Question

√5 is irrational prove that

Answer 1

let root 5 be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
root 5=p/q
=> root 5 * q = p
squaring on both sides
=> 5*q*q = p*p  ------> 1
p*p is divisible by 5
p is divisible by 5
p = 5c  [c is a positive integer] [squaring on both sides ]
p*p = 25c*c  --------- > 2
sub p*p in 1
5*q*q = 25*c*c
q*q = 5*c*c
=> q is divisble by 5
thus q and p have a common factor 5
there is a contradiction
as our assumsion p &q are co prime but it has a common factor
so √5 is an irrational

Answer 2

$\huge \red{\bf Answer}$

We need to prove that √5 is irrational

$\pink{\rm Proof:}$

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

ANSWER

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

$\green{ \bf \sqrt{5} \: is \: a \: irrational \: number}$