt{2}}$ is irrational.
(ii) $7 \sqrt{5}$

Let $7 \sqrt{5}$ is rational.
Therefore, we can find two co-prime integers $a, b(b eq 0)$ such that
$$
\begin{aligned}
7 \sqrt{5} & =\frac{a}{b} \
\Rightarrow \sqrt{5} & =\frac{a}{7 b}
\end{aligned}
$$
$\frac{a}{7 b}$ 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 co-prime integers $a, b(b eq 0)$ such that
$$
\begin{gathered}
6+\sqrt{2}=\frac{a}{b} \
\Rightarrow \sqrt{2}=\frac{a}{b}-6
\end{gathered}
$$

Since $a$ and $b$ are integers, $\frac{a}{b}-6$ 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, $6+\sqrt{2}$ is irrational.

\section*{Mathematics}
(Chapter - 1) (Real Numbers)
(Class X)

\section*{Exercise 1.4}

Accepted Answer

(No solution provided.)