Question

Prove that √2 is not rational number

Answer 1

Answer:

here proof

Step-by-step explanation:

$\sqrt{ 2 = a \div b } \:$

where a and b are co prime no.

$\sqrt{2 \times b = a}$

squaring both sides

$2b^{2} = {a}^{2}$

a^2 is divisible by 2

this implies, a is divisible by 2

let

$a = 2c$

where c is some integer

substitute from formula 3 after squaring

$2b ^{2} = 9c {}^{2}$

that is b^2 =2c^2

b^2 is divisible by 2

this implies b is divisible by 2

our assumption is incorrect so, it is not rational no.