NCERT Class 10 Maths – Exercise 1.2 Solutions

NCERT Class 10 Maths

Chapter 1 – Real Numbers | Exercise 1.2

(Rationalized Syllabus 2025-26)

💡 Key Concepts

Proof by Contradiction: To prove a number is irrational, we assume it is rational (can be written as $a/b$ where $a, b$ are coprime), and then show that this assumption leads to a mathematical contradiction.
Fundamental Theorem: If a prime number $p$ divides $a^2$, then $p$ divides $a$.

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

Step 1: Assumption

Let us assume, to the contrary, that $\sqrt{5}$ is rational.

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

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

Step 2: Squaring and Rearranging

Rearranging the equation, we get $\sqrt{5}b = a$.

Squaring both sides:

$$5b^2 = a^2 \quad \dots(i)$$

Since $a^2$ is divisible by 5, it follows that $a$ is also divisible by 5.

Step 3: Substitution

Since $a$ is divisible by 5, we can write $a = 5c$ for some integer $c$.

Substitute $a = 5c$ into equation $(i)$:

$$5b^2 = (5c)^2$$

$$5b^2 = 25c^2$$

$$b^2 = 5c^2$$

This means $b^2$ is divisible by 5, and hence $b$ is also divisible by 5.

Step 4: Conclusion

From Steps 2 and 3, we conclude that $a$ and $b$ have at least 5 as a common factor.

But this contradicts the fact that $a$ and $b$ are coprime.

This contradiction arose because of our incorrect assumption that $\sqrt{5}$ is rational.

✔️ Therefore, $\sqrt{5}$ is irrational.

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

Step 1: Assumption

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

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:

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

Step 2: Rearranging

We rearrange the equation to isolate $\sqrt{5}$:

$$2\sqrt{5} = \frac{a}{b} – 3$$

$$2\sqrt{5} = \frac{a – 3b}{b}$$

$$\sqrt{5} = \frac{a – 3b}{2b}$$

Step 3: Conclusion

Since $a$ and $b$ are integers, $\frac{a – 3b}{2b}$ is rational.

Therefore, $\sqrt{5}$ must also be rational.

However, this contradicts the known fact that $\sqrt{5}$ is irrational.

✔️ Hence, our assumption is false and $3 + 2\sqrt{5}$ is irrational.

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

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

Assume $\frac{1}{\sqrt{2}}$ is rational, equal to $\frac{a}{b}$.

$$\frac{1}{\sqrt{2}} = \frac{a}{b} \Rightarrow \sqrt{2} = \frac{b}{a}$$

Since $a, b$ are integers, $\frac{b}{a}$ is rational. So $\sqrt{2}$ is rational.

This contradicts the fact that $\sqrt{2}$ is irrational.

✔️ $\frac{1}{\sqrt{2}}$ is irrational.

(ii) $7\sqrt{5}$

Assume $7\sqrt{5}$ is rational, equal to $\frac{a}{b}$.

$$7\sqrt{5} = \frac{a}{b} \Rightarrow \sqrt{5} = \frac{a}{7b}$$

Since $a, b$ are integers, $\frac{a}{7b}$ is rational. So $\sqrt{5}$ is rational.

This contradicts the fact that $\sqrt{5}$ is irrational.

✔️ $7\sqrt{5}$ is irrational.

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

Assume $6 + \sqrt{2}$ is rational, equal to $\frac{a}{b}$.

$$6 + \sqrt{2} = \frac{a}{b} \Rightarrow \sqrt{2} = \frac{a}{b} – 6 = \frac{a – 6b}{b}$$

Since $a, b$ are integers, $\frac{a – 6b}{b}$ is rational. So $\sqrt{2}$ is rational.

This contradicts the fact that $\sqrt{2}$ is irrational.

✔️ $6 + \sqrt{2}$ is irrational.

🎉 Exercise 1.2 Completed | Chapter 1 Real Numbers