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We all know that Commutative property and Associative property of multiplication is always saved for the real and complex numbers. I know that if I recalculate it a million times, the result will be the same. But why does it work?

Is there an explanation for this, or is it just a coincidence that later made an axiom?

I am not an expert in mathematics, my level of knowledge is high school. Thank for you answer.

Honza Prochazka
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    Because we constructed these extended number systems with the requirement that these properties persist, so that results proved in the extended number system remain true when specialized, e.g. that's how we obtain simple ("linearized") proofs of properties of reals by using complex numbers, and why results like the Binomial Theroem hold true universally (in all commutative rings). More generally we axiomatize the sought algebraic structure (e.g. a ring) . – Bill Dubuque May 16 '20 at 20:20
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    For example, when constructing the ring $,\Bbb Z[x],$ of polynomials with integer coef's we assume that $x$ commutes with the coeffs $, nx = xn,$ (which is needed to prove the Binomial Theorem, Factor Theorem, Difference of Squares factorization etc). When we evaluate $x$ at some integer $n$ the results remain true for integers, because we only used (ring) laws in the proof of said polynomial theorems, and these ring laws remain true when we specialize to integers. This is a prototypical example of why we desire to preserve such (algebraic) structure / laws when we extend the structure. – Bill Dubuque May 16 '20 at 20:27
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    my level of knowledge is high school --- Since the answers seemed to have overlooked this, I'll mention that commutativity and associativity for fractions, real numbers, and complex numbers can be obtained from the fact that these hold for the natural numbers, and in this case it's essentially a matter of setting things up so that they'll be true in our "technical formulations", since it seems empirically evident that they hold in the "real world". For example, $3$ chairs in each of $5$ rows gives $3\cdot 5$ many chairs. (continued) – Dave L. Renfro May 16 '20 at 21:04
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    If we now rotate the floor the chairs are on by $90$ degrees (or move to an appropriately different location when looking at the chairs), then we'll see $5$ chairs in each of $3$ rows which gives $5 \cdot 3$ many chairs. Since our experience is that the number of chairs will not change when this is done, we conclude that $3 \cdot 5 = 5 \cdot 3.$ An interesting "picture proof" for associativity of multiplication is given on p. 13 of A Text Book of Algebra by William Steadman Aldis (1887). – Dave L. Renfro May 16 '20 at 21:09

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It is a theorem once we make a formal definition and can prove the property from the definition. Ultimately, these properties for complex numbers $\Bbb C$, are inherited from the properties for $\Bbb R$, which are inherited from them for $\Bbb Q$, then $\Bbb Z$ and in the end $\Bbb N$, whose properties we try to grasp with the Peano axioms.

The proofs of commutativity an associativity in $\Bbb N$ from a recursive definition using only the constant $0$ and the successor function $S$ a la $ n\cdot 0:=0$, $n\cdot Sm:=n\cdot m+n$ (and also $n+0:=n$, $n+Sm:=S(n+m)$) are somewhat technical (and perhaps surprisingly long if one really starts ab ovo) and it seems like a giant portion of good luck that we obtain such nice properties in the end.

So let's go back to a suitable motivation of multiplication: If we arrange pebbles in a rectangular grid of $n$ rows and $m$ columns, then the number of pebbles does not change if we look at the rectangle from a different point so that it appears as $m$ rows and $n$ columns. Hence if multiplication is to mimic the operation of "number of pebbles in a rectangle", then commutativity of multiplication is evident.

For associativity, consider a three-dimensional equivalent arranged in a cuboid grid $n$ long, $m$ wide, and $k$ high. We could rearrange the vertical columns in a line, then this line is $nm$ long and we obtain a (vertical) rectangle of $(n\cdot m)\cdot k$ pebbles. If we first rotate the cuboid to make $n$ the vertical extension, we arrive at $n\cdot (m\cdot k)$ pebbles. So in the end, via the "pebbles in a rectangle" definition of multiplication, commutativity and associativity are consequences of spatial symmetries and of the count of objects being invariant under movement in (abstract) three-dimensional space.

J. W. Tanner
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Hagen von Eitzen
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    (+1) When I wrote my 2 comments above (no more room to add more than a handful of characters, and about out of time anyway), saying "Since the answers seemed to have overlooked this", I didn't notice that your 2nd and 3rd paragraphs describe pretty much exactly what I included in my comments. – Dave L. Renfro May 16 '20 at 21:11
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The standard system of axioms for arithmetic is the Peano axioms, which are defined in terms of the number $0$ and a 'successor' function: that is, a function that adds $1$ to a natural number, to give the next (or successive) natural number.

Using these axioms, associativity and commutativity of multiplication are theorems that can be proved. See On the commutative property of multiplication (domain of integers, possibly reals) or http://math.ucsd.edu/~nwallach/peano.pdf.

Prasiortle
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