Why does multiplication always have the associative and commutative property

abstract-algebraarithmeticaxiomseducation

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.

Best Answer

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.