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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?

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    $\begingroup$ This will depend wildly on how you have defined multiplication in the first place, and which axioms you're willing you let your proof rely on. $\endgroup$ – Henning Makholm Oct 31 '18 at 1:21
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    $\begingroup$ Is this a stupid question? Or is it ok to ask this.... I honestly can't tell $\endgroup$ – Dude156 Oct 31 '18 at 1:22
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    $\begingroup$ You can do this because the distributive property works and is inviolate. Now why the universe (or our brains) work this way? Who knows? $\endgroup$ – fleablood Oct 31 '18 at 1:29
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    $\begingroup$ It's not a dumb question. Investigating structures where this isn't the case is a part of abstract algebra. If you take your question seriously you will be able to find out exactly what rules (in this algebraic structure) allow you to make these leaps. $\endgroup$ – Mason Oct 31 '18 at 1:40
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    $\begingroup$ To those, who voted as off-topic - you need to understand, that OP most probably doesn't have enough background to even understand what you're asking. Yet this is a perfectly valid question. By putting on hold you may discourage OP from further research. For the time just assume (as I do), that OP has some very basic level of math skills, maybe only recently learned about things like distributive property. $\endgroup$ – Ister Oct 31 '18 at 15:19
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Formally, this is a property of rings. Rings have a multiplication operation that distributes with respect to addition, meaning for any 3 numbers $a$ $b$ and $c$: $$a\cdot(b + c) = (a\cdot b) + (a\cdot c)$$ $$(b + c)\cdot a = (b\cdot a) + (c\cdot a)$$ The real numbers are a ring (they're actually a field, which is a special kind of ring), so that's half of your answer.

The other half is Eric's answer which addresses the natural numbers. Natural numbers are not a ring, but they do have a distributive property.

From a philosophical perspective, we could define anything we wanted, but what's interesting about this particular pattern is that it's so useful. Nothing prevents me from making $x+\frac{3}{2}$ means a triple gainer with a twist, but outside of diving, that particular pattern isn't all that useful. We tend to find that fields and rings show up rather often in physically meaningful scenarios.

Now from a philosophical perspective, it makes sense to point out that there are also lots of other really useful patterns that show up. For example, if you look at how we define rotations, such as using yaw, pitch, and roll to describe the orientation of an aircraft, those don't seem to add the way we want them to. The rotations form a pattern known as a group, which doesn't even have a concept of addition at all! They only have multiplication. And by that I mean mathematicians decided to call the one operation in this pattern "multiplication" because its rules are a generalization of matrix multiplication.

We also have all sorts of oddball cases which may or may not actually be philosophically relevant. For example, we can consider the ordinal numbers, which explore labeling objects as 1st, 2nd, 3rd, and so on. Ordinals grapple with the concept of infinity, which generally means they've got some quirks. One of the quirks of ordinals is that they are left distributive but not right distributive. That means I can use the distributive property in $a\cdot(b + c)$ but not $(b + c)\cdot a$! So that shows that we've come up with some really strange systems which look sane, but where the distributive law starts to get a little strange. (For what it's worth, part of the reason this law acts so strange is that multiplication isn't commutative in ordinals: $2\cdot\omega \neq \omega\cdot 2$)

So in the end, what makes this distributive law so interesting philosophically is that real numbers and natural numbers seem to be terribly good at describing the world around us, and both of them have distributive properties. But that doesn't mean that everything interesting has a distributive law, or even that the distributive law will make intuitive sense to you! Now the question for why real numbers and natural numbers are so useful in reality is a really interesting philosophical question which has lead some people to argue that mathematics is the language upon which reality sits.

I say it sits on a turtle. But who am I to judge?

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There are two ways I see to answer this question. One is from an axiomatic standpoint, where numbers are merely symbols on paper that are required to follow certain rules. The other uses the interpretation of multiplication as computing area. The former would take a good 5-10 pages to build up from the Peano axioms.

For the latter, you can draw a rectangle 13 units by 34 units. Break one side into 10 units and 3 units, and the other side into 30 units and 4 units. At this point, you should see that this decomposes the rectangle into four pieces, corresponding to the four terms you get from the distributive rule. The various was to compute $13 \cdot 34$ are all just ways to decompose the rectangle, and at the end of the day they all compute the same number: the area of the rectangle.

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    $\begingroup$ But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation? $\endgroup$ – Dude156 Oct 31 '18 at 1:24
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    $\begingroup$ @Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property. $\endgroup$ – RghtHndSd Oct 31 '18 at 1:31
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    $\begingroup$ There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition .... $\endgroup$ – fleablood Oct 31 '18 at 1:33
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    $\begingroup$ Proving why $13 \times 34=442$ is the same as proving $1+1=2$. It just proves how our undestanding of (mathematical) language works. It does not prove any empirical facts. $\endgroup$ – rexkogitans Oct 31 '18 at 7:23
  • $\begingroup$ @Dude156: My answer should address the question in your comment. In particular, if you define multiplication as counting the number of objects in a rectangular array with the given number of rows and columns, then that is enough to understand distributivity of multiplication over addition of natural numbers. Absolutely no notion of area is needed. Furthermore, if you want a more fundamental approach, you can define multiplication as iterated iteration, as explained in the second half of my answer. $\endgroup$ – user21820 Oct 31 '18 at 8:04
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Well, that is literally what the distributive law tells you. It tells you that $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$ and so whenever you break up the two factors of a product as a sum, you can use the "pieces" to compute the product.

So what you are really asking for is a proof of the distributive law itself. What constitutes a "proof" of such a basic fact depends heavily on what definitions of "numbers" and the operations on them that you are using (for some definitions, the distributive law is simply an axiom that you assume). But here is an intuitive explanation that works for natural numbers (and this can be turned into a rigorous proof if you define arithmetic of natural numbers in terms of cardinalities of sets).

We want to prove that $(a+b)c=ac+bc$. What does a product $xy$ of natural numbers mean? Well, it means you draw a grid of dots with $x$ rows and $y$ columns, and count up the total number of dots. So, to compute $(a+b)c$, you draw a grid with $a+b$ rows and $c$ columns. Now we observe that we can split such a grid into two pieces: the top $a$ rows and the bottom $b$ rows. The top $a$ rows form a grid with $a$ rows and $c$ columns, so they have $ac$ dots. The bottom $b$ rows form a grid with $b$ rows and $c$ columns, so they have $bc$ dots. In total, then, we have $ac+bc$ dots, so $(a+b)c=ac+bc$.

(In the computation of $(a+b)(c+d)$ above I used the distributive law in two different versions, one with the sum on the left side of the product and one with the sum on the right side of the product. You can prove the version with the sum on the right side of the product in the same way (you just split up the columns instead of the rows), or you can deduce it from the other version using commutativity of multiplication.)

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    $\begingroup$ I'm personally fond of this explanation. When I was rather young, my older brother explained a bit about the distributive law to me. I puzzled over it for a bit, and then came on my own to "ah - because if you draw a rectangular grid of dots, you can divide it into two smaller rectangles..." This was exciting as a discovery of a generalized rule (more interesting than "obvious" ones like $a + 0 = a$) which justified why we can say things like $14 \times 2 = (10 \times 2) + (4 \times 2)$. It definitely advanced my love of mathematics. $\endgroup$ – aschepler Oct 31 '18 at 10:38
  • $\begingroup$ I confess I never had the "draw rectangles" realization But I had the "paper bag" realization that $a + b$ is just putting $k=a+b$ into two different bags and $n(a + b)$ means we can $n$ copies of bags with $a$ items and $n$ copies of bags with $b$ items and we are clumping them together. To me that was utterly concrete but I came to realize it was too many intricate steps away for others and too abstract. .. I kind of wish I had a breaking rectangle moment. $\endgroup$ – fleablood Oct 31 '18 at 16:26
  • $\begingroup$ This can also be formalized if your formalization "defines" $a \cdot b$ to be the cardinality of the Cartesian product $\{ 1, \ldots, a \} \times \{ 1, \ldots, b \}$ and similarly $a + b$ is the cardinality of the disjoint union $\{ 1, \ldots, a \} \sqcup \{ 1, \ldots, b \} = (\{ 1 \} \times \{ 1, \ldots, a \}) \cup (\{ 2 \} \times \{ 1, \ldots, b \})$, and then you prove for any finite sets $A$ and $B$, $|A \times B| = |A| \cdot |B|$ and $|A \sqcup B| = |A| + |B|$, prove (or define) two finite sets have equal cardinality if there is a bijection between them, etc. $\endgroup$ – Daniel Schepler Oct 31 '18 at 18:16
  • $\begingroup$ @DanielSchepler: I suppose you would know what you are doing, but your comment would be misleading to students. If you define just the notion of equinumerosity, then you do not have actual cardinalities without picking a canonical representative (such as von Neumann ordinals). Even for finite cardinalities, you first have to prove that you can in fact find some natural $n$ such that $\{1..n\}$ is in bijection with $\{1..a\} \times \{1..b\}$, before you can define $a·b$ the way you said. But you cannot prove that without induction, so it is very much more involved than your comment suggests. $\endgroup$ – user21820 Nov 1 '18 at 8:40
  • $\begingroup$ for some definitions, the distributive law is simply an axiom that you assume --- but then you would still have to prove that there exists a well-defined operation satisfying these axioms. $\endgroup$ – Federico Poloni Nov 1 '18 at 10:40
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The simplest and historical answer is that we observed that certain things in the world can be counted, in the sense that we can put them in a line and label each one with their position in the line, and that these positions satisfy certain empirical facts, and hence we invented axioms to capture these facts. I specifically single out the axioms in that linked section (discrete ordered semi-ring axioms plus induction), rather than the other more commonly known axiomatization of PA, for two reasons.

Firstly, the ring axioms have been observed or proven to hold for a very wide variety of structures, so it is no surprise that people invent them, and they include distributivity of multiplication over addition.

Secondly, that specific axiomatization I linked to accurately captures our empirical knowledge of counting numbers. For example, from our understanding of spatial motions (translations and rotations) we readily grasp that addition is commutative: $a+b$ is the total count of $a$ objects followed by $b$ objects in one line, which is the same as $b+a$ by looking at the line from the other side. Similarly for the commutativity of multiplication: $a·b$ is the total count of a rectangular array of $a$ times $b$ objects, which is equal to $b·a$ by rotating the array by $90$ degrees. And distributivity ($a·(b+c) = a·b+a·c$) is clear by splitting the array. Associativity is also intuitive.

Note that the above axioms are sufficient for us to recover general distributivity of the form you are using: Observe that $(a+b)·c = c·(a+b) = c·a+c·b = a·c+b·c$ by commutativity and distributivity. Thus $(a+b)·(c+d) = (a+b)·c + (a+b)·d = (a·c+b·c) + (a·d+b·d)$, and the order of addition in the last expression does not matter by associativity.


There is a more complicated answer, that also goes a significant way to explaining why counting numbers are the way they are. If you have an operator $f$ on some collection $S$ that can be iterated (i.e. $f : S \to S$), then you can define $f^0 = \text{id}_S$ and $f^1 = f$, and define $f^{m+n} = f^m \circ f^n$ and $f^{m·n} = (f^m)^n$ for every counting numbers $m,n$. The last definition is valid because (by induction) $f^m$ is an operator on $S$ that can be iterated. Then (again by induction), you can actually prove the basic properties of $+,·$. For example:

Associativity of $+$

Take any naturals $k,m,n$. Then $f^{k+(m+n)} = f^k \circ ( f^m \circ f^n ) = ( f^k \circ f^m ) \circ f^n = f^{(k+m)+n}$. Here we are using the fact that function composition is associative, which is a core reason for almost every instance of associativity in mathematics.

Commutativity of $+$

First we show that $f^{m+1} = f^{1+m}$ for every natural $m$. $f^{0+1} = f^0 \circ f = f = f$ $\circ f^0 = f^{1+0}$ by definition. For any natural $n$ such that $f^{n+1} = f^{1+n}$, we also have $f^{(n+1)+1} = f^{n+1} \circ f = f^{1+n} \circ f$ $= f^{(1+n)+1} = f^{1+(n+1)}$. Therefore by induction we are done.

Now take any natural $m$. Then $f^{m+0} = f^m \circ f^0 = f^m$ $= f^0 \circ f^m = f^{0+m}$. And for any natural $n$ such that $f^{m+n} = f^{n+m}$, we also have $f^{m+(n+1)} = f^{(m+n)+1} = f^{m+n} \circ f = f^{n+m} \circ f$ $= f^{(n+m)+1} = f^{n+(m+1)}$ $= f^n \circ f^{m+1} = f^n \circ f^{1+m}$ $= f^{n+(1+m)} = f^{(n+1)+m}$. Therefore by induction we are done.

As you can see from the above proofs, induction is enough for us to bootstrap from very rudimentary notions of iteration to obtain addition and its properties. We can do the same for multiplication. But it should be clear that this was not how the axioms were originally invented.

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  • $\begingroup$ Good answer, but your second half uses notation and concepts that a high schooler isn't going to be familiar with. That's definitely university level maths. $\endgroup$ – Tam Coton Nov 2 '18 at 10:35
  • $\begingroup$ @TamCoton: Well, the mathematics used is accessible to any high-school student who is sufficiently interested. Yes it's more complicated, but that is the best way to explain why the natural numbers are a very naturally arising structure rather than an incidental one. $\endgroup$ – user21820 Nov 2 '18 at 13:43
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    $\begingroup$ A high schooler probably isn't going to know that f: S -> S means a function mapping every member of a set to another member of that set, nor are they going to know what to search to find that out. I suspect that id_S and ∘ are also likely to be confusing. $\endgroup$ – Tam Coton Nov 2 '18 at 13:50
  • $\begingroup$ @TamCoton: Actually, in many countries such notation is taught and used regularly in high-school. And, of course, when you read something on the internet and find unfamiliar terms, Google is just a few clicks away... $\endgroup$ – user21820 Nov 2 '18 at 14:09
  • $\begingroup$ @TamCoton: For reference, The Free Dictionary gives a definition of "operator" in Mathematics as "A function, especially one from a set to itself, such as differentiation of a differentiable function or rotation of a vector.". Googling either "id_S" or "id function" immediately brings up Wikipedia's article on identity function. Likewise Googling "∘" brings up Wikipedia's article on function composition. $\endgroup$ – user21820 Nov 2 '18 at 14:24
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The "reason" it works is because in this universe quantities stay the same no matter how we group or arrange them. Therefore because things don't magically appear and disappear we can count them. And if we wanted to combine things by counting one group and then another group and combing them with another we can add them.

Thus we no $10 + 3 = 7+ 6 = 13$ are consistent ways of grouping and identifying values.

And we can do multiple adding to define multiplication. So $6 +6 + 6+6 =6\times 4 = 24$. And as it quantities don't change when we group them then $n(a + b) = \underbrace{(a+b)+... + (a+b)}_{n} = \underbrace{a + a...+a}_{n} + \underbrace{b + b...+b}_{n}=n\times a + n\times b$.

That is the way the world works.

And that is what we teach children.

But what kind of unimaginative simpleton cares about how the real world works? Certainly not mathematicians.

Math is modeling and systemizing.

There are two standards:

1) Axiomatic definitions. The definition is a field, $F$ includes the axiom that for any $a,b,c \in F$ that $a(b + c) = ab + ac$. That is the algebrists saying "I don't care why this is true in the real world, but it is this way in my world because I say it is."

2) Construction: Develop the concept of natural numbers via the Peano Postulates and defining the idea of a first element $0$ and the ability to find a unique successor and a few basic axioms. ... And then prove distribution. In essence it is the same as how the real works, except we have defined the concept of number purely and abstractly, and not by giving Fred a bunch of apples.

This is the constructionist way of saying "I've extracted the pure essence of the real world into consistent abstract concepts".

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    $\begingroup$ Actually you don't even need reordering if you do it the other way round: $n(a+b)=\underbrace{n+\dots+n}_{a+b} = \underbrace{n+\dots+n}_a+\underbrace{n+\dots+n}_b = na+nb$. $\endgroup$ – celtschk Oct 31 '18 at 8:31
  • $\begingroup$ True... but I was hiding the issue that for the definition $n\times a= (a + a + a+....+a)$ that $a$ need not be an integer. $\endgroup$ – fleablood Oct 31 '18 at 16:17
  • $\begingroup$ In the other case, $n$ does not need to be an integer. And it also works for ordinals, where your version doesn't (because for ordinals, addition is not commutative). $\endgroup$ – celtschk Oct 31 '18 at 20:03
  • $\begingroup$ Perhaps. But then we are no longing in the real world where we are giving apples to Fred. Thing is we can do this because multiplication distributes. But why multiplicate distributes depends on... how we systemize our definitions and axioms. $\endgroup$ – fleablood Oct 31 '18 at 20:14
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    $\begingroup$ Apple(1/2 + 2) = (Apple + Apple +... )1/2 times plus (Apple + Apple + ... ) 2 times doesn't work. You cant add something to itself a 1/2 time. And $Apple\times k$ means $k$ Apples is not intuitively defined as clearly as all that.... $\endgroup$ – fleablood Oct 31 '18 at 21:17
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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.

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For any natural numbers $a$ and $b$ and for any two ordered quadruplets of natural numbers (c, d, e, f) and (g, h, i, j) where $a = c + d$; $b = e + f$; $a = g + h$ and $b = i + j$, $(c + d)(e + f) = (g + h)(i + j)$ can be proven as follows.

$(c + d)(e + f) = ab = (g + h)(i + j)$

This question can also be interpreted as follows: how can I prove that $ce + ef + de + df = gi + gj + hi + hj?$

This is how to do it. It is a theorem that for any natural numbers $k$, $l$, $m$, $n$, $(k + l)(m + n) = km + kn + lm + ln$ so

$ce + cf + de + df = (c + d)(e + f) = ab = (g + h)(i + j) = gi + gj + hi + hj$

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