### Chapter 1 Real Numbers Ex 1.3

**Question 1.** **Prove that √5 is irrational.** **Solutions:**

Let us assume that is rational.

∴ There exists co-prime integers a and b (b ≠ 0) such that

√5 = ab ⇒ √5b= 0

Squaring on both sides, we get

5b^{2}= a^{2}…… (i)

⇒ 5 divides a^{2} ⇒ 5 divides a

So, we can write a = 5c for some integer c.

From (i) and (ii)

5b^{2} = 25c^{2}

⇒ b^{2} = 5c^{2}

⇒ 5 divides b^{2}

⇒ 5 divides b

∴ 5 is a common factor of a and b.

But this contradicts the fact that a and b are co-primes.

This contradiction has arisen because of our incorrect assumption that √5 is rational.

Hence, √5 is irrational.

**Question 2.** **Prove that 3 + 2√5 is irrational.** **Solutions:**

Let us assume that 3 + 2√5 is rational.

∴ There exists co-prime integers a and b(b ≠ 0) such that

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is rational. Hence, we conclude that 3 + 2√5 is irrational.

**Question 3.** **Prove that the following are irrationals.**

**Solutions:**