How do I prove that an irrational number is indeed irrational?

[NOTE: An irreducible fraction (or fraction in lowest terms) is a fraction in which the numerator and denominator are integers that have no other common divisors than 1.]

Given that p is a prime, positive integer and not a square number, we know that √p is irrational. Let's prove this:

Proof: We shall use Proof by Contradiction;

Let's suppose towards a contradiction that √p is in fact rational.

This implies that there exists two non-negative integers, call them m and n, such that:

√p = m / n , where n is not zero and m/n is an irreducible fraction.

<=> p = m2 / n (Squared both sides)

<=> pn= m2....(#)  (Multiplied through equation by n2)

From eqn(#), it follows that, since p is prime, p | m(i.e. p divides m2which thus implies also that p | m. 

This means that there exists some natural number, call it k, such that:

pk = m....(##).

Now, sub eqn(##) into eqn(#), we thus obtain:

pn2 = p2k2

<=> n= pk2   (Divided through by p)

This last implies that p | n2, which further implies:

p | n.

BUT since p divides both m and n, this contradicts the fact that m and n were chosen to be irreducible, so our original assumption was incorrect.

It thus follows that p is in fact irrational, as required. 

Liam D. GCSE Maths tutor, A Level Maths tutor

2 years ago

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