Polygons and Divisibility

In the last post we introduced cyclic groups. The cyclic group Cn is generated by S, where S represents clockwise rotation through 360°/n (or 1/n of a revolution). It includes S2, which is 2/n of a revolution, and S3, which is 3/n of a revolution, and so on, all the way up to Sn-1. Notice that Sn = I is a complete revolution, which gets us back where we started.

More generally, suppose p is a whole number. Divide p by n to obtain the quotient q and remainder r. We can write p ÷ n = q R r or p = q ⋅ n + r. So Sp = Sqn+r = (Sn)q ∘ Sr. In other words, Sp consists of q complete revolutions composed with r/n of a revolution. It follows that Sp is the same as Sr. So, if we want to see what Sp does, we really only need to pay attention to the remainder of p ÷ n.

Suppose n = 8 for example. The group C8 is the group of rotations of the regular octagon, generated by a 45°-rotation S.

octagon symmetry

What transformation is (say) S29? Well, 29 ÷ 8 = 3 R 5, so S29 amounts to 3 complete revolutions composed with 5/8 of a revolution, or rotation through 5 ⋅ 45° = 225°. In other words, S29 = S5.

Now, we’ve noted that, if m is a divisor of n, then Cm is contained within Cn. This can be seen in the following way. Write n = m ⋅ q. Then Sq represents 1/m of a revolution. For instance, if n = 8, then we can take m = 4. Writing 8 = 4 ⋅ 2, we see that S2 represents one quarter of a revolution, i.e., a rotation through 90°.

polygon inscribed 2

This observation amounts to the following fact: a regular polygon with m sides can be inscribed in a regular polygon with n sides if and only if m is a divisor of n. (We regard a line segment as a “polygon” with two sides.) To draw the polygon with m sides, connect every qth vertex, where n = m ⋅ q.

For instance, the divisors of 6 are 2 and 3, so we can inscribe an equilateral triangle and a bisecting line segment in a regular hexagon. The triangle is constructed by connecting every second vertex. Or again, the divisors of 12 are 2, 3, 4, and 6, so we can inscribe a hexagon, a square, an equilateral triangle, and a bisecting line segment in the regular dodecagon. The square is constructed by connecting every third vertex, since 12 = 4 ⋅ 3.

polygon inscribed

The divisors of 15 are 3 and 5, so we can inscribe a regular pentagon and an equilateral triangle in the regular pentadecagon. The pentagon is constructed by connecting every third vertex, since 15 = 5 ⋅ 3.

So, if m is a divisor of n with n = m ⋅ q, then 〈Sq〉 = Cm. We’d like to answer the more general question now: If p is any whole number, then what does the group 〈Sp〉 amount to? In other words, if we keep composing p/n of a revolution with itself, what cyclic group do we obtain?

Let’s consider the example of n = 8 again. Take p = 3, and write T = S3. Then T is a 135°-rotation, or 3/8 of a revolution. Let’s start composing T with itself. To begin with, T2 is 6/8 of a revolution, or S6. Then T3 is 9/8 of a revolution, which is the same as 1/8 of a revolution. So T3 = S. Next, T4 is 12/8 of a revolution, which is the same as 4/8 of a revolution, or S4. Continuing like this, we find that the group generated by T consists of I, S3, S6, S, S4, S7, S2, and S5, in that order. So 〈T〉 = C8.

Now let’s take p = 6. Write U = S6. The group generated by U consists of S6, S4, S2, and I. We don’t obtain the entire group in this case. In fact, since S6 is a 270°-rotation, S4 is a 180°-rotation, and S2 is a 90°-rotation, we see that 〈U〉 = C4.

If we check each of the elements of C8, what we’ll find is the following. The group can be generated by each of S, S3, S5, and S7. The transformations S2 and S6 only generate C4. The transformation S4 generates C2. And the transformation I generates the trivial group C1.

Now, if p and n are whole numbers not both zero, then their greatest common divisor d is the largest whole number that divides both p and n. For instance, the greatest common divisor of 6 and 8 is 2, while the greatest common divisor of 12 and 30 is 6. In general, the group generated by Sp is the same as the group generated by Sd where d is the greatest common divisor of n and p. So, writing n = m ⋅ d, we find that 〈Sp〉 = Cm. If p and n share no common divisors larger than 1, then they are said to be relatively prime; in this case, m = n, and Sp generates the whole cyclic group.

Consider the case n = 12. Then S, S5, S7, and S11 each generate C12 since 12 shares no common divisors larger than 1 with 1, 5, 7, or 11. The transformations S2 and S10 each generate C6 since the greatest common divisor of 12 and 2, or 12 and 10, is 2. The transformations S3 and S9 each generate C4 since the greatest common divisor of 12 and 3, or 12 and 9, is 3. The transformations S4 and S8 generate C3. The transformation S6 generates C2. And I generates C1.

Or again, consider n = 15. Then S, S2, S4, S7, S8, S11, S13, and S14 each generate C15. Next, S3, S6, S9, and S12 each generate C5. The transformations S5 and S10 generate C3. And I generates C1.

Pick a single vertex of the n-sided polygon. Imagine repeatedly applying a rotation Sp to it (where p is less than n) and connecting each pair of consecutive points with a line segment. This amounts to connecting every pth vertex. If p happens to be 1 or n – 1, then we obtain the polygon itself. But if p is between 1 and n – 1, then we obtain either an inscribed polygon or a star. Also, the figure produced by p is the same as that produced by n – p; the direction is just reversed. It should also be clear that an inscribed star has n points if and only if p and n are relatively prime.

Everyone knows that there’s one 5-pointed star; this corresponds to p = 2 (or p = 3), because we draw it by connecting every second (or third) vertex.

star 5

Next, there are no 6-pointed stars because each of 2, 3, and 4 share common divisors with 6. But there are two 7-pointed stars, corresponding to p = 2 (or p = 5) and p = 3 (or p = 4).

star 7

We draw the first by connecting every second vertex, and the second by connecting every third vertex. There is only one 8-pointed star, corresponding to p = 3 (or p = 5).

star 8

It’s drawn by connecting every third vertex. If we connect every second vertex, we wind up with a square; if we connect every fourth vertex, we obtain a bisecting line segment. Next, there are two 9-pointed stars, corresponding to p = 2 (or p = 7) and p = 4 (or p = 5).

star 9

In general, if n happens to be a prime number, hence has no divisors other than 1 and itself, then there are (n – 3)/2 different n-pointed stars. For instance, there are (11 – 3)/2 = 4 different 11-pointed stars.

star 11

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Unity and Infinity

In recent weeks, Yahoo! News, that worthy and reliable news source, ran an article purporting to list the eleven most beautiful mathematical equations (ten having been too few). Alas, of the list, five are not mathematical, and one is not an equation. However, among the remaining five is the following:

1=.999\ldots

where the 9’s are understood to keep going forever. In my experience, this equation prompts varied responses. The most common is sheer disbelief. “How could these two be equal?” people ask. The answer to this is that, if they aren’t equal, then we should be able to find their difference:

1 - .999\ldots = \;?

But these doubters of the mysteries of unity and infinity can’t state the answer. Generally they’ll admit that it must be a number smaller than every other number, but they still insist that it can’t be zero. What’s interesting is that they’re generally comfortable with the equation

\frac{1}{3} = .333\ldots

After all, a calculator tells us it’s so. But if we multiply both sides by 3, then don’t we get the original?

It isn’t surprising that our equation seems a little dubious. What this really stems from is not understanding what infinite decimal expansions like .333… and .999… actually mean. Now, terminating decimal expansions are easy to understand. For instance,

.9 = \frac{9}{10} \;\;\;\;\;\; .99 = \frac{9}{10} + \frac{9}{100} \;\;\;\;\;\; .999 = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000}

But by what right can we write down something like

.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots

where the sum just keeps going and going? How can we add infinitely many numbers? Does it even make sense to talk about something like that?

To answer this, let’s consider the sequence of partial sums:

.9 = \frac{9}{10} \;\;\;\;\;\; .99 = \frac{9}{10} + \frac{9}{100} \;\;\;\;\;\; .999 = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000}

and so on. Notice that each term in the sequence is a bit larger than the preceding, and that all are less than 1:

.9 < .99 < .999 < .9999 < \cdots < 1

Modern mathematics defines the meaning of the infinite decimal expansion as follows: the value of the expansion is the number x such that, for any positive number ϵ, no matter how small, our sequence of partial sums is eventually in between x – ϵ and x. Put in another way, the value is x if, for any small number ϵ, the difference between x and the nth partial sum is smaller than ϵ if n is large enough.

So, you see, the admission that the difference between 1 and .999… is a number “smaller than every other number” means that .999… is in fact equal to 1.

Counting without Numbers

Imagine a person living at the dawn of civilization, a goatherd, let’s say, dwelling somewhere in the Fertile Crescent. Every day the goatherd lets his animals out of their pen into the pasture so they can graze. When evening comes, he opens the gate and calls to his goats, and they return.

One day the goatherd notices that the herd seems to take up less space in the pen. He begins to worry that he may be losing some goats to thieves or wild animals while they’re out in the field grazing. How is he to make certain?

One obvious suggestion might be to count the goats. That’s what you or I would do. But our goatherd is living at a time when there was no systematic way to count.

Think about this. The English language has proper names for the first twelve counting numbers: one, two, three, and so on, up to twelve. Beyond that, we use the base ten numeration system to label the numbers. For instance, twenty-seven is two tens and seven ones. Three hundred and forty-five is three hundreds, four tens, and five ones.

This machinery originated in India in fairly recent times, only one or two thousand years ago. Our goatherd has no such system. If he wants to label the numbers, he just has to make up proper names for them, and there’s only so many proper names you can come up with. For all we know, his culture may not even have a word for two; the aborigines of Australia are said to have words only for one and many. It would be about as reasonable to ask our goatherd to invent a numeration system on the spot as it would be to ask him to build a computer from scratch. So, how is he to keep track?

Here’s an idea. He could gather a big heap of pebbles and get a large basket. As the goats go out in the morning, he puts one pebble in the basket for each animal that passes him. Once the pen is empty, he knows he has exactly as many pebbles in the basket as goats in the pasture. In other words, he knows that he could pair off the goats and the pebbles without leaving anything out.

goats

Then, when the herd returns in the evening, he can remove one pebble for each goat that passes. If he runs out of goats first, he knows he has a problem. If he runs out of pebbles first, well, he knows that nature has taken its course.

This assignment of pebbles to goats is known as a one-to-one correspondence. Various peoples of antiquity actually did use such methods to keep track of amounts. The ancient Sumerians are said to have used baked clay tokens rather than pebbles for their accounting. They would then seal the tokens in a clay pouch, and put as many marks on the pouch as there were tokens inside. Eventually they decided to do away with the tokens and just use the marks. And the first numeration system was born.

cantorYou see, whenever we count, we are establishing a one-to-one correspondence between a list of numbers and a group of objects. The set of counting numbers may thus be viewed as a universal, abstract set of “pebbles.” Instead of pairing goats with pebbles, we pair goats with numbers, and pebbles with numbers. This involves a profound leap in human thought. The same idea forms the foundation of the modern theory of number as formulated by the great German mathematician Georg Cantor (1845 – 1918). It is to Cantor that we owe the knowledge that there are different kinds of infinities, and that the set of real numbers is more “numerous” than, say, the set of counting numbers.

The child psychologist Jean Piaget (1896 – 1980) studied the role of one-to-one correspondence in early childhood development. In The Child’s Conception of Number, he describes several stages. First, the child compares groups of objects by noting their spacial arrangement or extension, much as our goatherd did when he observed the size of his herd in the pen. This frequently leads to incorrect responses. Later, the child may be brought to recognize the equivalence of two sets through observing a pairing. But it is not until the child realizes that anything done respectively to the two groups can be undone, thus restoring them to the paired arrangement, that they arrive at a true grasp of counting. In group-theoretic terms, we would say that the child has to recognize that the operations performed on the sets are invertible.

So here we have a remarkable parallel between the origins of counting at the dawn of civilization, the theoretical foundation of sets and numbers, and the development of the conception of number in the human mind.