$ab = ba$
This is, inherently, true. Some texts drop it like an axiom without any justification. But I'm a bit curious where it stems from or basically why/how it works. If anyone could enlighten me a bit further, I'd be most grateful. Peano axioms? From set theory? Help.
What bugs me with this definition is it's real world "application":
Let's say I have a power output of 5 W ( $kg\cdot m^2 \cdot s^{-3}$) and I want 30% of that power output (to have some units and natural context).
Naturally, 30% is 30 1/100 (% is the unit). For simplicity, let's express that as:
$30$%$ = 30/100 = 3/10 = 3d$ where $d = 1/10$, a simpler unit (because %W would look… Wrong?).
So 30% of power is then $5 W \cdot 3d$, and we can force it to "make sense" if we associate d with W or basically scale the unit of power by d (which is intuitively understood as being 10 times smaller then the output of 1 W) $dW$. And we want to scale these 5 units of $dW$ three times.
$5 dW \cdot 3$
But the commutative property says it is the same as scaling 3 units of $dW$ 5 times or:
$3 dW \cdot 5$
And the result is the same. This is the bit that hurts my head, the fact it is the same. I try to interpret the same as, for example 50 J of work, it's either applying 50 N over 1 m of distance or 1 N over 50 m of distance.
Down here is additional stuff I think I have (you don't have to read it if you know how to answer immediately). I don't claim it's correct, if anyone parses this information, feel free to correct me.
$m\sum\limits_{i=1}^{n}{1} = n\sum\limits_{i=1}^{m}{1}$
This is how I've been trying to break it down, as it is repeated addition. It is repeated $n$ times and multiplied by $m$, and reverse on the right side. There is $x = m – n$, so when I multiply $n$ $m$ times, each repeated addition of $n$ lacks exactly $x$ to $m$.
$mn = nm$
$x = m-n$
$m(m-x) = (m-x)m$
$m^2 – mx = m^2 – mx$
$ true $
And yes, sadly, I realise that I am using the very property I'm trying to prove ($xm$ shifted as $mx$ on the right-hand side.)
And it even calls in the distributive property of multiplication over addition which I derive (for myself, informally) from the very nature of multiplication and the way we "process" numbers:
For example, number $55$ times $2$
$55 = 5 \cdot 10^1 + 5 \cdot 10^0 = 5 \cdot 10 + 5 \cdot 1$
$(5 \cdot 10 + 5 \cdot 1) \cdot 2$
This is the basis of our positional notation which appends digits of varying orders of magnitude or units together in a way it makes "sense". Each one is b times bigger than the one to the right. Basic stuff. It is why I expressed 10 and 1 explicitly, those are what I consider units in this case.
Now, from the definition of multiplication, which is at its heart just repeated addition, it is truly the same if you add together $2$ $50$ times and then again $5$ times or "all at once" (figuratively, we are always doing the former mentally) $2$ $55$ times.
That added together gives 110, which is true.
To me, the distributive property (in case of integer multiplication) is a repercussion of the very definition of multiplication which is repeated addition (evading strict concepts). I am not certain whether this is a good way to look at it.
Best Answer
The Peano axioms provide an axiomatic basis for the natural numbers, including addition and multiplication of them. In short, they define $0$ and a successor function $S$ which is used to define the natural numbers by $1:=S(0),2:=S(1),\ldots$ in the obvious manner. Addition is defined recursively by $a+0=a$ and $a+S(b)=S(a+b)$, while multiplication is defined by $a\cdot 0=0$ and $a\cdot S(b)=a+(a\cdot b)$. From these axioms and induction we can prove that addition and multiplication are commutative. For my proof I will assume that addition is both associative and commutative, but if you want you can prove these facts from the axioms as well.
Lemma 1: $0\cdot b = 0$ for all $b$.
Proof: For the base case, $0\cdot 0=0$ by the axioms. If $0\cdot b=0$, then $0\cdot S(b)=0+(0\cdot b)=0+0=0$. Thus by induction $0\cdot b=0$ for all $b$.
Lemma 2: $a+(b\cdot a)=S(b)\cdot a$ for all $a,b$.
Proof: We shall induct on $a$. For the base case, $0+(b\cdot 0)=b\cdot 0=0=S(b)\cdot 0$. Suppose $a+(b\cdot a)=S(b)\cdot a$. Then $$\begin{eqnarray} S(a)+(b\cdot S(a))&=&S(a)+(b+(b\cdot a))\\ &=&(S(a)+b)+(b\cdot a)\\ &=&(b+S(a))+(b\cdot a)\\ &=&S(b+a)+(b\cdot a)\\ &=&S(a+b)+(b\cdot a)\\ &=&(a+S(b))+(b\cdot a)\\ &=&(S(b)+a)+(b\cdot a)\\ &=&S(b)+(a+(b\cdot a))\\ &=&S(b)+(S(b)\cdot a)\\ &=&S(b)\cdot S(a) \end{eqnarray}$$ so by induction this holds for all $a,b$.
Proof of Theorem: For the base case, $a\cdot 0 = 0 = 0\cdot a$. If $a\cdot b=b\cdot a$, then $a\cdot S(b)=a+(a\cdot b)=a+(b\cdot a)=S(b)\cdot a$. Thus by induction $a\cdot b=b\cdot a $ for all $a,b$.