Explaining $P \Rightarrow Q$ and $\lnot Q \Rightarrow P$ to non-mathematicians I am currently a teaching assistant in a course of mathematics for economists (I am an economics student myself, at the beginning of my own math studies) and found that many of the students had problems fully understanding logical implications. Specifically, there were problems with the following question:

Assume that the statement $P \Rightarrow Q$ is true. Which of the following statements are true (or false). Give examples.
(a) $Q \Rightarrow P$ (b) $\lnot Q \Rightarrow P$ (c) $\lnot Q \Rightarrow \lnot P$ (d) $\lnot P \Rightarrow \lnot Q$

Now obviously, the only thing we can say just by looking at the truth value of $P \Rightarrow Q$ is that the statement (c) is always true. For the other ones, we do not know whether they are true or not. To me this is clear just by looking at the truth table, but some of the students misinterpreted this and claimed that (b) is definitely false, if $P \Rightarrow Q$ is true, as they tried to use 'real-world examples', such as "If it rains ($P$), then the street is wet ($Q$).
I tried to give them a mathmatical example, to make things less ambiguous, so I told them to think about the following example

$P$: $x,k \in \mathbb{N},\space x= 4k \space$ (i.e. $x$ is a multiple of 4)
$Q$: $x,n \in \mathbb{N}, \space x= 2n \space$ (i.e. $x$ is even)

Now, obviously, $P \Rightarrow Q$ is true, if either 1) both $P$ and $Q$ are true, 2) $P$ is false and $Q$ is true, or if both $P$ and $Q$ are false. Now the statement $\lnot Q \Rightarrow P\space$ is then true in the cases 1) and 2), but not in case 3) (as $\lnot Q$ is true and $P$ is false, which makes the implication false). Using my example, this would be the same as making the conclusions 1) "If $x$ is not not even (i.e. even), then $x$ is a multiple of 4" ($\lnot Q \Rightarrow P\space$ is true), 2) "If $x$ is not not even (i.e. even), then $x$ is not a multiple of 4" ($\lnot Q \Rightarrow P\space$ is true) and 3) "If $x$ is not even, then $x$ is not a multiple of 4" ($\lnot Q \Rightarrow P\space$ is false).
Now I had thought that I had understood logics, but especially the last one puzzles me a bit. In the cases 1) and 2) it makes sense that our conclusion might be both false or true. But the last bit, which makes the statement $\lnot Q \Rightarrow P\space$ different from $P \Rightarrow Q$ in the truth table is false. Is it because of that apparent contradiction, that the truth value of $\lnot Q \Rightarrow P\space$ cannot be inferred by looking at the truth value of the implication? Or is my example flawed?
I hope someone can give me a more intuitive explanation. As I said, it's one of these things that I know must work like that, but finding and making sense of this on my one apparently made me realize that I am still lacking true understanding of this fact.
Thank you in advance.
 A: Hint
Try with the false case :

$P \to Q$ is false only when $P$ is true and $Q$ is false.

Applying it to e.g. $\lnot Q \to P$, this means that it is false only when both $Q$ and $P$ are false.
But this is not the only case that "makes" $P \to Q$ true.
Conclusion : from the fact that $P \to Q$ is true, we cannot conclude that $\lnot Q \to P$ is false.
A: There are a couple of things going on here. First of all, you made a mistake in your example. You defined $P$ as 'x is a multiple of 4' and $Q$ as 'x is even', and then continued:

Now, obviously, P⇒Q is true, if either 1) both P and Q are true, 2) P is false and Q is true, or if both P and Q are false. Now the statement ¬Q⇒P  is then true in the cases 1) and 2), but not in case 3) (as ¬Q is true and P is false, which makes the implication false). Using my example, this would be the same as making the conclusions 1) "If x is not not even (i.e. even), then x is a multiple of 4" (¬Q⇒P  is true), 2) "If x is not not even (i.e. even), then x is not a multiple of 4" (¬Q⇒P  is true) and 3) "If x is not even, then x is not a multiple of 4" (¬Q⇒P  is false).

However, applying these 3 cases to the expression ¬Q⇒P, you should have gotten:
1) (P and Q both true) "If x is not even, then x is a multiple of 4"
2) (P false and Q true): "If x is not even, then x is not a multiple of 4"
3) (P and Q both false): "If x is even, then x is not a multiple of 4"
Now, any normal person would look at these claims and say:
1) is false (if x is not even ... it certainly cannot be a multiple of 4!)
2) is true (yes, what we just said!)
3) is false (if x is even ... it could be, but doesn't have to be, a multiple of 4. So false!)
So: if you are given that P⇒Q is true ... then we can't tell the truth-value of ¬Q⇒P
Aha! So this example does seem to work to demonstrate that we cannot figure out the truth-value of b) ... (which is what you tried to do) ... Cool! That clears it all up then, right?
Well, you could try and use this explanation on your students, but if you have a real smart one, they might say: 'Wait, not so fast!' ... and here is why:
Go back to your original claim P⇒Q. Again, you said:

Now, obviously, P⇒Q is true, if either 1) both P and Q are true, 2) P is false and Q is true, or if both P and Q are false. 

Really, 'obviously'?!
Let's go through the 3 cases, and apply them to the P⇒Q claim itself:
1) (P and Q both true) "If x is a multiple of 4, then x is even"
2) (P false and Q true): "If x is not a multiple of 4, then x is even"
3) (P and Q both false): "If x is not a multiple of 4, then x is not even"
What does any normal person say to these claims?
1) true! (of course!)
2) false! (Just because x is not a multiple of 4 doesn't mean x is even (e.g. x could be 5)
3) false! (Just because x is not a multiple of 4 doesn't mean x is not even (e.g. x could be 6)
Oh man! So P⇒Q is not true in cases 2 and 3?!  Is our truth-table incorrect?!
OK, so what we're dealing here is what's called the "Paradox of the Material Implication" (look it up ... there is a ton of literature on this topic), which points out that the way we use 'if ... then' sentences in real life often doesn't quite match with the 'material conditional', which is a mathematically defined truth-functional operator, written as $\rightarrow$ (or, as you do $\Rightarrow$), and which is informally expressed as 'if ... then'.
This case of the 'x is multiple of 4' and 'x is even' provides a perfect example of this. Go back to the claim of P⇒Q in the case of P and Q both being false, i.e. the claim 'If x is not a multiple of 4, then x is not even'. Again, any normal person would say that this is a false claim, and yet according to the truth-table, P⇒Q in that case is true: the mismatch is that we interpret the claim as a logical implication within the realm of mathematics (in which the implication does indeed not hold*), but the material conditional is purely truth-functional, i.e. it makes the truth-value of P⇒Q a simple function of the truth-values of $P$ and $Q$ (hence the word 'truth-functional, by the way) without looking for any kind of further connection. Indeed, an even simpler example is 'If grass is green, then bananas are yellow'. Many people would say that that claim wouldn't make much sense: yes, grass is green, and bananas are yellow, but there is no further connection between those two claims; grass being green doesn't make or force or cause bananas to be yellow. In fact, this lack of any connection will make many people say that that claim is false, and yet, using the truth-table definition (and given that grass is green and bananas are yellow), the claim would be considered true.
OK, so what does all this mean? It means that you have to be very careful with these conditionals, especially when you try to use real life examples to explain some mathematical property of them, because sometimes they can really point you in the wrong direction (indeed, I don't blame your and your students' confusion at all!). 
