# How to explain the commutativity of multiplication to middle school students?

It's easy for natural numbers: $3\times 5=5\times 3$

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but how do you explain that $x.y=y.x$ for any real numbers $x$ and $y$.

Moreover, in $\Bbb{N}$, do you prefer to define $n\times m=\underbrace{n+n+\cdots+n}_{m\text{ times}}$ or $n\times m=\underbrace{m+m+\cdots+m}_{n\text{ times}}$.

• How will you explain real numbers to middle school students? – GregRos Jul 2 '13 at 16:55
• This and this. – Git Gud Jul 2 '13 at 17:07
• @Greg: Can't you take your ruler and use it to measure the diagonal of a square? – MJD Jul 2 '13 at 17:17
• @GregRos I explain the existence of irrational numbers (in fact algebraic numbers and $\pi$) and then I define the real numbers to be their union. – llllllllllllllllllllllllllllll Jul 2 '13 at 17:30
• @GregRos I discuss integers and rational numbers a little before going into square roots and the Pythagorean theorem. Then after proving the Pythagorean theorem, I draw an axis and ask them if there's numbers on this axis other than the rationals? Then I draw an isosceles right triangle ABC such that $AB=AC=1$ and the hypotenuse $[BC]$ is on the axis. They prove that $BC=\sqrt{2}$ and then I prove (by contradiction) that $\sqrt{2}$ isn't rational.... – llllllllllllllllllllllllllllll Jul 2 '13 at 17:48

First do it with integers: make a rectangular array of dots, then turn the rectangle ninety degrees. Now instead of an array of $n$ rows, each with $m$ dots, it's an array of $m$ rows, each with $n$ dots. But the number of dots didn't change, only the way they were arranged.

More generally, a rectangle with real-length sides doesn't change its area when you rotate it a quarter-turn.

• That's pretty neat. I guess you don't really need to explain real numbers at all. – GregRos Jul 2 '13 at 17:00
• That's probably a good thing, since real numbers are incredibly complicated and bizarre. – MJD Jul 2 '13 at 17:00
• Yeah, direct analogy works really well in this case, and I think that a lot of them have a picture of real numbers as "like a measurement you get off a ruler" which meshes well with this. – rschwieb Jul 2 '13 at 17:05
• While intuitively is a very nice explanation, isn't in reality backwards: area of rectangles makes sense just because multiplication is commutative. – N. S. Jul 2 '13 at 18:36
• I strongly disagree. Multiplication is a model of the areas of rectangles, and not vice versa; there were rectangles before there was multiplication. We define multiplication the way we do because we want it to have certain properties, and the reason we want it to have those particular properties is that those are the properties that we observe of certain objects in the physical universe, in particular rectangular objects. – MJD Jul 2 '13 at 19:29

One option, which my teacher did with us (8th grade, so may not work so well with 6th graders), is to show the class a system where multiplication (or perhaps even addition) is not commutative, like matricies or 3D vectors. After showing them that that type of multiplication is not commutative, it becomes much easier to understand that there is something special about multiplication on the reals.

The entire rest of the field axioms can be handled the same way, if you can just show systems that don't have all the properties that they take for granted.

• distributivity of multiplication over addition
• trichotomy (for any $a,b$, exactly one of $a < b$, $a = b$, or $a> b$)
• substitution ($a=b \oplus c \wedge c=d \to a=b \oplus d$)
• Well, 8th graders don't know anything about 3D vectors much less about matrices so I think it should be the other way around: students will appreciate the commutativity of multiplication on $\mathbb{R}$ when they learn 3D vectors and the cross product not learning 3D vectors and the cross product to appreciate the commutativity of multiplication on $\mathbb{R}$. – llllllllllllllllllllllllllllll Jul 2 '13 at 20:32
• @metacompactness "8th graders dont know anything about 3D vectors much less about matricies." That's what you are for. "Students will appreciate the commutativity of multiplication on $\mathbb{R}$ when they learn 3D vectors and matricies." That is exactly the point I was making. – AJMansfield Jul 2 '13 at 20:43
• @BlueRaja-DannyPflughoeft Maybe you're right; I can explain why the multiplication of two decimal numbers $\frac{a}{10^n}$ and $\frac{b}{10^m}$ is commutative (as a consequence of the commutativity of the multiplication on $\mathbb{Z}$). When $n,m\to +\infty$, we get this property on $\mathbb{R}$. – llllllllllllllllllllllllllllll Jul 7 '13 at 16:34