Not understanding valid argument form with implies, modus ponens I understand that if you have
if p then q
p
$\therefore$ q
that when "if p then q" is true, and you know p to be true, then it follows that q is true.
What I don't get is: wouldn't you have to know the truth of p and q beforehand? How else would you know if the implication "if p then q" is true or not?
For example:
If the sum of the digits of 371,487 is divisible by 3, then 371,487 is divisible by 3
The sum of the digits of 371,487 is divisible by 3
$\therefore$ 371,487 is divisible by 3
I could also have
If the sum of the digits of 371,487 is divisible by 3, then 17 is divisible by 3
The sum of the digits of 371,487 is divisible by 3
$\therefore$ 17 is divisible by 3
Clearly, in the second one, the conclusion is certainly not true, but the premise is, making the implication not true. In other words, you would have to know that either statement in the implication is true. But that would make the rest of the statements in the sequence a moot point.
Clearly, I don't understand how this works!
 A: 
What I don't get is: wouldn't you have to know the truth of p and q beforehand? How else would you know if the implication "if p then q" is true or not?

No.  Here are some examples of statements where "if p then q" is obviously true, but we know nothing about p or q:


*

*If Bill is taller than me, then Bill is taller than everyone shorter than me.

*If I live next door to Emily, and Joe lives next door to me, then Joe does not live next door to Emily.

*If $f$ is a continuous function, then $f$ has a well-defined Riemann integral on every compact interval.


Since you don't know who Bill is, you obviously don't know if he's taller than me, or if he's taller than anyone else for that matter.  That is, we don't know whether p or q are true, but do we know that p implies q.
A: A valid argument form is valid, irregardless of the truth values of the relevant variables. 
What we have, with modus ponens, is the argument form:
$$\begin{align} & \;p\rightarrow q \\ &\;p \\ & \hline \\ \therefore & \;q\end{align}$$
This is valid, whatever the truth values of $p$ and $q$.
What this means is that we essentially have an argument that goes as follows:

IF $\;(p \rightarrow q)\;$ is true, and IF $\;(p)\;$ is true, then necessarily, $q$ is true. 

A valid argument form doesn't mean that the guarantee that $p\rightarrow q$ is necessarily true, nor that $p$ is necessarily true, but rather, it guarantees that whenever it is the case that both premises are true, then the conclusion is necessarily true, as well.
