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.