Prove $\sqrt{k}$ is not a rational number. Suppose $k>1$ is an integer, and k is not a square number, then $\sqrt{k}$ is not a rational number.
Proof:
Let $\sqrt{k}=\frac{p}{q}$, and $(p,q)=1$,So $q^2|p^2$, $p\neq 1$, $k$ is not an integer.When $q=p=1$, and $k>1$.
And $q=1$, then $\sqrt{k}=p$, and $k$ is a square number.
I do think it's so easy...Where is wrong?
How to do it.
 A: In short: I think you started correctly, but made some odd jumps.  I'm going to offer some critique of the proof--please don't take anything here personally. :)
A proof is more than just a string of symbols--rather, it must be a clearly (yet concisely) written work that allows the reader a peak inside your mind when you proved the proposition.
So, I would suggest:


*

*Whenever you start a proof by contradiction, make sure to denote it.

*Always "introduce" your variables.  Are you assuming $k$ is an integer?  A complex number?  What about $p$ and $q$?

*Don't use notation unless it really helps.  Sometimes it's easier to say (and nearly always easier to read) "$p$ and $q$ coprime," rather than "$(p, q) = 1$".  

*Show your algebra, or at least mention that you're doing some.  Don't make me think about why $\sqrt{k} = \frac{p}{q}$ implies $q^2|p^2$--show me from the definition of divides.


So, a start to the proof could be:

Proof:
  Assume proposition is false.  That is, there exists a $k \in \Bbb{Z}^+$ such that $k>1$ is not a perfect square and $\sqrt{k}$ is rational.
As $\sqrt{k}$ is rational, there exist coprime integers $p$ and $q$ ($q\ne0$) such that:
  $$\sqrt{k} = \frac{p}{q}$$
Rearranging the above, we find that:
  $$p = q\sqrt{k}$$
  It follows:
  $$p^2 = q^2k$$
  ...

