SEBA Class 10 Mathematics Chapter 1.2 Solutions (Real Numbers) for New Book 2026

Here you will get all the solutions of SEBA Class 10 Mathematics Chapter 1.2 Real Numbers from the New SCERT Textbooks 2026. For better understanding of this chapter, first visit Class 10 Maths Chapter 1.1 Solutions.

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

Solution:

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

Let, $\sqrt{5} = \frac{a}{b}$, where $a$ and $b$ are coprime and $b \ne 0$

So, $b\sqrt{5} = a$.

Squaring on both sides and rearranging, we get:

$5b^2 = a^2 \quad \text{— (1)}$

$\Rightarrow b^2 = \frac{a^2}{5}$

∴ $a^2$ is divisible by $5$

So, $a$ is divisible by $5$

Let $a = 5c$ for some integer $c$ and substituting the value of $a$ in equation (1), we get:

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

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

$\Rightarrow b^2 = 5c^2$

$\Rightarrow \frac{b^2}{5} = c^2$

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

so $b$ is also divisible by $5$.

Therefore, $a$ and $b$ have at least $5$ as a common factor.

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

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

So, we conclude that $\sqrt{5}$ is irrational.

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

Solution:

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

That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that:

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

So, $2\sqrt{5} = \frac{a}{b} – 3$

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

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

Since $3$, $2$, $a$, and $b$ are integers, $\frac{a – 3b}{2b}$ is rational, and so $\sqrt{5}$ is rational.

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

This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.

So, we conclude that $3 + 2\sqrt{5}$ is irrational.

3. Prove that the following are irrationals:  

(i) $\frac{1}{\sqrt{2}}$  

(ii) $7\sqrt{5}$  

(iii) $6 + \sqrt{2}$  

(i) $\frac{1}{\sqrt{2}}$  

Solution:

Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational.

That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that:

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

So, $\sqrt{2} = \frac{b}{a}$

Since $a$ and $b$ are integers and $a \ne 0$, thus $\frac{b}{a}$ is rational, and so $\sqrt{2}$ is rational.

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

This contradiction has arisen because of our incorrect assumption that $\frac{1}{\sqrt{2}}$ is rational.

So, we conclude that $\frac{1}{\sqrt{2}}$ is irrational.

(ii) $7\sqrt{5}$  

Solution:

Let us assume, to the contrary, that $7\sqrt{5}$ is rational.

That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that:

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

So, $\sqrt{5} = \frac{a}{7b}$

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

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

This contradiction has arisen because of our incorrect assumption that $7\sqrt{5}$ is rational.

So, we conclude that $7\sqrt{5}$ is irrational.

(iii) $6 + \sqrt{2}$  

Solution:

Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational.

That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that:

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

So, $\sqrt{2} = \frac{a}{b} – 6$

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

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

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

This contradiction has arisen because of our incorrect assumption that $6 + \sqrt{2}$ is rational.

So, we conclude that $6 + \sqrt{2}$ is irrational.

4. The product of a non-zero rational number and an irrational number is  

(A) always irrational

(B) always rational

(C) always Integer

(D) rational or irrational  

Solution: (A) always irrational

The product and quotient of a non-zero rational and irrational number is always irrational.

Example:
2 (rational) × √3 (irrational) = 2√3 (irrational)

5. $\sqrt{5} + \sqrt{3} + 2$ is     

(A) a natural number

(B) an integer

(C) a rational number

(D) an irrational number  

Solution: (D) an irrational number  

Explanation:

The sum of two irrational numbers $(\sqrt{5} + \sqrt{3})$ is irrational, and the sum of an irrational number and a non-zero rational number is always irrational.

So, the correct option is (D)

6.

Assertion (A): $\sqrt{2} + \sqrt{5}$ is an irrational number  

Reason (R): If $p$ and $q$ are prime positive integers, then $\sqrt{p} + \sqrt{q}$ is an irrational number

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for (A)

(C) Assertion (A) is true but Reason (R) is false

(D) Assertion (A) is false but Reason (R) is true  

Solution: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)

If $p$ and $q$ are prime positive integers, then $\sqrt{p} + \sqrt{q}$ is an irrational number. Since $2$ and $5$ are primes, $\sqrt{2} + \sqrt{5}$ is an irrational number.

Hence, both (A) and (R) are true, and Reason (R) is the correct explanation for (A).

So, the correct option is (A)

7. Assertion (A): $\sqrt{a}$ is an irrational number, when $a$ is a prime number.  

Reason (R): Square root of any prime number is an irrational number.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)

(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation for A.

(C) Assertion (A) is true but Reason (R) is false.

(D) Assertion (A) is false but Reason (R) is true.  

Solution: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)

We know that if $p$ is a prime number, then $\sqrt{p}$ is an irrational number. Since $a$ is a prime number, $\sqrt{a}$ is irrational.

Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for (A).

So, the correct option is (A)

8. $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{10}$ are all irrationals  

Which pair among them is like irrationals?  

(A) $\sqrt{3}, \sqrt{6}$  

(B) $\sqrt{8}, \sqrt{10}$  

(C) $\sqrt{2}, \sqrt{8}$  

(D) $\sqrt{7}, \sqrt{8}$  

Solution: (C) $\sqrt{2}, \sqrt{8}$  

Two irrational numbers are called like irrationals if their simplest radical parts are identical.

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$

Both $\sqrt{2}$ and $2\sqrt{2}$ have $\sqrt{2}$ as their irrational part.

So, the correct option is (C)