Question

prove the following -

(a) √2 is irrational

By long and clear methods ​

Answer 1

Solution:

Let us assume that, √2 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

              √2 = $\frac{a}{b}$

On squaring both sides, we get ;

        2 = $\frac{a^{2}}{b^{2}}$

   ⇒ a² = 2b²

Clearly, a² is divisible by 2.

So, a is also divisible by 2.

Now, let some integer be c.

⇒ a = 2c

Substituting for a, we get ;

⇒ 2b² = 2c

Squaring both sides,

⇒ 2b² = 4c²

⇒ b² = 2c²

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

Therefore, a and b have at least 2 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 √2 is rational.

So, we conclude that √2 is irrational.