Exercise 1.2 (Revised) - Chapter 1 - Real Numbers - Ncert Solutions class 10 - Maths

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NCERT Class 10 Maths Solutions: Chapter 1 - Real Numbers

Ex 1.2 Question 1.

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

Answer.

Let us prove $\sqrt{5}$ irrational by contradiction.
Let us suppose that $\sqrt{5}$ is rational. It means that we have co-prime integers $\boldsymbol{a}$ and $\boldsymbol{b}$ ( $b \neq 0$ ) such that $\sqrt{5}=\frac{a}{b}$
$
\Rightarrow b \sqrt{5}=a
$

Squaring both sides, we get
$
\Rightarrow 5 b^2=a^2
$

It means that 5 is factor of $a^2$
Hence, 5 is also factor of $a$ by Theorem.... (2)
If, $\boldsymbol{5}$ is factor of $\boldsymbol{a}$, it means that we can write $\boldsymbol{a}=5 \boldsymbol{c}$ for some integer $\boldsymbol{c}$.
Substituting value of $\boldsymbol{a}$ in (1),
$
5 b^2=25 c^2 \Rightarrow b^2=5 c^2
$

It means that 5 is factor of $b^2$.
Hence, 5 is also factor of $\boldsymbol{b}$ by Theorem. ... (3)
From (2) and (3), we can say that 5 is factor of both $\boldsymbol{a}$ and $\boldsymbol{b}$.
But, $\boldsymbol{a}$ and $\boldsymbol{b}$ are co-prime.
Therefore, our assumption was wrong. $\sqrt{5}$ cannot be rational. Hence, it is irrational.

Ex 1.2 Question2.

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

Answer.

We will prove this by contradiction.
Let us suppose that $(3+2 \sqrt{5})$ is rational.
It means that we have co-prime integers $\boldsymbol{a}$ and $\boldsymbol{b}(b \neq 0)$ such that
$
\begin{aligned}
& \frac{a}{b}=3+2 \sqrt{5} \Rightarrow \frac{a}{b}-3=2 \sqrt{5} \\
& \Rightarrow \frac{a-3 b}{b}=2 \sqrt{5} \\
& \Rightarrow \frac{a-3 b}{2 b}=\sqrt{5} \ldots(1)
\end{aligned}
$
$\boldsymbol{a}$ and $\boldsymbol{b}$ are integers.
It means L.H.S of (1) is rational but we know that $\sqrt{5}$ is irrational. It is not possible. Therefore, our supposition is wrong. $(3+2 \sqrt{5})$ cannot be rational.

Hence, $(3+2 \sqrt{5})$ is irrational.

Ex 1.2 Question3.

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

Answer.

(i) We can prove $\frac{1}{\sqrt{2}}$ irrational by contradiction.

Let us suppose that $\frac{1}{\sqrt{2}}$ is rational.
It means we have some co-prime integers $a$ and $b(b \neq 0)$ such that
$
\begin{aligned}
& \frac{1}{\sqrt{2}}=\frac{a}{b} \\
& \Rightarrow \sqrt{2}=\frac{b}{a} \ldots
\end{aligned}
$
R.H.S of (1) is rational but we know that $\sqrt{2}$ is irrational.

It is not possible which means our supposition is wrong.
Therefore, $\frac{1}{\sqrt{2}}$ cannot be rational.
Hence, it is irrational.
(ii) We can prove $7 \sqrt{5}$ irrational by contradiction.

Let us suppose that $7 \sqrt{5}$ is rational.
It means we have some co-prime integers $a$ and $b(b \neq 0)$ such that
$
\begin{aligned}
& 7 \sqrt{5}=\frac{a}{b} \\
& \Rightarrow \sqrt{5}=\frac{a}{7 b} \cdots
\end{aligned}
$
R.H.S of (1) is rational but we know that $\sqrt{5}$ is irrational.

It is not possible which means our supposition is wrong.

Therefore, $7 \sqrt{5}$ cannot be rational.
Hence, it is irrational.

(iii) We will prove $6+\sqrt{2}$ irrational by contradiction.

Let us suppose that $(6+\sqrt{2})$ is rational.
It means that we have co-prime integers $\boldsymbol{a}$ and $\boldsymbol{b}(b \neq 0)$ such that
$
\begin{aligned}
& 6+\sqrt{2}=\frac{a}{b} \\
& \Rightarrow \sqrt{2}=\frac{a}{b}-6 \\
& \Rightarrow \sqrt{2}=\frac{a-6 b}{b} \cdots
\end{aligned}
$
$\boldsymbol{a}$ and $\boldsymbol{b}$ are integers.
It means L.H.S of (1) is rational but we know that $\sqrt{2}$ is irrational. It is not possible. Therefore, our supposition is wrong. $(6+\sqrt{2})$ cannot be rational. Hence, $(6+\sqrt{2})$ is irrational.