# On the commutative property of multiplication (domain of integers, possibly reals)

$ab = ba$

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.

• 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

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 $a,b$.

• +1 for introducing me to Peano’s axioms. The OP specifically asked for “domain of integers” $\mathbb{Z}$ but the Wikipedia entry deals only with Naturals $\mathbb{N}$. I came looking for proof of commutativity of multiplication for negative integers but found I hadn’t actually reached a solid basis for commutativity of positive integers. Peano’s axioms gave me that, but I am still left wondering, like the OP, how to extend this proof to negative integers, or even $\mathbb{Q}$ and $\mathbb{R}$. – lukejanicke Dec 29 '18 at 14:15

These properties came from geometry, where numbers are represented as lengths of lines and products as areas of rectangles.

If a rectangle has sides a and b, the area is defined as ab. If the rectangle is rotated 90 degrees, the area stays the same but the sides are now b and a, so ab = ba.

Other rules are similarly derived (e.g., a+b=b+a, a(b+c) = ab + ac).

I do not know everything that is going on on this page, but I just wanted to say that Euclid, when he proves the commutativity of multiplication in VII - 16 of the elements uses the way we can alternate a proportion, so that A:B::C:D becomes A:C::B:D. Thus when you do multiplication like Descartes, where you make the product by taking it such that 1:factor1::factor2:product, it will be true that factor2 x factor1 = product because, by alternating the last proportion, 1:factor2::factor1:product. I think that the alternation of proportion is the fundament of the commutation of multiplication - prove that however you think figure out.

You can prove the commutative property in a much simpler way.

$let$ $a=3, b=5$

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is obviously the same number of cells as

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The first one is $3 \times 5$, and the second one is $5 \times 3$. Both examples have 15 cells.

Here is a proof for all non-negative integers. We are attempting to show that ab = ba. Let’s introduce a new equivalence, b+e = a (i.e. e is defined as the difference between a and b; note that if e is 0 then the proof becomes trivial). Now we write: (1) ab = b + b +… +b (where the summation contains ‘a’ terms) This is nothing more than stating the definition of ab, ie b summed a times. We can also write (2) ba = b(b+e) since b + e = a, by our own definition. We now try to show that equation (2) can be rewritten in the form of equation (1). We expand on equation (2) by writing: (3) b*(b+e) = (b+e) + (b+e) +…(b+e) (where the summation contains ‘b’ terms) This is very similar to what we did regarding equation (1), ie b*(b+e) is just (b+e) summed b times. Using some properties of addition, we can transform the right hand side of (3) to read: (4) (b+e) + (b+e) +…(b+e) = b+b+…b + e + e + … e (where the summation contains ‘b’ b-terms and ‘b’ e-terms) Now what we are going to do is assume the very thing we set out to prove! That is usually a big no-no unless you are using induction, which is basically where this is going. If you compare the right hand side of equation (1) to the right hand side of equation (4), you will see they are similar; all we have to do is show that e summed b times is equal to b summed e times and we will have shown that the two sides are equal: we will have b summed a times in (1), and b summed b times plus b summed e times in (4), and since b+e = a, the right hand side of (4) is just b summed a times, and as this is identical to (1) we are done. The key to the proof is showing that e summed b times is equal to b summed e times, but that is just saying (5) be = eb which is literally what we set out to prove. The advantage that we have now (after going through all that work) is that we have reduced the number space of the original problem; e by definition is less than a (in the case where e is equal to a, b is identically 0, and the whole proof becomes trivial). We can continue in this way to reduce the number space of the problem until we eventually get to a base case which can be shown to be trivially true (namely when e = 0); this is the nature of inductive proof . I know this isn’t as formal as a textbook proof, but it is a cute little intuitive proof that I had not yet seen presented in such a way on the internet, so I thought I would submit it. I hope it helps someone!

• here is a mathjax reference and you might like to improve on the spacing for readability as well. – Siong Thye Goh Aug 21 '18 at 2:02
• I enjoyed trying to follow this (which you presently nicely on Quora) but it is not induction and therefore assuming $be=eb$ to prove $ab=ba$ is an error. – lukejanicke Dec 29 '18 at 14:31