Question

how to prove that √5 is irrational number

Answer 1

hey friend

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

If √5 is rational, then it can be expressed by some number a/b (in lowest terms). This would mean:
(a/b)² = 5. Squaring,
a² / b² = 5. Multiplying by b²,
a² = 5b².

If a and b are in lowest terms (as supposed), their squares would each have an even number of prime factors. 5b² has one more prime factor than b², meaning it would have an odd number of prime factors.

Every composite has a unique prime factorization and can't have both an even and odd number of prime factors. This contradiction forces the supposition wrong, so √5 cannot be rational. It is, therefore, irrational.


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