You wouldn't. It's not true: the square root of 4 is rational.
 I think this proof follows pretty much the same lines as for the square root of 2.
I know to prove the irrationality of we simply minipulate the following:
until we hit the contradiction of not being in reduced terms. I also know that we must take into account that all number can be written in the form or for ; and the fact that if is an even number then is an even number. Heres my question, how do you prove the irrationality of a certain opertation on numbers in general? And for odd square roots?
Two examples of what I mean:
 How would you prove that if that is always irrational?
 How would you prove is irrational? Would you find a way to define numbers in terms of 3's ? Just like with the ? So all numbers would be of the form , , and ; and would you some how use that fact to prove the irrationality of ?
Thanks in advance
Assuming that is rational, then it takes the form , where and m and n do not have common factors.
Since is an integer, it follows that m/3 is an integer too. Thus, 3 is a factor of m.
Let where ,
We see that both m and n have a common factor of 3 which contradicts our assumption.
Okay, your right, I think I've figured out the proof for all even numbers, but I'm alittle lost for the odd numbers. Here is what I have, please point out any mistakes I made.
-> We want to prove that is irrational, as long as no numbers satisfy .
[i] If is rational, then it can be written in the following form:
[ii] This leads to the following equation:
is either even or odd. Lets start with even.
Say is even, then for some we have
Thus, if is even, [of the form ]
then is even, and of the form
for some natural number .
Thus, we have and :
This proves to be even.
If is even, then we'd arrive at the form (for some )
Thus and have a common divisor, and this is a contradiction.
Is this correct? For this one case where for some and for some ? I don't know exactly how to go about proving it with the odd numbers, I tried on pen and paper using the form for but the "common factor" never seemed to show up. Can somebody give me a hint or show me the right direction?
When you use the Unique Factorization Theorem, you get a simple proof. How would you prove that if that is always irrational?
Let be a positive integer. We can show that if is rational, then must be a perfect square. From here it follows that if is not a perfect square, then is irrational.
Suppose that for positive. Then . For any prime , let and be the highest powers of that divide and , respectively.
Now, because and using the Unique Factorication Theorem, we must have and then . This means that is a perfect square.
This way you can even prove a more general result: For any positive integers ; if is rational, then must be a perfect th power.
There is a nice proof I saw once. It can be generalized to any positive integer that is not a perfect sqaure.
Soppose that , where are postive integers that are relatively prime. Then for some integers . (Bézout's Identity)
Because , we have and . Then , and is clearly an integer.
This means that is an integer, but ; contradiction.