Understanding Vacuously True (Truth Table) I don't know very much about formal logic, and I'm trying to understand the concept of vacuously true statements. Consider the truth table below:
$$\begin{array} {|c|}
\hline
P & Q & P\implies Q & Q\implies P & P\iff Q  \\ \hline
T & T & T & T & T  \\ \hline
T & F & F & \color{blue} T & F\\ \hline
F & T & \color{blue}T & F & F\\ \hline
F & F & \color{blue} T & \color{blue}T & T
\end{array}$$.
The blue letters are definitions. To see why these definitions are the correct choices (as opposed to $F$s), suppose we changed the lower left $\color{blue} T$ to an $F$ (this would then force the lower right $\color{blue}T$ to an $F$), so $P\iff Q$ would be $F$ for $P$ and $Q$ both $F$, which isn't want we want. So I see why this makes sense to choose the lower  entries as $\color{blue}T$.
However, it isn't clear to me why $P\implies Q$ true for $P$ false and $Q$ true is the sensible choice (same for $Q\implies P$ true for $Q$ false and $P$ true). For if the $\color{blue}T$s in the middle rows were switched to $F$s, then $P\iff Q$ would still be $F$. So I don't see what the problem would be. Can someone please explain?

Here is why we say $P \implies Q$ is true if $P$ is false and $Q$ is true:
Let $P$ be the statement "it's raining outside" and $Q$ be the statement "the car is wet".
In order for $P \implies Q$ to be true, what needs to happen is: every time it rains outside, it better follow that the car is wet.  That's all you need to check.
So the only time $P \implies Q$ is false is when we get that it's raining outside and the car isn't wet.
When it's not raining, we don't have that it's raining outside and the car is not wet.  Since we don't have this, that means $P \implies Q$ isn't false, since the only time it is false is when we get that it's raining outside and the car isn't wet.  If it's not raining outside, we don't get this, so the statement is not false, so it must be true.
