1. Assume that sqrt(2) is a rational number, meaning that there exists an integer 'a' and an integer 'b' in general such that

a/b = sqrt(2).

2. Then sqrt(2) can be written as an irreducible fraction a/b such that a and b are co-prime integers.

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

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 and multiples of 2 are even.)

5. It follows that 'a' must be even (as squares of odd integers are never even).

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

7. Substituting 2k from step 6. for a in the second equation of step 3. 2 b^2 = (2k)^2 is equivalent to 2b^2 = 4k^2, which 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 steps 5. and 8. a and b are both even, which contradicts that a/b is irreducible as stated in step 2.

We get contradiction thus the assumption 1. that sqrt(2) is a rational number must be false. This means that sqrt(2) is not a rational number i.e. sqrt(2) is irrational.