Why is $-a \times -b = ab$? This might be a bit of a meta-mathematical question (if there is such a thing as meta-mathematics) but it confuses me slightly. 
I think of multiplication (whether rightly or wrongly) as recursive adding. For example, $2 \times 4$ is just adding 2 four times i.e $+2 +2 +2 +2 = 8$. 
I think of multiplying negative numerals e.g $2 \times -4$ as subtracting 2 four times i.e $-2 -2 -2 -2 = -8$. Unfortunately this analogy seems to break down when multiplying to negative numbers e.g $-2 \times -4$. I don't understand how $-2 \times -4$ can result in $+8$. 
What is the reason behind this, if any? Is my analogy incorrect? Is there a flaw in my fundamental understanding of multiplication?
 A: Receiving 4 gifts of 5 dollars gives you a result of $4 \times 5 = +20$ dollars. This means you benefit by 20 dollars.
"Receiving" $-4$ gifts of 5 dollars means giving four gifts of 5 dollars. This results in a loss to you of 20 dollars, so $(-4) \times 5 = -20$.
Receiving 4 "gifts" of $-5$ dollars means someone is making you pay them 5 dollars, and doing it four times. That gives you a result of $4 \times (-5) = -20$ dollars.
"Receiving" $-4$ "gifts" of $-5$ dollars again means giving someone four of these (undesirable) gifts. Thus you are charging someone 5 dollars, and doing it four times. The result is of course good for you (if you don't care about karma), since you benefit by 20 dollars. So $(-4) \times (-5) = 20$. 
Here is a second approach. 
$4 \times (-5)$ is obtained by adding four copies of $-5$:
$$4 \times (-5) = (-5) + (-5) + (-5) + (-5) = -20.$$
So it is natural that $4 \times (-5)$ should be -20, by the "repeated addition" understanding of multiplication that you mentioned. Of course, we can't add "-4 copies" of -5 together, but pay attention to the pattern in the following list.
$4 \times (-5) = -20$
$3 \times (-5) = -15$
$2 \times (-5) = -10$
$1 \times (-5) = -5$
$0 \times (-5) = 0$
You will surely agree that if we are to multiply negative numbers at all, then the pattern should continue this way:
$(-1) \times (-5) = 5$
$(-2) \times (-5) = 10$
$(-3) \times (-5) = 15$
etc.
A: $$a \times b = ab$$
$$a \times b + a \times (-b) = a \times (b + (-b)) = a \times 0 = 0 \quad (\therefore a 
\times (-b) = -ab)$$
$$a \times (-b) + (-a) \times (-b) = (a + (-a)) \times (-b) = 0 \times (-b) = 0 \quad (\therefore (-a) \times (-b) = -(-ab) = ab)$$
A: I am posting this, maybe, you can be back in the future and see it when you get additional course. In Mathematics, we introduce a structure which is called a Group. You can search through web to find out many many points about this beautiful structure as well as its application in other fields like Chemistry. We call a non-empty set $G$ a Group when there is an operation , like a map in Calculus, acting as $G\times G\to G$. If this map can satisfy some points for $G$ then (G,*) is called a Group.


*

** is associative.

*There is an element $e\in G$ such that a*e=e*a=a $a\in G$.

*For every $a\in G$, there is an element $b\in G$ such that a*b=b*a=e.


Now the integers $\mathbb Z$ is an additive group. It means that it is a group under $+$. In fact this group let  High school students to do what ever they wants by integers easily. Here, is a fundamental definition in which you can get why we do that addition (as you asked).

Defenition: $\underbrace{a+a+\cdots+a}_{n}=na$ while $n>0$. $na=0$ while $n=0$ and $$(-n)*a=n*(-a)=\underbrace{(-a)+(-a)+\cdots+(-a)}_{n}$$ while $n<0$.

Now try to do the latter identity for $(-2)(-4)$. Note that in additive group $\mathbb Z$, the identity element is $0$ and the inverse of $4$ is $-4$ and vice versa.
A: First a one-line answer: Taking away $x$ times a debt of $y$ is the same as giving $x$ times $y$. Below is a much more detailed explanation with the same underlying reason, which also relates directly to ring theory should you learn it in the future.
I don't find the answer to the other (duplicate) question convincing enough (I mean, how can you convince any ordinary person that such a thing as a real-line exists? It is in fact merely an axiom, the Cantor-Dedekind axiom, that a geometric line is isomorphic to the real line.) I would also say that we define the axioms for a group or field the way we do so precisely because of our intuition, so the question remains as to what that intuition is. As such I'm going to give a different kind of answer. user118829's answer is also a good complementary way to look at it.
We can think of addition and multiplication in a different way than as binary operations. Let us use nice little blocks. The most important thing we can do with them is to do absolutely nothing! Let us denote that by $[]$.
For addition we can easily understand the procedure of adding $x$ blocks to the current set of blocks. Let us denote that as $[+x]$, which can be considered a function that takes a number of blocks and returns a new number of blocks. We can now understand $x$ as "0 blocks with x blocks added", and $x+y$ as "$0$ blocks with $x$ blocks added and then $y$ blocks added". Let us write that as $0[+x][+y]$ (with the procedures done in order from left to right). Now we can easily accept that addition (of blocks) is associative and commutative by common sense applied to blocks. Notice that we could have understood addition without talking about starting with $0$ blocks, because these procedures work with any number of blocks. What makes $0$ special is that $[+0] = []$; adding $0$ blocks doesn't do anything. That is why $0$ is called the additive identity.
Then now we can talk about undoing addition. To undo $[+x]$ is to take away $x$ blocks, which we shall denote as $[-x]$. Note that $[-0] = [+0] = []$. By common sense, $[+x][-x] = []$ (because doing something and undoing it completely is equivalent to doing nothing). Also $[-x][+x] = []$ (because adding blocks undoes taking away blocks). This is why the negative numbers are called inverses of the positive ones.
Now for multiplication, we can understand $x×c$ as repeatedly adding $x$ blocks $c$ times to $0$ blocks, and we shall denote this by $[+x]^c$ where $^c$ means "repeat $c$ times". Note that repeating anything $0$ times is the same as doing nothing. In particular, $[+x]^0 = []$. We also see that repeating something $a$ times, and then $b$ more times, is the same as repeating it $a+b$ times. So we get $[+x]^a [+x]^b = [+x]^{a+b}$. Furthermore, we could repeatedly undo adding $x$ blocks $c$ times, which we shall denote by $[+x]^{-c}$. Notice that the above identity still holds for negative $a$ or negative $b$ because repeats of a procedure and repeats of its inverses stack together and cancel just like adding and taking away blocks. In particular $[+x]^c [+x]^{-c} = [+x]^0 = []$.
The previous paragraph applies to taking away blocks too, and so $[-x]^c [-x]^{-c} = [-x]^0 = []$. In general, doing something $c$ times cancels undoing it $c$ times. But adding $x$ blocks $c$ times also cancels taking away $x$ blocks $c$ times, which is $[+x]^c [-x]^c = []$. So actually adding $x$ blocks $c$ times is the same as undoing taking away $x$ blocks $c$ times, which is the main point. In symbols, $[+x]^c = [+x]^c [] = [+x]^c [-x]^c [-x]^{-c} = [] [-x]^{-c} = [-x]^{-c}$, which gives $x×c = (-x) × (-c)$ when you start with $0$ blocks.
Up to here we are done for integer multiples, since we can replace blocks by water to handle arbitrary real quantities for $x$ (assuming water is infinitely divisible...). Also, the ideas developed up to here applies directly to ring theory. If we want arbitrary real quantities for $c$, we can first look at adding $x/c$ as adding an amount such that when repeated $c$ times it is the same as adding $x$. In short, $(x/c)×c = x$. Clearly not every procedure can be split up into identical procedures like this, so this only applies to some kinds of procedures such as adding real quantities. Moreover, we also want multiplying by $c$ and dividing by $c$ to be inverses, so we also need $(x×c)/c = x$, which makes division by $0$ not well-defined. By common sense we can then obtain all the basic arithmetic facts about fractions. And then we use fractions with bigger and bigger numerator and denominator to approximate arbitrary real numbers, and all the facts will carry over.
All this is to me a perfectly intuitive reasoning to obtain the axioms for rings and fields, as well as the special field of the reals. It is a bit long but I hope it is a sufficiently complete explanation that can be easily understood by anyone. Do leave a comment if you spot any typographical errors or think any part can be better explained!
A: If you find natural that $2 \times (-4) = (-4) \times 2 = (-4) + (-4)$ you would agree that $a \times (-b) = - (a\times b)$. This can be made true even if $a<0$ and/or $b<0$ resulting in $(-a)\times(-b) = - ((-a)\times b) = -(-(a \times b)) = a \times b$. 
