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This is a pattern even school kids could discover (when gently pointed to). I never did conciously, and cannot remember to have been pointed to explicitly, neither at school nor later:

$$\color{red}{\mathbf{2}}\cdot 9 = 1\color{red}{\mathbf{8}}$$ $$\color{red}{\mathbf{8}}\cdot 9 = 7\color{red}{\mathbf{2}}$$

$$\color{blue}{\mathbf{3}}\cdot 9 = 2\color{blue}{\mathbf{7}}$$ $$\color{blue}{\mathbf{7}}\cdot 9 = 6\color{blue}{\mathbf{3}}$$

$$\color{green}{\mathbf{4}}\cdot 9 = 3\color{green}{\mathbf{6}}$$ $$\color{green}{\mathbf{6}}\cdot 9 = 5\color{green}{\mathbf{4}}$$

which may come as kind of a miracle when first discovering it.

In mathematical terms

$$\boxed{a\cdot (10-1) \equiv b \mod 10\ \ \ \ \Leftrightarrow\ \ \ \ \ b\cdot (10-1) \equiv a \mod 10 \\ a\cdot (10-1) \equiv b \mod 10\ \ \ \ \Leftrightarrow\ \ \ \ \ a + b = 10 \equiv 0 \mod 10}$$

This holds not only for $10$ but for every $p \in \mathbb{N}$, i.e. in every "number system":

$$\boxed{a\cdot (p-1) \equiv b \mod p\ \ \ \ \Leftrightarrow\ \ \ \ \ b\cdot (p-1) \equiv a \mod p \\ a\cdot (p-1) \equiv b \mod p\ \ \ \ \Leftrightarrow\ \ \ \ \ a + b = p \equiv 0 \mod p}$$

and is responsible for the fact that the graphical multiplication tables of $\mathbb{Z}/p\mathbb{Z}$ always looks the same for $p-1$:

enter image description here

I wonder if there are attempts (in educational research and literature) to make use of the simple observability of the pattern above to explain to (clever) school kids that the observed regularity is not by pure coincidence, why it is so, and what it does "mean".

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I would say probably not; the work (algebra mod n) and the payoff (those digits come out in a neat pattern) are not in a good ratio.

If your goal is to get students to do some of the work (modular arithmetic), you should choose a different payoff. I recommend Pascal's Triangle mod n.

Outline:

  • Introduce modular arithmetic as a "what if:" What if 4 = 0?
  • Practice adding and multiplying numbers under the assumption that 4 = 0.
  • Make a multiplication table where 4 = 0.
  • Show them how Pascal's Triangle is generated with regular old integers.
  • Give them hexagonal graph paper and have them use it to generate Pascal's Triangle where 4 = 0.

From the linked MAA page

For a better graphical effect, tell them not to write any of the 0's (leave them blank). Then when you look at the picture from farther away, you will see all the cool patterns. If your students are super-diligent, you could demand different colors for different values, but I really recommend at least leaving the 0's blank; negative space is a great artistic technique.

Then let the students pick a different n and make their own. Bring multiple sizes of hexagonal graph paper so that students who are fired up can try mod 10 or mod 8 or mod 11 or whatever while some others will just make a little bit of mod 4. They will find mod 2 to be hilarious.

To clarify why I think this answers the question: I think the payoff here is pretty great -- the students get to play in a weird field of math, they get the idea that math is huge and mysterious and cool, and they walk away with a pretty picture. The work here doesn't even look like work to them. But if they want to later do the application you suggested, they will be more able to do so after this introduction anyway.

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  • $\begingroup$ Ah crap; I already basically wrote this answer here: matheducators.stackexchange.com/questions/13264/… .. should I just delete this and turn it into a comment? $\endgroup$ – Chris Cunningham Nov 12 '18 at 16:09
  • $\begingroup$ No, please leave it! $\endgroup$ – Hans-Peter Stricker Nov 12 '18 at 16:16
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    $\begingroup$ I'm talking about school kids not (college?) students. Giving school kids a hexagonal graph paper is already quite advanced (far away from what they know). On the other side, payoff is not all that counts, resp. it depends on what counts as a payoff. The fun of discovering a simple pattern may be a smaller payoff than of discovering an intricate pattern. The payoff may otherwise be: the ease of future calculations or a deeper inisght/better understanding. $\endgroup$ – Hans-Peter Stricker Nov 12 '18 at 16:21
  • $\begingroup$ Anyway: Thanks for your answer! $\endgroup$ – Hans-Peter Stricker Nov 12 '18 at 16:22

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