Logic Problem with truth tables According to a truth table, if "p is false, and q is false" then "p implies q" is true. However, when studing inverses, we see that the inverse of a conditional statement may or may not be true.
For example, 
Statement:  If a quadrilateral is a rectangle, then it has two pairs of parallel sides.
Inverse     If a quadrilateral is not a rectangle, then it does not have two pairs of parallel sides. (FALSE!)
Does this not contradict the truth table? 
In the inverse, p and q are both false; however, the inverse of the true statement is not true.
Please help!
 A: Ok, so you know that $p \rightarrow q$ is true when $p$ and $q$ are false. The inverse $\neg p \rightarrow \neg q$ is also true when $p$ and $q$ are both false. 
Your confusion seems to be that you are conflating the above situation with a different claim: $(p\rightarrow q) \rightarrow (q\rightarrow p)$. Notice that this is false in general, but given that $p$ and $q$ are both false, it is true.
So in your example it is indeed true that if a quadrilateral is a rectangle there are parallel sides, and indeed the converse of this statement is false, so we don't have $(p\rightarrow q) \rightarrow (q\rightarrow p)$ where $p$ is "is a rectangle" and $q$ is "has parallel sides". However, if I am given a quadrilateral which is neither a rectangle, nor has parallel sides, both the conditionals are true, so in this situation they imply one another (as a true thing is implied by everything).
A: $P\rightarrow Q$ is equivalent to saying that,


*

*If $P$ is true then $Q$ is true.


*If $Q$ is false then $P$ is false.

In other words, $P\rightarrow Q$ is equivalent to $\neg Q\rightarrow \neg P$.
In your example we take $$P:=\text{a quadrilateral is a rectangle}\\Q:=\text{a quadrilateral has two pairs of parallel sides}$$
It is not clear what you mean by inverse. However, if you want to mean converse then there is a mistake in your converse. It should be,

If a quadrilateral has doesn't have two pairs of parallel sides then it is not a rectangle.

And this is consistent with the truth table.
A: You are confused because probably the concept of truth tables was not explained clearly enough to you.

According to a truth table, if "p is false, and q is false" then "p implies q" is true.

This is correct, but it is not the only thing that "p implies q" means. There are four conditions listed in the whole truth table that have to be satisfied by "p implies q".
The truth table also says:

If "p" is true and "q" is false, then "p implies q" is false.
If "p" is false and "q" is true, then "p implies q" is true.

These mean that "q implies p" is not always the same as "p implies q", because when "p" is true and "q" is false, they are different. However, they are the same when "p" and "q" are both true or both false.
When two (compound) propositions are always the same regardless of the truth values of the atomic propositions, we say that they are equivalent. So "p implies q" and "q implies p" are not equivalent if "p" and "q" are atoms.
On the other hand, it is possible for "p implies q" and "q implies p" to be equivalent if "p" and "q" are not atoms, even if "p" and "q" are different. For example, "(a and b) implies (b and a)" and "(b and a) implies (a and b)" are equivalent.
