Question

Prove that root 5 is irrational.​

Answer 1

$\huge{\underline{\underline{\sf{PrOoF}}}}$

Let √5 be a rational number.

then it must be in form of

$\frac{a}{b} \: \: \: \: where \: b≠0$

( where a and b are co-prime)

$\sqrt{5} = \frac{a}{b}$

$\sqrt{5} \times b = a$

$\sqrt{5} b = a$

SQUARING BOTH SIDES WE GET

$( \sqrt{5} b) {}^{2} = a {}^{2}$

$5b {}^{2} = a {}^{2} - - - - - (1)$

a² will be divisible by 5

So a is also divisible by 5

Let a=5c( for some integer c)

Also SQUARING BOTH SIDES we get

$a {}^{2} = 25c {}^{2} - - - - - (2)$

PUT The value of a² in Eq (2)

$5b {}^{2} = 25c {}^{2}$

$b {}^{2} = 5c {}^{2}$

So b is divisible by 5

Therefore a and b have a common factor as 5

so ,our assumption is wrong (because we have assumed that a and b are coprime but they have common factor as 5)

This statement contradicts our assumption

Therefore √5 is an irrational number