Proofs By Contradiction

This is also called reductio ad absurdum.

Reductio ad absurdum, which Euclid loved so much, is one of a mathematician's finest weapons. It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game

G.H. Hardy, A Mathematician's Apology, 1940.

To prove a theorem forall n P(n), assume that the theorem does not hold. I.e, exists n P(n) and prove that a contradiction (or absurditity results).

Consider a simple example.

Theorem For every n, If n >  2 and n is prime then n is odd.

Proof We will prove by contradiction.

The original statement is

 (forall n in mathbb{N} left( (n > 2 mbox{AND} Prime(n)) Rightarrow n mod 2 = 1 right))

The negation of this statement is

 (exists n in mathbb{N} left( n > 2 mbox{AND} Prime(n) mbox{AND}  n mod 2 not= 1 right))

Proof by Contradiction (Example)

Let us assume that the original statement is false. It's negation must be true for some n. Therefore, there is a n such that n > 2, n is prime and n is even, all at the same time.

  • Since n is even, we can write n = 2m for some m.

  • Now, since n > 2, as well, m cannot be equal to 1.

  • As a result, n can be written as the product of 2 and another number m that is not 1.

  • Therefore, n is composite.

  • However, we assumed that n was prime.

  • A number cannot be prime and composite at the same time. Therefore, we have a contradiction.

Assuming the negation of the theorem leads to a contradiction. Therefore, the theorem is true.


Proof By Contradiction

Let us look at how such proofs look like.

Universal Statement

To prove: forall n in ... P(n).


Assume, for the sake of contradiction, that the statement does not hold. In other words, there is a number n such that neg P(n). We then use this to prove by contradiction.

Universal Implication

This is a special case of universal statements.

To prove: forall n  P(n) Rightarrow Q(n).


Assume, for the sake of contradiction, that the statement does not hold. Therefore, there is a number n such that P(n) holds but neg Q(n) holds (or Q(n) does not hold). Starting from this, we derive a contradiction.

Existential Statement

To prove exists n in ...  P(n), assume that forall n in ... (neg P(n)) and derive the contradiction.

Examples of Proof By Contradiction

Here are some famous reductios.

Theorem There are infinitely many prime numbers.


Proof is by contradiction. Let us assume that there are finitely many (let us say n) primes.

  • Therefore, the prime numbers are p_1,ldots,p_n and every other number (except 1) is composite.

  • Consider the number  p = 1+ p_1 times p_2 times cdots times p_n.

  • We know that it is composite since the number of primes is finite and p not= p_i.

  • We showed using strong induction that every number is divisible by some prime number.

  • Therefore, p is divisible by one of p_1,ldots,p_n.

    • However, dividing p by p_i leaves a remainder of 1 for each i = 1,ldots,n.

    • As a result, we conclude that p is a prime too.

  • This means that p_1,ldots,p_n cannot be all the primes. Therefore, we have a contradiction.

Here is another one from Euclid.

Theorem The number sqrt{2} is irrational.


Proof is again by contradiction. Let us assume that sqrt{2} = frac{p}{q} for integers p and q. We also assume that p,q are at their lowest terms. In other words, p,q have no common factors.

We know that frac{p}{q} = sqrt{2}. Therefore, p^2 = 2 q^2.

  • This means that p^2 is even. Therefore, p is even.

  • However, q cannot be even since frac{p}{q} would then not be in their lowest terms.

  • Therefore, q is odd.

  • Since p is even, we can write p = 2 k. Therefore p^2 = 4 k^2 = 2 q^2.

  • We conclude that q^2 = 2 k^2. In other words, q^2 is even and hence q is even.

This is a contradiction since q cannot be odd and even at the same time. QED.