NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2#
Welcome to our NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2. This exercise asks students to prove that numbers such as $\sqrt{5}$, $3 + 2\sqrt{5}$, $\dfrac{1}{\sqrt{2}}$, $7\sqrt{5}$ and $6 + \sqrt{2}$ are irrational, using the method of proof by contradiction.
Theorem 1.2: Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.
- $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ are irrational.
- The sum, difference, product or quotient of a non-zero rational number and an irrational number is irrational.
1. Prove that $\sqrt{5}$ is irrational. #
Proof
Proof by contradiction.
Assume that $\sqrt{5}$ is rational. Then we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$$\sqrt{5} = \frac{a}{b}$$Squaring both sides:
$$5 = \frac{a^2}{b^2}$$$$\Rightarrow a^2 = 5b^2 \quad \cdots (1)$$
So 5 divides $a^2$. By Theorem 1.2, 5 divides $a$.
Let $a = 5c$ for some integer $c$. Substituting $a = 5c$ in equation (1):
$$a^2 = 5b^2$$$$(5c)^2 = 5b^2$$
$$\Rightarrow 25c^2 = 5b^2$$
$$\Rightarrow b^2 = 5c^2$$
So 5 divides $b^2$. By Theorem 1.2, 5 divides $b$.
So 5 divides both $a$ and $b$. This contradicts the fact that $a$ and $b$ are co-prime.
∴ Our assumption is wrong. Hence, $\sqrt{5}$ is irrational.
2. Prove that $3 + 2\sqrt{5}$ is irrational. #
Proof
Proof by contradiction.
Assume that $3 + 2\sqrt{5}$ is rational. Then we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$$3 + 2\sqrt{5} = \frac{a}{b}$$$$\Rightarrow 2\sqrt{5} = \frac{a}{b} - 3$$
$$\Rightarrow 2\sqrt{5} = \frac{a - 3b}{b}$$
$$\Rightarrow \sqrt{5} = \frac{a - 3b}{2b}$$
$a$ and $b$ are integers, so $\dfrac{a - 3b}{2b}$ is rational, and therefore $\sqrt{5}$ is rational.
But this contradicts the fact that $\sqrt{5}$ is irrational (proved in Question 1).
∴ Our assumption is wrong. Hence, $3 + 2\sqrt{5}$ is irrational.
3. Prove that the following are irrationals: #
(i) $\dfrac{1}{\sqrt{2}}$
Proof by contradiction.
Assume that $\dfrac{1}{\sqrt{2}}$ is rational. Then we can find co-prime integers $a$ and $b$ ($a \neq 0$, $b \neq 0$) such that:
$$\frac{1}{\sqrt{2}} = \frac{a}{b}$$$$\Rightarrow \sqrt{2} = \frac{b}{a}$$
$a$ and $b$ are integers, so $\dfrac{b}{a}$ is rational, and therefore $\sqrt{2}$ is rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
∴ Our assumption is wrong. Hence, $\dfrac{1}{\sqrt{2}}$ is irrational.
(ii) $7\sqrt{5}$
Proof by contradiction.
Assume that $7\sqrt{5}$ is rational. Then we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$$7\sqrt{5} = \frac{a}{b}$$$$\Rightarrow \sqrt{5} = \frac{a}{7b}$$
$a$ and $b$ are integers, so $\dfrac{a}{7b}$ is rational, and therefore $\sqrt{5}$ is rational.
But this contradicts the fact that $\sqrt{5}$ is irrational.
∴ Our assumption is wrong. Hence, $7\sqrt{5}$ is irrational.
(iii) $6 + \sqrt{2}$
Proof by contradiction.
Assume that $6 + \sqrt{2}$ is rational. Then we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$$6 + \sqrt{2} = \frac{a}{b}$$$$\Rightarrow \sqrt{2} = \frac{a}{b} - 6$$
$$\Rightarrow \sqrt{2} = \frac{a - 6b}{b}$$
$a$ and $b$ are integers, so $\dfrac{a - 6b}{b}$ is rational, and therefore $\sqrt{2}$ is rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
∴ Our assumption is wrong. Hence, $6 + \sqrt{2}$ is irrational.