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$ab = ba$

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.

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3  
One way to think of the commutative property of multiplication is that the area of a rectangle ($ab$) doesn't change when you rotate it ninety degrees ($ba$). –  Alex Becker Feb 21 '12 at 7:37
    
Yes, that is one the interpretations I am currently running in my head, I've tried analyzing it by getting the Cartesian product (AxB) of the set A and B whose members are decrements of its cardinality, which could be interpreted on a coordinate system to get the same amount of points, differing only by rotation. But I would love a rigorous proof. –  Wannaknow Feb 21 '12 at 7:43
    
A rigorous proof of the commutative property? I know of a proof from the Peano axioms, but it all depends on what you're willing to start from. –  Alex Becker Feb 21 '12 at 7:52
    
Please, if you could relay it out. I know it's all dependent on the basis we agree to, peano axioms sound good, please continue. :D –  Wannaknow Feb 21 '12 at 7:55
    
possible duplicate of Commutativity of multiplication in $\mathbb{N}$ –  joriki Feb 21 '12 at 8:44

1 Answer 1

up vote 3 down vote accepted

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 $a,b$.

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