Proof by contradiction looks like, in this we assumed the contradiction and prove that what we assumed is contradict itself by using theorem that already proved or have the tautology.
Theorem 1: Statement S is true. Proof by contradiction
Proof: Assume that the statement S is false. Then, derive a contradiction (such as 1 + 1 = 3).
In other words, show that the statement “¬S ⇒ false” is true. This is sufficient, because the contrapositive of the statement “¬S ⇒ false” is the statement “true ⇒ S”. The latter logical formula is equivalent to S, and that is what we wanted to show.
Theorem 2: Let n be a positive integer. If n2 is even then n is even.
Proof: We will prove the theorem proof by contradiction. So we assume that n2 is even, but n is odd. Since n is odd, we know in our part 1 Theorem 1 that n2 is odd. This is a contradiction, because we assumed that n2 is even.
Theorem 3: √2 is irrational, i.e., √2 cannot be written as a fraction of two integers m and n.
Proof: We will prove the theorem by contradiction. So we assume that √2 is rational. Then √2 can be written as a fraction of two integers, √2 = m / n, where m ≥ 1 and n ≥ 1. We may assume that m and n do not share any common factors, i.e., the greatest common divisor of m and n is equal to one; if this is not the case, then we can get rid of the common factors. By squaring √2 = m/n, we get 2n2=m2. This implies that m2 is even.Then by Theorem 2, m is which means that we can write m as m = 2k, for some positive integer k. It follows that 2n2 = m2 = 4k2, which implies that n2 = 2k2. Hence, n2 i
s even. Again by Theorem 2, it follows that n is even.
We have shown that m and n are both even. But we know that m and n are not both even. Hence, we have contradiction. Our assumption that √2 is r
ational is wrong. Thus, we can conclude that √2 is irrational.
There is the nice discussion of this proof in the book My Brain is Open: The Mathematical Journeys of Paul Erdos.