TITLE: Why can we multiply by breaking up the factors as sums in different ways?
QUESTION [33 upvotes]: My friend and I were discussing some mathematical philosophy and how the number systems were created when we reached a question. Why can we multiply two different numbers like this?
Say we had to multiply $13\times 34$. One may break this up like $(10+3)\times (30+4)$. Applying distributive property here will give us the answer of 442. We can also choose to multiply this as $(6+7)\times (22+12)$. Intuitively, we can hypothesize that this should give us the same answer as $13\times 34$. How can we prove that our answer will be equal regardless of how we break up the numbers?

REPLY [3 votes]: How can we prove that our answer will be equal regardless of how we break up the numbers?

I may be not getting the gist of your question, other answers seem deep; but on some (shallowish) level you can proceed as follows.
If you have $A$ and $B$ and you choose $a$ and $b$ freely, then
$$A \equiv a+(A-a)$$
$$B \equiv b+(B-b)$$
Note I use "$\equiv$" symbol that is stronger than "$=$"; see this answer of mine (its part named "definitions vs. equations or conditions") to get the difference. In the above expressions I used "$\equiv$" to emphasize they are true regardless of $a$ and $b$ (compare $A=A+b$ which may or may not be true (depending on $b$), hence only "$=$", not "$\equiv$").
In your example $A=13$ and $B=34$; initially you choose $a=10$ and $b=30$ and you calculate $(A-a)=3$ and $B-b=4$. Whatever numbers you choose for $a$ and $b$, your answer is expressed as
$$W(A,B,a,b) \equiv (a+(A-a))\times(b+(B-b)) \tag{*}$$
Now use
$$x-y \equiv x+(-1)\times y$$ to turn subtraction into addition; and use distributive property to get rid of the parentheses one by one. You don't want to cancel out $a$ and $-a$ right away, this would be trivial (as if you didn't break up the number); instead apply distributive property first in the same way you did with your numbers, observe that some expressions cancel out eventually. The outcome is:
$$W(A,B,a,b) \equiv A \times B \tag{**}$$
I don't know if this is what you asked for. Still on some level it shows that the answer (i.e. $W$) for fixed $A$ and $B$ is equal (i.e. the same) regardless of how we break up the numbers (i.e. regardless of what $a$ and $b$ we choose).

Formally for all this to work $\times$ and $+$ must denote some operations (with certain properties) that you can perform on members of some set (a set $A$, $B$, $a$, $b$, $0$, $1$ and $-1$ must belong to). It all comes to the concept of a ring this answer has already introduced. It's a ring with unity. The concept requires $0$, $1$ and $-1$ to be somewhat special:
$$0+x \equiv x$$
$$1 \times x \equiv x$$
$$1+(-1)=0$$
Plus (in our case) there is commutativity of $+$ and $\times$. If you go from (*) to (**) step by step and in a formal way then you will notice you need to use almost all these properties.