Question

Prove
$\sqrt{5}$
is a irrational number ?

Answer 1

let is suppose that sqrt 5 is a rational number.

sqrt5=p/q (where p,q are co prime and q isn't equal to 0)

squaring both sides
$5 = \frac{ {p}^{2} }{{q}^{2} }$
$5 {q}^{2} = {p}^{2}$
______(1)
therefore, 5 divides p sq.
so 5 divides p.

take c an integer such as
5c=p

squaring both sides
$25 {c}^{2} = {p}^{2}$
from (1)

$25 {c}^{2} = 5 {q}^{2}$
$5 {c}^{2} = {q}^{2}$
therefore, 5 divides q sq.
so 5 divides q.

therefore, p and q aren't co prime because they have a common factor 5.

so ,our assumption was wrong.
therefore, sqrt 5 is an irrational no.

HOPE IT HELPS YOU '_'

Answer 2

let √5 be a rational no
let 5 be p/q where p and q has no common factor
on squaring both sides
5=Psq/Qsq
since Psq is divisible by 5
Qsq is divisible by 5
Qsq is divisible by 5
Let Qsq be 5m
25Msq=5Qsq
25/5Msq=Qsq
5Msq=Qsq
since Msq is divisible by 5
Qsq is divisible by 5
Q. is divisible by 5
HENCE OUR ASSUMPTION IS WRONG √5 IS AN IRRATIONAL NUMBER
where sq=square
HOPE IT HELPED