prove that √5 is irrational
Answers
Answered by
2
Hey friend here is your answer
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 number]
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
May be or hope it will help u
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 number]
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
May be or hope it will help u
czy:
please learn me the theoram of proving irrational the numbers
Answered by
17
We need to prove that √5 is irrational
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
Similar questions