# Prove that √5 is irrational

**Solution:**

Let us prove that √5 is an irrational number.

This question can be proved with the help of the contradiction method. Let's assume that √5 is a rational number. If √5 is rational, that means it can be written in the form of a/b, where a and b integers that have no common factor other than 1 and b ≠ 0. i.e., a and b are coprime numbers.

√5/1 = a/b

√5b = a

Squaring both sides,

5b^{2} = a^{2} ... (1)

This means 5 divides a^{2}.

From this, 5 also divides a.

Then a = 5c, for some integer 'c'.

On squaring, we get

a^{2} = 25c^{2}

Put the value of a^{2} in equation (1).

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

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

This means b^{2} is divisible by 5 and so b is also divisible by 5. Therefore, a and b have 5 as common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √5 is a rational number. So, we conclude that √5 is irrational.

**☛ Check:** NCERT Solutions Class 10 Maths Chapter 1

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**Video Solution:**

## Prove that √5 is irrational

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.3 Question 1

**Summary:**

√5 is irrational by the method of contradiction.

**☛ Related Questions:**

- Prove that 3 + 2√5 is irrational.
- Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
- Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: (i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/2352 (vii) 129/225775 (viii) 6/15 (ix) 35/50 (x) 77/210
- Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. (i) 13/3125 (ii) 17/8 = 2.125 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/ (23 x 52) (vii) 129/(22 x 57 x 75) (viii) 6/15 (ix) 35/50 (x) 77/210

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