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,
5b2 = a2 ... (1)
This means 5 divides a2.
From this, 5 also divides a.
Then a = 5c, for some integer 'c'.
On squaring, we get
a2 = 25c2
Put the value of a2 in equation (1).
5b2 = 25c2
b2 = 5c2
This means b2 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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