#### Question: Prove that \\sqrt5) is irrational.

#### Explanatory Answer

Let us assume the contrary. i.e., \\sqrt5) is rational.

So, \\sqrt5) = \\frac{a}{b}) where a and b are co-prime integers and b ≠ 0

∴ a = b\\sqrt5)

Squaring both sides, a^{2} = 5b^{2} ----------- (1)

Inference 1: So, we can infer that 5 divides a^{2}

Because ‘a’ is a positive integer, if 5 divides a^{2}, __5 will also divide a__.

∴ ‘a’ can be written as a multiple of 5 i.e., a = 5c for some integer c

Substitute a = 5c in (1)

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

Inference 2: 5 divides b^{2}.

Because ‘b’ is a positive integer, if 5 divides b^{2}, __5 will also divide b__.

5 divides a. 5 divides b

a and b have 5 as common factor.

But we picked ‘a’ and ‘b’ as co-prime integers. So, 'a' and 'b' cannot have 5 as a common factor.

The contradiction is because of our incorrect assumption that \\sqrt5) is rational.

So, we can conclude that \\sqrt5) is irrational.