Question

Prove that √5 is an irrational number .

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Answer 1

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Answer 2

Solution:

Let us assume that, √5 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

√5 = $\frac{a}{b}$

On squaring both sides, we get ;

5 = $\frac{a^{2}}{b^{2}}$

⇒ a² = 5b²

Clearly, a² is divisible by 5.

So, a is also divisible by 5.

Now, let some integer be c.

⇒ a = 5c

Substituting for a, we get ;

⇒ 5b² = (5c)²

Squaring both sides,

⇒ 5b² = 25c²

⇒ b² = 5c²

This means that, 5 divides b², and so 5 divides b.

Therefore, a and b have at least 5 as a common factor. But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arises because of our assumption that √5 is rational.

So, we conclude that √5 is irrational.