SEBA Class 10 Maths Chapter 1 Exercise 1.2 Solutions – Real Numbers

Q1. Prove that √5 is irrational.

Answer:
Let us assume, to the contrary, that √5 is a rational number.

Then it can be expressed in the form: \[ \sqrt{5} = \frac{a}{b} \] where a and b are integers having no common factor other than 1 (i.e., they are coprime), and \(b \neq 0\).

Squaring both sides, we get: \[ 5 = \frac{a^2}{b^2} \] \[ a^2 = 5b^2 \] This shows that \(a^2\) is divisible by 5, and hence a is also divisible by 5.

So, let \(a = 5c\) for some integer c.

Substituting this value in the equation: \[ (5c)^2 = 5b^2 \] \[ 25c^2 = 5b^2 \] \[ b^2 = 5c^2 \] This implies that \(b^2\) is divisible by 5, and hence b is also divisible by 5.

Thus, both a and b are divisible by 5, which contradicts our assumption that a and b are coprime.

This contradiction arises because our initial assumption is incorrect.

Therefore, √5 is irrational.

Q2. Prove that 3 + 2√5 is irrational.

Answer:
Let us assume, to the contrary, that \(3 + 2\sqrt{5}\) is a rational number.

Since 3 is a rational number, subtracting 3 from both sides, we get: \[ 2\sqrt{5} = (3 + 2\sqrt{5}) - 3 \] The right-hand side is rational because it is the difference of two rational numbers.

Therefore, \(2\sqrt{5}\) is rational.

Dividing both sides by 2, we get: \[ \sqrt{5} = \frac{2\sqrt{5}}{2} \] Hence, √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction arises due to our incorrect assumption.

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

Q3. Prove that the following are irrational:

(i) \( \frac{1}{\sqrt{2}} \)
Answer:
Let us assume, to the contrary, that \( \frac{1}{\sqrt{2}} \) is a rational number.

Then it can be written in the form: \[ \frac{1}{\sqrt{2}} = \frac{a}{b} \] where a and b are integers having no common factor other than 1 and \(b \neq 0\).

Taking reciprocal on both sides, we get: \[ \sqrt{2} = \frac{b}{a} \] This shows that √2 is rational.

But we know that √2 is irrational.

This contradiction arises due to our incorrect assumption.

Therefore, \( \frac{1}{\sqrt{2}} \) is irrational.

(ii) \( 7\sqrt{5} \)
Answer:
We know that 7 is a rational number and √5 is an irrational number.

Let us assume that \(7\sqrt{5}\) is rational.

Dividing both sides by 7, we get: \[ \sqrt{5} = \frac{7\sqrt{5}}{7} \] This shows that √5 is rational.

But we know that √5 is irrational.

This contradiction arises due to our incorrect assumption.

Therefore, \(7\sqrt{5}\) is irrational.

(iii) \( 6 + \sqrt{2} \)
Answer:
Let us assume, to the contrary, that \(6 + \sqrt{2}\) is a rational number.

Since 6 is a rational number, subtracting 6 from both sides, we get: \[ \sqrt{2} = (6 + \sqrt{2}) - 6 \] The right-hand side is rational as it is the difference of two rational numbers.

Thus, √2 becomes rational.

But we know that √2 is irrational.

This contradiction arises due to our incorrect assumption.

Therefore, \(6 + \sqrt{2}\) is irrational.

Q4. 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

Answer: (A) always irrational

Q5. √5 + √3 + 2 is

(A) a natural number
(B) an integer
(C) a rational number
(D) an irrational number

Answer: (D) an irrational number

Q6. Assertion (A): √2 + √5 is an irrational number
Reason (R): If p and q are prime positive integers, then √p + √q is an irrational number

Choose the correct option:

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

Answer: (A)

Q7. Assertion (A): √a is an irrational number, when a is a prime number.
Reason (R): Square root of any prime number is an irrational number.

Choose the correct option:

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

Answer: (A)

Q8. √2, √3, √5, √6, √7, √8, √10 are all irrationals. Which pair among them is like irrationals?

(A) √3, √6
(B) √8, √10
(C) √2, √8
(D) √7, √8

Answer: (C) √2, √8