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

Last Updated : 03 Apr, 2024

Content of this article has been updated in the article Class 10 NCERT Solutions- Chapter 1 Real Numbers â€“ Exercise 1.2 as per the revised syllabus of NCERT.

### Question 1. Prove that âˆš5 is irrational.

Solution:

Let âˆš5 be a rational number.

âˆš5 = p/q be a rational number, where p and q are co-primes and qâ‰ 0.

Then, âˆš5q = 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.

âˆš5 is not a rational number, it is irrational.

### Question 2. Prove that 3+2âˆš5 is irrational.

Solution:

Let 3+2âˆš5 be a rational number.

i.e., 3+2âˆš5= p/q be a rational number, where p and q are co-primes and qâ‰ 0.

Subtracting 3 from both sides,

2âˆš5=p/q-3

2âˆš5=(p-3q)/q

Now dividing both the sides by 2, we get

âˆš5=(p-3q)/2q

Here, p and q are integers so (p-3q)2q is a rational number. It implies, âˆš5 should be a rational number but âˆš5 is an irrational number and hence, there is a contradiction.

It is because of the wrong assumption.

3+2âˆš5 is an irrational number.

### (i) 1/âˆš2                  (ii) 7âˆš5                  (iii) 6+âˆš2

Solution:

(i) Let us assume that 1/âˆš2 is a rational.

Therefore, there exist co-prime integers p and q (q â‰  0) such that

1/âˆš2=p/q â‡’ âˆš2=q/p

Since p and q are integers, we get q/p is rational and so âˆš2 is rational.

But this contradicts the fact that âˆš2 is irrational.

This contradiction has arisen because of our incorrect assumption that 1/âˆš2 is rational.

Hence, 1/âˆš2 is irrational.

(ii) Let us assume that 7âˆš5 is a rational.

Therefore, there exist co-prime integers p and q (q â‰  0) such that

7âˆš5=p/q â‡’ âˆš5=q/7p

Since p and q are integers, we get q/7p is rational and so âˆš5 is rational.

But this contradicts the fact that âˆš5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7âˆš5 is rational.

Hence, 7âˆš5 is irrational.

(iii) Let us assume that 6+âˆš2 is a rational.

Therefore, there exist co-prime integers p and q (q â‰  0) such that

6+âˆš2 = p/q â‡’ âˆš2 = q/p – 6

Since p and q are integers, we get q/p – 6 is rational and so âˆš2 is rational.

But this contradicts the fact that âˆš2 is irrational.

This contradiction has arisen because of our incorrect assumption that 6+âˆš2 is rational.

Hence, 6+âˆš2 is irrational.

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