Question

Prove that
2 is irrational.​

Answer 1

Answer:

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Answer 2

Correct Question:-

Prove that $\sqrt{2}$ is irrational number.

SoLution :-

let us assume that $\sf\sqrt{2}$ is rational number.

So, it can be written in the form of $\sf\frac{a}{b}$

Such that, $\sf\sqrt{2} \; = \; \frac{a}{b}$

$\implies \sf\ b \sqrt{2} \: = \: a$

Squaring both sides:-

$\sf\ 2b^2 \; = \; a$

$\sf\ 2 \; divides\; a^2$

So, 2 is also divisible by a.

We, can write a = 3c for some integer c.

$\implies\sf\ 2b^2 \; = \; 4c\; that \; b^2 \; = \; 2c^2$

This Means, 2 divides b² and 2 is also divisible by b.

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

It arises contradiction because of our wrong assumption that $\sf\sqrt{2}$ is rational.

Hence, $\sf\sqrt{2}$ is irrational number.

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