Two Advanced Topics and Other Issues

This is the first of two planned spring break posts. As usual, I like to mix in a variety of topics during these spring break posts. Today I'll write about two topics too advanced for high school (much less middle school), and then it's back to my favorite vacation time topic -- traditionalists -- which will be in the second post.

Contents

1. Theoni Pappas and Quaternions
2. Glen Van Brummelen and Spherical Trigonometry
3. A Few Loose Ends

Theoni Pappas and Quaternions

This is what Theoni Pappas writes on page 100 of her Magic of Mathematics:

"It is hard for the layperson to comprehend how can numbers be described as having various dimensions. It would seem that a number is just a number -- something describing a particular quantity."

Pappas goes on to explain that the real numbers are one-dimensional, while the complex numbers are in fact two-dimensional. This explains why we must graph them on the complex plane. So a natural question to ask is, can there exist numbers with three -- or even more -- dimensions? Indeed, as soon as we learn that the real numbers can be extended to form the complex numbers, we might wonder whether the complex numbers can be extended to form higher-dimensional numbers.

As it turns out, the 19th century British mathematician Sir William Hamilton was the one who first invented some higher-dimensional numbers -- the quaternions. As you might guess by their name, the quaternions are four-dimensional. As Hamilton discovered, there are no three-dimensional numbers -- we must skip up to the fourth dimension. The set of quaternions is often denoted H in honor of their inventor Hamilton.

So far in my Pappas posts, I've only briefly written about what she writes on each page. But for this week's section, "Quaternions & The Games Numbers Play," I'll take advantage of spring break to write extensively about these new numbers.

We begin by noting that quaternions are four-dimensional. The first dimension is the real dimension, so we can use 1 to denote this dimension. We already have a symbol for the second dimension -- we can just use the letter i that also denotes the second dimension of the complex numbers, as our quaternions will be an extension of the complex numbers. It's traditional then to use the next two letters of the alphabet for the remaining two dimensions, j and k.

Adding quaternions is easy -- just as with the complex numbers, we just add like terms:

-1 + 5i - 3j + 5k
+3 - 3i        - 4k
-2 + 2i - 3j + k

So the tricky operation is quaternion multiplication. In Algebra II, we learn how to multiply complex numbers -- we basically use FOIL, then make the substitution i^2 = -1 to complete the operation. As it turns out, Hamilton came up with the following equations that will help us multiply quaternions:

i^2 = j^2 = k^2 = ijk = -1

We already know i^2 = -1 from the complex numbers -- and now we see that j^2 and k^2 are both equal to -1 as well. We also see that multiplying all three terms together give -1 as well. But what if we want to multiply just two of the letters as well? That is, what are the answers to:

1. ij = ?
2. ji = ?
3. jk = ?
4. kj = ?
5. ki = ?
6. ik = ?

Let's try the first one. We want to find ij. We don't know what it is, but we do know what ijk is:

ijk = -1

At this point we'd love to divide both sides by k, but we don't know what -1/k is. So instead, let's just multiply both sides by k:

ijk = -1
ijkk = -1k
ijk^2 = -k
-ij = -k
ij = k

Ok, so we solved #1 above -- ij is just k. Now we look at #2, and determine what ji is. At this point you might ask, didn't we just solve it to be k?

Well, let's play around a little:

ijk = -1
iijk = -i
-jk = -i
jk = i (That solves #3 above!)
jjk = ji
-k = ji

So apparently, ji is -k, not k. You may wonder whether we made a sign error at some point, but actually we didn't. In fact, if we solve all six problems above, we obtain:

(1-2) ij = k, yet ji = -k
(3-4) jk = i, yet kj = -i
(5-6) ki = j, yet ik = -j

Hamilton discovered that quaternion multiplication is not communicative! Quaternions -- unlike real or complex numbers -- do not satisfy all the postulates of operations (mentioned in Lesson 1-7 of the U of Chicago text).

Some readers may notice that we assumed that multiplication is associative above, since we assumed that (ijk)k = ij(kk) earlier. Also, we wrote i(-1) = -i above, even though we showed later on that multiplication isn't commutative. It's usual in algebra that when given a choice of what to give up, we choose commutativity of multiplication before either associativity or the -1 rule.

After all, we've seen several noncommutative operations before. The most obvious such operation is matrix multiplication. And here's another example we've seen here on the blog -- composition of isometries isn't commutative either! In fact, we've seen that with reflections:

-- If the composite of two reflections RS is a translation with vector v, then the reverse composite SR is a translation with vector -v.
-- If the composite of two reflections RS is a rotation with angle theta, then the reverse composite SR is a rotation with angle -theta.

Thus the composite of reflections is always anticommutative -- RS and SR aren't merely different -- they're inverses!. The multiplication of quaternions isn't always anticommutative, though -- ij and ji are opposites, but if both quaternions have j- and k- parts equal to zero, then it reduces to the multiplication of complex (or real) numbers, which is definitely commutative!

Let's perform a few multiplications:

(-1 + 5i - 3j + 5k)(3 - 3i - 4k) = -3 + 3i + 4k + 15i - 15i^2 - 20ik - 9j + 9ji + 12jk + 15k - 15ki - 20k^2
                                               = -3 + 3i + 4k + 15i + 15 + 20j - 9j - 9k + 12i + 15k - 15j + 20
                                               = 32 + 30i - 4j + 10k

(3 - 3i - 4k)(-1 + 5i - 3j + 5k) = -3 + 15i - 9j + 15k + 3i - 15i^2 + 9ij - 15ik + 4k - 20ki + 12kj - 20k^2
                                               = -3 + 15i - 9j + 15k + 3i + 15 + 9k + 15j + 4k - 20j - 12i + 20
                                               = 32 + 6i - 14j + 28k

As you can see, multiplication isn't commutative.

As it turns out, there's another connection between quaternions and isometries. Even though quaternions are 4D, they can be used to model 3D rotations. This is complicated though. If P is the point whose rotation image P' we're trying to find, then we write P as a quaternion p in the obvious manner -- the i-, j-, and k-parts are the coordinates of P, and the real part is zero. The rotation itself is represented as a quaternion q. Basically, the real part of q corresponds to the angle of the rotation, while the imaginary parts correspond to the axis of rotation. (Recall that 3D rotations have axes and not centers.) This correspondence isn't exact and requires some trigonometry.

Then the rotation image is calculated as:

p' = qpq^-1

Here's it's significant that quaternion multiplication isn't commutative -- otherwise qpq^-1 would be equal to pqq^-1 or p, which would map every point to itself rather than rotate it.

The composite of two 3D rotations isn't generally a rotation, But the composite of two rotations whose axes pass through the origin is another such rotation. Of course this quaternion trick only works for axes through the origin, since q0q^-1 is zero. (We saw something similar happen with rotation matrices earlier.)

The fact that quaternions and rotations are related once again demonstrates that rotations, along with the other transformations, aren't mere inventions of the Common Core writers.

In her book, Pappas tells a story whose protagonist is a quaternion. It ends up going to the numbers convention, where it meets elements of N (natural numbers), Z (integers), Q (rational numbers), R (real numbers). and C (complex numbers). The others must decide whether to admit the quaternion to the numbers convention.

Glen Van Brummelen and Spherical Trigonometry

This past weekend was the final weekend of the Barnes and Noble Education Week. As it turned out, the high school from which I graduated was also hosting a book fair at the store. A portion of all profits during that weekend would be donated to my alma mater. So it was the perfect weekend for me to purchase another math book.

The book I purchased was Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry, by Glen Van Brummelen, a Canadian mathematician, in 2013. Here's the Table of Contents:

1. Heavenly Mathematics
2. Exploring the Sphere
3. The Ancient Approach
4. The Medieval Approach
5. The Modern Approach: Right-Angled Triangles
6. The Modern Approach: Oblique Triangles
7. Area, Angles, and Polyhedra
8. Stereographic Projection
9. Navigating by the Stars

In a way, spherical trigonometry is a natural extension of the spherical geometry we studied the last two summers on the blog, just as ordinary plane trig is an extension of plane geometry. I chose this book because it's logical to make this my side-along reading book for this summer. (Yes, I dropped side-along reading during the school year, but the summer is another matter.) Thus this is a natural follow-up to the last two summers of Legendre.

Inside the dust jacket, it reads:

"Spherical trigonometry was at the heart of astronomy and ocean-going navigation for two millennia. The discipline was a mainstay of mathematics for centuries, and it was a standard subject in high schools until the 1950's. Today, however, it is rarely taught."

So unlike quaternions (which were never a high school topic), we see an argument that spherical trig was once taught to teenagers. Again, it makes sense that since earth is (approximately) a sphere, it follows that spherical geometry and trig would be useful to learn.

I've only glanced at the book so far, since I'll be reading it in more detail this summer. But I do notice that many formulas from plane trig will have analogs in spherical trig. There is a deeper relationship between lengths and angles, since every length is an arc, and arcs have angle measures. Indeed, we know that if the radius of the underlying sphere is 1 and angles are measured in radians, then length and arc measure are identical.

Again, more and more I like the idea of teaching a geometry that incorporates both plane (Euclidean) and spherical geometry. I've mentioned several times before that the so-called "neutral geometry" doesn't incorporate spherical geometry -- instead it incorporates Euclidean and this mysterious hyperbolic geometry. But hyperbolic geometry has no natural model, whereas the earth itself is an approximate model of spherical geometry.

In the past, I wanted to propose a "normal geometry" that incorporates Euclidean and spherical geometry, just as neutral geometry includes both Euclidean and hyperbolic geometry. In fact, I like the name "natural geometry" even better than "normal geometry," since the word "natural" sounds more like "neutral." Some desired theorems of natural geometry would be theorems of both Euclidean and spherical geometry, including:

-- SSS, SAS, ASA
-- The sum of the angles of a triangle is at least 180 degrees.
(Contrast this with neutral geometry: the sum of the angles of a triangle is at most 180 degrees.)

The parallel postulate would actually hold in natural geometry -- because the way it's written, it holds vacuously in spherical geometry. (Recall vacuous truth from my last post.) Instead, the postulate "through any two points there is exactly one line" must be dropped, since through the poles there are infinitely many lines.

Well, we definitely have something to look forward to this summer!

A Few Loose Ends

Lately, I've been thinking about reflections on the coordinate plane. We already know how to reflect points in the x- and y-axes, but what if we want to reflect in a line other than the axes -- in particular, an oblique mirror (that is, a mirror that is neither horizontal nor vertical)?

Unlike oblique rotations (which usually requires some trig), oblique reflections can be performed without trig. Let's recall the definition of reflection -- the point A reflected over the mirror l is the point A' if and only if l is the perpendicular bisector of AA'.

Okay, so l and AA' are perpendicular. Now suppose l is given by an equation y = mx + b. So the slope of l is m. This means that we know exactly what the slope of AA' must be -- its slope and m must be opposite reciprocals.

But just knowing the slope of AA' doesn't make it easy to find A'. We'd need to find the point L on l such that AL is perpendicular to l. This can be accomplished by using Point-Slope using the coordinates of A and the slope -1/m -- this gives us an equation for AL. The point L is determined by solving a system of equations (the equation for AL and the equation for l). Finally, we find A' as the point on AL that's the same distance from L as A is but on the other side of l -- the easiest way to do this is to add the vector AL to the coordinates of L.

No trig is required, but this is very complicated. But I'm considering a special case where the reflection is easy to find. Consider the following situation:

Mirror: y = 2x - 4
Preimage: A(-2, -3)

The slope of the mirror is 2, so the slope of a line perpendicular to it is -1/2. So we start at the point A(-2, -3) and count down 1, right 2 to the point (0, -4). This point is clearly on the mirror -- and so we count down 1, right 2 again to the point A'(2, -5), which is the desired image.

If the original preimage had been B(-4, -2), then we must count two "slope steps" (where each slope step is down 1, right 2) to get to the point (0, -4) on the mirror. So then we count an equal number of slope steps to reach B'(4, -6), which is the desired image.

If the original preimage had been C(1, 3), then we count one slope step to reach (3, 2), which is clearly on the mirror. (Note that we're not doing this in our heads -- we're doing it on a graph where we've already drawn the mirror before beginning the reflection. It should be obvious once we've drawn the graph that (3, 2) is on the mirror.) So we count an equal number (1) of slope steps to reach C'(5, 1), which is the desired image.

Of course, not every point is easy to reflect in this manner. If the preimage is (0, 0), then one slope step takes us to (2, -1), which isn't on the mirror -- in fact it's on the other side of the mirror from the original preimage (the origin).

But I can easily see a question directing a high school student to reflect a triangle across an oblique line, where all of the vertices are carefully chosen such that each one is a whole number of slope steps from the mirror (for example, where the mirror and points A, B, and C are given exactly as listed above). In fact, I recently saw this sort of question somewhere.

This question requires the use of slope properties to perform oblique reflections. Again, I've pointed out that if we're using transformations to prove the properties of the coordinate plane (including slope properties, derived from similarity, dilations, and rotations), then we shouldn't be using those properties to perform transformations. But we've already seen that the Common Core tests have this chicken-and-egg problem anyway.

This concludes this first spring break post. The other post will be in a few days.