# Class 10 NCERT Solutions- Chapter 1 Real Numbers – Exercise 1.3

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

**Solution:**

Let √5 be a rational number.

√5 = p/q be a rational number, where p and q are co-primes and q≠0.

Then, √5q = p

=> 5q

^{2 }= p^{2}(by, squaring both the sides) ….(i)Therefore, 5 divides p

^{2}, and according to the theorem of rational number, for any prime number p which divides a^{2}also divides a. So, we can writep = 5k

Putting the value of p in equation

(i), we get5q

^{2 }= (5k)^{2}5q

^{2 }= 25k^{2}Dividing by 25,

q

^{2}/5 = k^{2}Similarly, we can conclude that q will be divisible by 5, and we already know that p is divisible by 5

But p and q are co-prime numbers. So there is a contradiction and it is because of the wrong assumption we made in the first place.

√5 is not a rational number, it is irrational.

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

**Solution: **

Let 3+2√5 be a rational number.

i.e., 3+2√5= p/q be a rational number, where p and q are co-primes and q≠0.

Subtracting 3 from both sides,

2√5=p/q-3

2√5=(p-3q)/q

Now dividing both the sides by 2, we get

√5=(p-3q)/2q

Here, p and q are integers so (p-3q)2q is a rational number. It implies, √5 should be a rational number but √5 is an irrational number and hence, there is a contradiction.

It is because of the wrong assumption.

3+2√5 is an irrational number.

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

### (i) 1/√2 (ii) 7√5 (iii) 6+√2

**Solution:**

(i)Let us assume that 1/√2 is a rational.Therefore, there exist co-prime integers p and q (q ≠ 0) such that

1/√2=p/q ⇒ √2=q/p

Since p and q are integers, we get q/p is rational and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.

Hence, 1/√2 is irrational.

(ii)Let us assume that 7√5 is a rational.Therefore, there exist co-prime integers p and q (q ≠ 0) such that

7√5=p/q ⇒ √5=q/7p

Since p and q are integers, we get q/7p is rational and so √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7√5 is rational.

Hence, 7√5 is irrational.

(iii)Let us assume that 6+√2 is a rational.Therefore, there exist co-prime integers p and q (q ≠ 0) such that

6+√2 = p/q ⇒ √2 = q/p – 6

Since p and q are integers, we get q/p – 6 is rational and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 6+√2 is rational.

Hence, 6+√2 is irrational.