Month: April 2015

by

Literature Spotlight: Infinity in the Real World

I recently read a new-ish novel by one of my favorite authors, the incomparable Terry Pratchett, that provided me with some much-needed food for thought. The Long Earth, a collaboration between Pratchett and Stephen Baxter, centers around the invention and distribution of a simple contraption enabling its user to ‘step’ between an infinite number of parallel dimensions. Each of these dimensions is slightly different from every other, possibly depicting a series of ‘what if?’ alternate Earths, and the entirety together is referred to as ‘The Long Earth.’ One of the most curious things about the Long Earth, however, is that none of these alternate Earths have any humans on them – no cities, no civilizations, simply wild and beautiful vistas with plenty of local wildlife and a few enigmatic ‘humanoid’ races that are rarely seen. Forget space travel – mankind can simply step across the Long Earth and find millions of pristine new worlds to conquer!

The novel brings up quite a few intriguing issues stemming from the idea of the Long Earth being essentially infinite and available. It struck me as touching on the difficulty of pinning down the concept of infinity (see my Math Journey about Xeno’s Paradox). Difficulties like: in all those alternate Earths, is the footprint of the United States still US territory? Do the people who strike out to ‘colonize’ those Earths have to pay taxes to a government that’s not even in their reality? And if not, should the US government be sending taxpayer dollars to help fund these colonial expeditions? How do you police your citizens or exploit your cheap labor when disgruntled people can simply step out of reality and find another empty world? Add to that the fact that a small portion of the population are ‘natural steppers,’ who can travel the Long Earth without the nausea that usually accompanies stepping, and another small portion are so-called ‘phobics,’ who can’t step at all, no matter what, and you’ve got the makings of a class war, complete with religious cults and propaganda surrounding the ‘unnatural’ act of stepping.

I’ve always felt that a book doesn’t need to be “classical literature” in order to be ripe for essay-writing and English-class-style discussion. With summer break on the horizon, now’s a great time to make a list of lighter, easier-to-read books that still have enough substance to challenge you over the summer. I like to read fun books with my students over the summer and encourage them to think critically about everything they read, and this book has just been added to my list of popular fiction worthy of a summer unit. What does it mean to have all of infinity at your fingertips, to be able to step away from all responsibility and carve out a new life for yourself at any time?

by

Mathematical Journeys: A Gnarly System of Three Equations

I received this problem from a friend, who was having trouble while helping her nephew with it. It turned out to be quite a doozy, so I’m presenting it as today’s Math Journey to show how the process we used last time works even with a gnarly, complicated problem.

Solve using the Addition Method:

3x – 3y + 4z = – 15
3x + y – 3z = – 8
23x – y – 4z = 0

As we discussed last month, the basic idea behind solving a system of equations is to use one equation to solve another for a specific variable, and to do that enough times that you can eventually rewrite one of those equations with only one variable in it, and solve from there. The way I learned to do this is the “substitution” method, where you solve one equation for one variable, plug the expression in for that variable in a second equation, et cetera until you’re down to one variable. The addition method works a little differently, but it’s the same basic goal: eliminate enough of the variables that you can solve the system. The basic idea of the addition method is to figure out a way to get two of the equations to have opposite coefficients on one of the variables, so that if you were to add all the terms in those two equations, you’d cancel out one of the variables entirely.

Take a look at the first and third equations for now:

3x -3y +4z = -15
23x -y -4z = 0

The z’s have a positive 4 and a negative 4 as coefficients, so if we were to add the third equation to the first equation, which in practice means adding straight down each column in the equations, we’d get:

26x -4y + 0z = -15

So the z’s will cancel each other out. Now, our new first equation is:

26x -4y = -15

Good; we’ve eliminated one variable from the first equation. Go back to the original system. Look at the second and third equations. Do you see any way to eliminate the z’s using just those two equations in the system? I do. It’s not as immediately apparent, but it’s there.

3x +y -3z = -8
23x -y -4z = 0

This one we have to do some preliminary work for. If we’re going to add them, we need one equation to have a positive coefficient and the other a negative, and they need to have the same absolute value. They’re both negative right now, and the numbers are different. No problem – think about finding a Lowest Common Denominator back in middle school. Remember, we can multiply an entire equation by a single constant without changing the value of the equation, just like finding an equivalent fraction. We’ll have to do both cases for this one – sometimes you can get away with one, but not today.

Think LCD – what numbers are we going to multiply each equation by to eliminate the z’s?

The first one should get a 4, and the second one a 3, and one of them needs to be negative. Doesn’t matter which one, so let’s pick. I say negative 3 and positive 4.

4 (3x +y -3z = -8) → 12x +4y -12z = -32
-3 (23x -y -4z = 0) → -69x +3y+12z = 0

So now, when we add straight down, we get:

-57x +7y +0z = -32

Or:

-57x +7y = -32

Okay, now let’s go back to our original system. Which equation did we use in both of our additions? The third, right? Okay, so set that one aside for now. We’re going to replace the first and second equations with our new ones:

26x -4y = -15
-57x +7y = -32

Now, we can treat these two as a system of two variables in two equations, and repeat the above process to reduce it down to one variable. I’d say, since the coefficients are smaller and easier to handle, let’s try to eliminate y. Same process as before – what’s our LCD? 28. So:

7 (26x -4y = -15) → 182x -28y = -105
4 (-57x +7y = -32) → -228x +28y =-128

When we add those straight down, we’ll get:

-46x = -233

And THAT we can solve for x, since it’s only got one variable:

x = 233/46 (I’ll leave it as a fraction, since it’s ugly)

Now, we can do what’s sometimes called “back-solving” to get our final answer.

First, plug that into the first of our new (2-variable) equations to solve for y:

26x -4y = -15
26(233/46) -4y = -15
(3029/23) -4y = -15
-4y = -15 -(3029/23)
-4y = -(3374/23)
y = -(3374/23) / -4
y = 1687/46

Now that we have y, we substitute BOTH our x and our y into the third equation (the one we set aside at the beginning – the one with three variables still in it) to solve for z.

23x -y -4z = 0
23(233/46) – (1687/46) -4z = 0
(5359/46) – (1687/46) = 4z
3672/46 = 4z
z = (3672/46) * (¼)
z = 918/46

or, when you reduce it all the way: 459/23

So, our final values for all three variables are:

x = 233/46
y = 1687/46
z = 459/23

And yes, they’re ugly, but yes, that’s it.

Just to be sure, I solved this same problem using the substitution method, and got the same answers. It’s a bit more convoluted to use the addition method, but it still works. (My guess is that’s why the problem specifically required you to use the addition method – it would have been too easy the other way!) As with Lowest Common Denominator problems, if you don’t have compatible numbers given to you the fractions are going to get big and ugly really fast. This one was unfortunate in that 23 is prime, so all the numbers were forced into being pretty big. Don’t worry though; you’ll still get an answer eventually.

Incidentally, systems can be tricky to get the hang of at first because any given system has many different paths you could take to get to the end. You can start with whichever equation you want and solve for whichever variable you want. Ideally, the problem should work no matter what order you choose to use the equations in. Some paths are shorter or easier than others, though, so it pays to think briefly about how to tackle a new problem to save yourself a headache later on.