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. 

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