Question

prove that root 2 is irrational

Answer 1

hey dear you answer is here

see attachment
wherever you see 5 put 2
and with this method you can prove root of any prime number is irrational.

Answer 2

$\huge{\color{#3134e2}{\mathcal{H}}\color{#24af15}{\mathcal{E}}\color{#e0bb18}{\mathcal{Y}} \quad \color{#c40932}{\mathcal{F}}\color{#1b5430}{\mathcal{R}} \color{#c97676}{\mathcal{I}}\color{#0f2a63}{\mathcal{E}}\color{#90ba27}{\mathcal{N}}\color{#eded17}{\mathcal{D}}}$

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.