Question

prove that root 2 is irrational

Answer 1

Let √2 be. a rational number
Then by definition ,√2=a/b( where a and b are coprimes and b≠0)
By squaring on both sides
2=a^2/b^2
2b^2= a^2
Therefore,2 is a factor of a^2
2 is a factor of a
Let a=2m
2b^2=(2m)^2
2b^2=4m^2
b^2=2m^2
Therefore 2 is a factor of b^2
2 is a factor of b
But a and b are coprimes
This is a contradiction
Our assumption is wrong
√2 is irrational

Answer 2

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.