This is a proof by contradiction.
Assume that sqrt(2) is rational, so it is equal to the ratio of two positive integers a and b. That is, we suppose that sqrt(2) = a / b, where a and b are integers and the fraction is in lowest terms.
By the problem setup, we have 2 = a^2 / b^2.
This implies 2b^2 = a^2.
Since a^2 is equal to 2 times an integer, it must be even. Since the square of an even number is even and the square of an odd number is odd, we know that a must be even.
Since a is even, that means that a = 2c for some integer c, so a^2 = (2c)^2 = 4c^2.
This implies that 2b^2 = 4c^2. We can devide by 2 on both sides and get b^2 = 2c^2.
By the same logic used above, this implies that b is even.
Since a and b are both even, we can reduce the fraction a/b by dividing both the numerator and the denominator by 2.
However, this contradicts our assumption that the fraction was in lowest terms, which proves the original statement.
Would you rather see this proof in video form? Paul has kindly produced just such a video.