Negation of "If n is a perfect square, then all of it's p-levels, for a prime factor, is even" I am trying to negate the statement "If $n$ is a perfect square, then all of its p-levels, for a prime factor, is even"
My guess is "If $n$ is not a perfect square, then for at least one of its p-levels, for a prime factor, is odd"
What do you guys think? 
 A: "If $A$, then $B$" is logically equivalent to "not $A$, or $B$".  When negating an or statement, you negate both statements, and change it to an and statement.  So the negation of "if $A$, then $B$" is "$A$ and not $B$".  
A: Let $P$ be the statement "$n$ is a perfect square" and let $Q$ be the statement "all of $n$'s $p$-levels, for a prime factor, are even".  Your first statement is $P\Rightarrow Q$.  What you've done in making your second statement is you've negated $P$ and $Q$ separately, giving you $\neg P \Rightarrow \neg Q$.  But what you want is $\neg(P\Rightarrow Q)$, not $\neg P\Rightarrow \neg Q$.  These are not logically equivalent (write down a truth table if you don't see why).
The only way $P\Rightarrow Q$ can be false is if $P$ is true but $Q$ is false.  So $\neg(P\Rightarrow Q)$ is logically equivalent to $P \wedge \neg Q$.  I'll leave it to you to try and write that out in words.
A: "If $P$ then $Q$" is like a conditional promise. It is a true promise as long as it is not broken. The only way to break a promise is to have the condition $P$ be true and yet the promised $Q$ be false.
