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

  • Last Updated : 28 Dec, 2020

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

=> 5q2 = p2 (by, squaring both the sides) ….(i)



Therefore, 5 divides p2, and according to the theorem of rational number, for any prime number p which divides a2 also divides a. So, we can write

p = 5k

Putting the value of p in equation (i), we get

5q2 = (5k)2

5q2 = 25k2

Dividing by 25,

q2/5 = k2

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.

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