Question

Root 2 is irrational .prove it

Answer 1

Hello Mate!

$let \: the \: \sqrt{2} \: be \: a \: rational \:no. \\ \sqrt{2} = \frac{p}{q} \\ 2 = \frac{ {p}^{2} }{ {q}^{2} } \\ 2 {q}^{2} = {p}^{2}$

so, 2 is fator of p^2
2 is factor of p

$let \: some \: natural \: no \: be \: m \\ 2m = p \\ {2}^{2} {m}^{2} = {p}^{2} \\ 4 {m}^{2} = 2 {q}^{2} \\ 2 {m}^{2} = {q}^{2}$

Here, 2 is factor of q^2 as well as q

But p/q have no common factor accept 2 so its shows that our route is wrong and we get that
$\sqrt{2} \: is \: irrational \: no.$

Hope it helps!☺

Answer 2

Let root 2 be a rational number 
That is, we can find integers a and b (≠ 0) such that √2 = (a/b)

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

√2b = a

⇒ 2b^2=a^2 (Squaring on both sides) → (1)

Therefore, a^2 is divisible by 2

Hence ‘a’ is also divisible by 2.

So, we can write a = 2c for some integer c.

Equation (1) becomes,

2b^2 =(2c)^2

⇒ 2b^2 = 4c^2

∴ b^2 = 2c^2

This means that b^2 is divisible by 2, and so b is also divisible by 2

Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational.