Question

Prove that root 2
an irrational
number​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

let us assume that $\sqrt{2$ is a rational number

therefore, $\sqrt{2$ = a/b [here a and b are co-prime number]

=> b$\sqrt{2$= a [by simplifying we got]

on squaring both the sides,

$a^{2}$ = 2$b^{2}$ -- [1]

therefore ,$a^{2}$ is divisible by 2. hence a is also divisible by 2.

so, we can write a = 2m [m is any integer]

2$b^{2}$= 4$m^{2}$

=> $b^{2}$= 2$m^{2}$

hence , $b^{2}$ is divisible by 2 and b is also divisible by 2

this contradicts our fact that $\sqrt{2$ is rational no. [a and b are not co-prime]

therefore, $\sqrt{2$ is an irrational number