Question

Prove that √3 is a irrational no.?

Answer 1

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Answer 2

Solution:

Let us assume that, √3 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

√3 = $\frac{a}{b}$

On squaring both sides, we get ;

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

⇒ a² = 3b²

Clearly, a² is divisible by 3.

So, a is also divisible by 3.

Now, let some integer be c.

⇒ a = 3c

Substituting for a, we get ;

⇒ 3b² = 3c

Squaring both sides,

⇒ 3b² = 9c²

⇒ b² = 3c²

This means that, 3 divides b², and so 2 divides b.

Therefore, a and b have at least 3 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 √3 is rational.

So, we conclude that √3 is irrational.