Why aren't vacuous truths just undefined? I am struggling to understand this. According to truth tables, if $P$ is false, it doesn't matter whether $Q$ is true or not: Either way, $P \implies Q$ is true.
Usually when I see examples of this people make up some crazy premise for $P$ as a way of showing that $Q$ can be true or false when $P$ is something outrageous and obviously untrue, such as "If the moon is made of bacon-wrapped apple-monkey carburetors, then I am a better wakeborder than Gauss."
$P$ is clearly false, but $P \implies Q$ is true no matter what the state of $Q$ is, and I don't understand why.
Are we saying "If $P$ is false, then all bets are off and $Q$ can be anything, either true or false, and not contradict our earlier claim, and if it isn't false, it must be true"?
Otherwise why can't we say that if $P$ is false, then we can't make any claims one way or the other on whether or not it implies anything at all? 
 A: Consider the statement:

All multiples of 4 are even.

You would say that statement is true, right?
So let's formulate that in formal logic language:

$\forall x: 4|x \implies 2|x$

(Here "$a|b$" means "$a$ divides $b$", that is, $b$ is a multiple of $a$.)
Now a $\forall$ statement is true if it is true whatever you insert for the quantified variable (after all,that's "for all" means). So let's try to insert $3$:

$4|3 \implies 2|3$

But wait, $4|3$ is false! Moreover, $2|3$ is also false. So the only way for the original statement to be true is that the implication $\text{false}\implies\text{false}$ gives true.
A similar argument can be done for $\text{false}\implies\text{true}$.
A: Consider any proof by contradiction, for example Euclid's proof of prime infinity. 

If $x$ is the largest prime number ("P"), then $y=\prod_{p\,\,prime} p+ 1$ is a prime ("Q").

This is a valid implication that can be shown with perfectly valid simple arguments. However, the premise is never satisfied. But within the proof, the reader may not yet see that it is ''nonsense'' at this step. You just follow the logic step by step. The fact that the premise is never satisfied does not make the logic (used to derive the claim above) less valid. So it is reasonable to assume that the claim above is True.
A: Intuitively, consider $P \implies Q$ to be a promise. For example: I say, "If you ever jump over the moon, then I'll pay you a thousand dollars." 
Now, some time passes and you obviously haven't jumped over the moon. The question is now effectively: Have I kept my promise true, or broken it (been false)? 
It seems clear that I've kept that particular promise, granted that at no time was I obligated to give you the thousand dollars. In fact, even if I had given you a thousand dollars for some other service, it wouldn't have any bearing on that particular, empty offer. 
A: This is done so that classical propositional calculus follows some natural rules. Let's try to motivate this, without getting into technical details:
The expression "$P\Rightarrow Q$" should be read "$P$ implies $Q$", or "whenever $P$ is true, $Q$ is also true".
The negation of such an expression would be a counter-example, i.e., "there is some case in which $P$ is true but $Q$ is not".
So assume $P$ is not true. The negation "$\lnot(P\Rightarrow Q)$" is not true in this case, by our interpretation above, so "$P\Rightarrow Q$" must be true.
We are basically using the rules that either an expression or its negation should be true, and that the negation of the negation of an statement is the statement itself. These are basic rules which are natural and useful, even though as a consequence we have that "$P\Rightarrow Q$" is true whenever $P$ is false.
A: $p\to q$ is logically equivalent to $\neg p \vee q$.  Hence if $p$ is false, then $\neg p$ is true, so $\neg p \vee q$ is true.  This is the situation we call vacuously true.
A: One way to look at this is by set inclusion:
"$P \implies Q$" is the same as saying that "the set of mathematical statements that have $P$ as true (call it $S_P$) is a subset of the set of statements that have $Q$ as true (call it $S_Q$)." So if $P$ is false, then $S_P = \emptyset$. It follows that whatever $S_Q$ is, $\emptyset = S_P \subset S_Q$. Hence, $P \implies Q$ is vacuously true. $Q$ by itself is not vacuously true: you need the antecendent, $P \implies$, for the "vacuous" property to make sense.
A: In a more informal sense, I like to think that $P\implies Q$ means that "$Q$ is at least as true as $P$". Which means that if $P$ is something false, then anything is "more true"  than $P$, and thus the statement $P\implies Q$ is true.
A: OK, here's a crack at it, based on something we had going in comments.  It's an argument by analogy, which isn't really a good idea, but I think it can help you make sense of why it has to be this way:
I'm running a currency exchange and promise to give you one Cuban Convertible Peso for each US Dollar you give me.
You give me USD 10 on Monday, and I give you CUC 10 in exchange.
On Tuesday you give me nothing.  I give you nothing in exchange.
On Wednesday, you give me USD 5.  I give you CUC 5 in exchange.
On Thursday, you try to give me a Euro.  I could just accept EUR 1 from you as a gift, offer some amount of change for it, or push it back to you and point at the sign that says I trade USD for CUC.  No matter which of these I choose, I am still fulfilling my promise.  
On Friday, you didn't give me anything, but I give you CUC 0.25 out of the goodness of my heart.  Or maybe because you're amusing me and that's worth rewarding.
No amount of not giving me USD can change the truth of the statement "Monty is an honest currency broker who keeps his promise to give at least as much CUC for every USD you give him".  
When you gave me USD 0, I gave you at least CUC 0.  For the four days, you've given me USD 15 and I've given you CUC 15.25 (and maybe some more for the EUR 1).  I'm absolutely keeping my promise every single day, and if you claim I'm not keeping my promise on the days you gave me USD 0, then you're lying, because I gave you at least CUC 0 on those days.  You can't even argue that "the jury's out" on whether I gave you at least as many CUC each day as you gave me in USD, because 0=0.  To do so would itself be untrue, and an actual jury might rule you've libeled me by doing that.  
Similarly, to treat (P⟹Q) as anything but true when P is false would itself be false.  
A: A different point of view is to think of an argument as a chain of lights. A light is lit if you know it is true. 
P implies Q says that when you manage to ignite the P light then P ignites the Q light too. However, if P is not lit then it doesn't mean that you can't light Q in some other way.
A: This is what Terence Tao has to say:

This is discussed in Appendix A.2 of my book. The notion of
  implication used in mathematics is that of material implication, which
  in particular assigns a true value to any vacuous implication. One
  could of course use a different convention for the notion of
  implication, however material implication is very useful for the
  purpose of proving mathematical theorems, as it allows one to use
  implications such as “if A, then B” without having to first check
  whether A is true or not. Material implication also obeys a number of
  useful properties, such as specialisation: if for instance one knows
  for every x that P(x) implies Q(x), then one can specialise this to a
  specific value of x, say 3, and conclude that P(3) implies Q(3). Note
  though that by doing so a non-vacuous implication can become a vacuous
  implication. For instance, we know that $x \geq 5$ implies $x^2 \geq
> 25$ for any real number $x$; specialising this to the real number 3,
  we obtain the vacuous implication that $3 \geq 5$ implies $3^2 \geq 25$.
The way I like to think of material implication is as follows: the
  assertion that A implies B is just saying that “B is at least as true
  as A”. In particular, if A is true, then B has to be true also; but if
  A is false, then the material implication allows B to be either true
  or false, and so the implication is true no matter what the truth
  value of B is.

A: If we assume the principle of bivalence, and we want the law of identity to hold true, I answer the question by saying that the reason lies in that logical implication is not logical equivalence.
The law of identity for all propositions x, (x ⟹ x), ensures that if p is false and q is false, then (p ⟹ q) is true.
The true does not imply the false, thus if p is true and q is false, (p ⟹ q) is false.
Thus, our truth table looks like the following so far:
⟹     false  true
false  true     ?
true   false  true

Now if the question mark got filled in by false, then ⟹ would indicate logical equivalence.  But, we don't logical equivalence for logical implication.  We don't want to hold that {($\alpha$ ⟹ $\beta$), $\beta$} $\vdash$ $\alpha$.  And we want ⟹ to work out as truth-functional.  In other words (false ⟹ true) matches to some truth value.  Consequently, (false ⟹ true) holds true.
A: If the light is green, I can make a right turn. If the light is red, I might still be able to make the turn-it depends if the sign says "no turn on red" and if any cars are coming. But you don't know whether I can make the turn if all you know is that the light is red.
If you know that I can make the turn, you don't know what color the light is- it could either be green, it a red light that I'm allowed to make the turn at. But if you know that I couldn't make the turn, it's guaranteed that the light was red.
As another example, if I rob a bank, I go to jail. However, if I don't rob a bank, that doesn't guarantee that I don't get arrested- I could have committed a different crime. On the same note, if I am in jail, you don't know if I robbed a bank or not, but if I'm not in jail, I definitely haven't broken in to a bank.
