One proof of the number's irrationality is the following proof by infinite descent. It is also a proof by contradiction, which means the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, thereby implying that the proposition must be true.

1. Assume that √2 is a rational number, meaning that there exists an integer a and an integer b such that a / b = √2.

2. Then √2 can be written as an irreducible fraction a / b such that a and b are coprime integers and (a / b)^2 = 2.

3. It follows that a^2 / b^2 = 2 and a^2 = 2 b^2. ((a / b)^n = a^n / b^n)

4. Therefore a^2 is even because it is equal to 2 b^2. (2 b^2 is necessarily even because it is 2 times another whole number; that is what "even" means.)

5. It follows that a must be even (as squares of odd integers are themselves odd).

6. Because a is even, there exists an integer k that fulfills: a = 2k.

7. Substituting 2k from (6) for a in the second equation of (3): 2b^2 = (2k)^2 is equivalent to 2b^2 = 4k^2 is equivalent to b^2 = 2k^2.

8. Because 2k^2 is divisible by two and therefore even, and because 2k^2 = b^2, it follows that b^2 is also even which means that b is even.

9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).

Q.E.D.

Since there is a contradiction, the assumption (1) that √2 is a rational number must be false. The opposite is proven: √2 is irrational.