If p then q misunderstanding? The statement $P\rightarrow Q$ means: if $P$ then $Q$.
p | q | p->q
_____________
T | F |  F
F | F |  T
T | T |  T
F | T |  T

Lets say: if I'm hungry $h$ - I'm eating $e$.
p      | q      | p->q
_______________________
h      | not(e) |  F
not(h) | not(e) |  T
h      | e      |  T
not(h) | e      |  T    // ?????

If I'm not hungry, I'm eating? (This does not make any sense...)
Can you please explain that for me?

Rather than your example about food — which is not very good, there is a lot of people starving and not eating —, let consider a more mathematical one : if $n$ equals 2 then $n$ is even.
I — How to interpret the truth table ?
Fix $n$ an integer.
Let $p$ denote the assertion “$n = 2$”, and $q$ the assertion “$n$ is even”. These two assertions can be true or false, depending on $n$. What does mean that the assertion “$p \to q$” is true ?
This means precisely :


*

*$p$ true and $q$ false is not possible ;

*$p$ false and $q$ false is a priori possible (e.g. with $n = 3$) ;

*$p$ true and $q$ true is a priori possible (e.g. with $n = 2$) ;

*$p$ false and $q$ true is a priori possible (e.g. with $n = 4$).


II — Some common errors
The assertion “($n$ is divided by 2) $\to$ ($n$ is divided by 3)” is definitely not true for all $n$, but it can be true, for example for $n=3$, or $n=6$.
The fact that “false implies true” should not be read as “if not $p$ then $q$”. Indeed “false implies false” also holds, so if $p$ is false then either $q$ is false, either it is true, which is not a big deal.
