Question about conditional statements as applied to math? I was being bothered by the fact that $p \implies q$ is defined when $p$ is false, so I thought I would try an example in math terms to help me understand it; but I got a stuck:
Let's define
$p: x > 0$
$q:$ The equation $100 = \sqrt x$ has a solution in $\mathbb{R}$ 
Consider the statement
$$p \implies q$$
$$\begin{array}{|c|c|c|}
\hline
p&q&p\implies q\\ \hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T\\\hline
\end{array}$$
Now $(p \implies q) = T$ makes sense when $p$ and $q$ are both true, and this agrees with the truth table. But if we look at the third row of the truth table, we are told that $(p \implies q)$ should be true even if $p=F$ and $q=T$. However, mathematically the statement
$$x \le 0 \implies \text{The equation $100 = \sqrt x$ has a solution in }\mathbb{R} $$ 
is of course false. How do we reconcile this? My only idea is that somehow one of $p, q$ is not a "real" statement or that they are not independent from each other, but according to what I've learned so far they are valid.
 A: Well, the two statements p and q in your case are not independent. Therefore stating one automatically determines the truth value of the other. I'm your case, saying that $x \leq 0$ already predetermines the truth value of the statement $q$.
A: The truth value of the statement $x\le0$ depends on $x$, and the truth value of the statement $100 = \sqrt x$ also depends on $x$.  These statements would be independent if their truth did not depend on common state, for instance if you had  $y\le0 \implies 100 = \sqrt x$.
A: The issue you are facing is "vacuous truth", but also a misunderstanding of the syntax. 
The truth table for $p \Rightarrow q$ makes sense intuitively except in the the case when $p$ is false. 
Think about it this way... you have to go out and find "things" and test whether $p$ is true, and then test whether $q$ is true. Then you will make a determination on whether $p \Rightarrow q$ is true. When $p$ is actually false, you will never find any thing in the world which will allow you to determine it is true, hence the set of things where $p$ is true is empty. 
Hence, instead, you go out and look for things where $q$ is false, and lo and behold, everytime you find that, you will also find that $p$ is false, and hence conclude that $\neg q \Rightarrow \neg p$.
Also, as I noted in my comment, you have negated $p=\{x>0\}$, therefor $\neg p =\{ x\leq 0\}$. The latter is an altogether different statement. $\neg p$ is not involved in determining the truth value of $p \Rightarrow q$.
A: You are using the variable $x$ in both the statements $p$ and $q$. These two statements are both properties that $x$ might satisfy.  A better way to write them is as $P(x)$ and $Q(x)$.  
So let $P(x)$ denote the statement "$x > 0$" and let $Q(x)$ denote the statement "$100=\sqrt{x}$".  
You need to be rigorous and rewrite the implication $p \implies q$. Do you want to say "$\forall x \in \mathbb{R}, P(x) \implies Q(x)$"? Or is it "$\exists x \in \mathbb{R}$ such that $P(x) \implies Q(x)$"?  The first implication is false and the second implication is true.  To determine whether these implications are true or false, we use the truth table definition of an implication.  
