To add a proof to the existing answer:
A rational number is a real number which can be expressed as the ratio of two integers. That is, x is rational if and only if x = a/b where a and b are integers and b!=0.
Now, suppose sqrt(33) is rational. Then sqrt(33) = a/b for some integers a and b. Then, squaring both sides, we get (sqrt(33))^2 = (a/b)^2
=> 33 = a^2/b^2
=> 33b^2 = a^2
Now we consider how many times each side is divisible by 33. Suppose a is divisible by 33 m times, that is, a = j*33^m for some integers j and m where j is not divisible by 33, and similarly b = k*33^n for integers k and n where k is not divisible by 33.
Substituting those in, we get
33(k*33^n)^2 = (j*33^m)^2
=> 33*k^2*33^(2n) = j*33^(2m)
=> k*33^(2n+1) = j*33^(2m)
As neither k nor j are divisible by 33, we know that the left hand side is divisible by 33 exactly 2n+1 times, whereas the right hand side is divisible by 33 2m times.
But 2n+1 is odd and 2m is even, this meaning they are divisible by 33 a different number of times, and thus cannot be equal. Thus our initial assumption was wrong, and so sqrt(33) is not rational, meaning it is irrational.
As a side note, this method can show that sqrt(n) is irrational for any positive integer n unless n is a perfect square.
