Question

prove that √5 is a irrational number

Answer 1

et 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

Answer:

Step-by-step explanation:

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