## My 2018 Mathematics A To Z: Sorites Paradox

Today’s topic is the lone (so far) request by bunnydoe, so I’m under pressure to make it decent. If she or anyone else would like to nominate subjects for the letters U through Z, please drop me a note at this post. I keep fooling myself into thinking I’ll get one done in under 1200 words.

This is a story which makes a capitalist look kind of good. I make no vouches for its truth, or even, at this remove, where I got it. The story as I heard it was about Ray Kroc, who made McDonald’s into a thing people of every land can complain about. The story has him demonstrate skepticism about the use of business consultants. A consultant might find, for example, that each sesame-seed hamburger bun has (say) 43 seeds. And that if they just cut it down to 41 seeds then each franchise would save (say) $50,000 annually. And no customer would notice the difference. Fine; trim the seeds a little. The next round of consultant would point out, cutting from 41 seeds to 38 would save a further$65,000 per store per year. And again no customer would notice the difference. Cut to 36 seeds? No customer would notice. This process would end when each bun had three sesame seeds, and the customers notice.

Part of the paradox’s intractability must be that it’s so nearly induction. Induction is a fantastic tool for mathematical problems. We couldn’t do without it. But consider the argument. If a bun is unsatisfying, one more seed won’t make it satisfying. A bun with one seed is unsatisfying. Therefore all buns have an unsatisfying number of sesame seeds on them. It suggests there must be some point at which “adding one more seed won’t help” stops being true. Fine; where is that point, and why isn’t it one fewer or one more seed?

A certain kind of nerd has a snappy answer for the Sorites Paradox. Test a broad population on a variety of sesame-seed buns. There’ll be some so sparse that nearly everyone will say they’re unsatisfying. There’ll be some so abundant most everyone agrees they’re great. So there’s the buns most everyone says are fine. There’s the buns most everyone says are not. The dividing line is at any point between the sparsest that satisfy most people and the most abundant that don’t. The nerds then declare the problem solved and go off. Let them go. We were lucky to get as much of their time as we did. They’re quite busy solving what “really” happened for Rashomon. The approach of “set a line somewhere” is fine if all want is guidance on where to draw a line. It doesn’t help say why we can anoint some border over any other. At least when we use a river as border between states we can agree going into the water disrupts what we were doing with the land. And even then we have to ask what happens during droughts and floods, and if the river is an estuary, how tides affect matters.

We might see an answer by thinking more seriously about these sesame-seed buns. We force a problem by declaring that every bun is either satisfying or it is not. We can imagine buns with enough seeds that we don’t feel cheated by them, but that we also don’t feel satisfied by. This reflects one of the common assumptions of logic. Mathematicians know it as the Law of the Excluded Middle. A thing is true or it is not true. There is no middle case. This is fine for logic. But for everyday words?

It doesn’t work when considering sesame-seed buns. I can imagine a bun that is not satisfying, but also is not unsatisfying. Surely we can make some logical provision for the concept of “meh”. Now we need not draw some arbitrary line between “satisfying” and “unsatisfying”. We must draw two lines, one of them between “unsatisfying” and “meh”. There is a potential here for regression. Also for the thought of a bun that’s “satisfying-meh-satisfying by unsatisfying”. I shall step away from this concept.

But there are more subtle ways to not exclude the middle. For example, we might decide a statement’s truth exists on a spectrum. We can match how true a statement is to a number. Suppose an obvious falsehood is zero; an unimpeachable truth is one, and normal mortal statements somewhere in the middle. “This bun with a single sesame seed is satisfying” might have a truth of 0.01. This perhaps reflects the tastes of people who say they want sesame seeds but don’t actually care. “This bun with fifteen sesame seeds is satisfying” might have a truth of 0.25, say. “This bun with forty sesame seeds is satisfying” might have a truth of 0.97. (It’s true for everyone except those who remember the flush times of the 43-seed bun.) This seems to capture the idea that nothing is always wholly anything. But we can still step into absurdity. Suppose “this bun with 23 sesame seeds is satisfying” has a truth of 0.50. Then “this bun with 23 sesame seeds is not satisfying” should also have a truth of 0.50. What do we make of the statement “this bun with 23 sesame seeds is simultaneously satisfying and not satisfying”? Do we make something different to “this bun with 23 sesame seeds is simultaneously satisfying and satisfying”?

I see you getting tired in the back there. This may seem like word games. And we all know that human words are imprecise concepts. What has this to do with logic, or mathematics, or anything but the philosophy of language? And the first answer is that we understand logic and mathematics through language. When learning mathematics we get presented with definitions that seem absolute and indisputable. We start to see the human influence in mathematics when we ask why 1 is not a prime number. Later we see things like arguments about whether a ring has a multiplicative identity. And then there are more esoteric debates about the bounds of mathematical concepts.

Perhaps we can think of a concept we can’t describe in words. If we don’t express it to other people, the concept dies with us. We need words. No, putting it in symbols does not help. Mathematical symbols may look like slightly alien scrawl. But they are shorthand for words, and can be read as sentences, and there is this fuzziness in all of them.

And we find mathematical properties that share this problem. Consider: what is the color of the chemical element flerovium? Before you say I just made that up, flerovium was first synthesized in 1998, and officially named in 2012. We’d guess that it’s a silvery-white or maybe grey metallic thing. Humanity has only ever observed about ninety atoms of the stuff. It’s, for atoms this big, amazingly stable. We know an isotope of it that has a half-life of two and a half seconds. But it’s hard to believe we’ll ever have enough of the stuff to look at it and say what color it is.

That’s … all right, though? Maybe? Because we know the quantum mechanics that seem to describe how atoms form. And how they should pack together. And how light should be absorbed, and how light should be emitted, and how light should be scattered by it. At least in principle. The exact answers might be beyond us. But we can imagine having a solution, at least in principle. We can imagine the computer that after great diligent work gives us a picture of what a ten-ton lump of flerovium would look like.

So where does its color come from? Or any of the other properties that these atoms have as a group? No one atom has a color. No one atom has a density, either, or a viscosity. No one atom has a temperature, or a surface tension, or a boiling point. In combination, though, they have.

These are known to statistical mechanics, and through that thermodynamics, as intensive properties. If we have a partition function, which describes all the ways a system can be organized, we can extract information about these properties. They turn up as derivatives with respect to the right parameters of the system.

But the same problem exists. Take a homogeneous gas. It has some temperature. Divide it into two equal portions. Both sides have the same temperature. Divide each half into two equal portions again. All four pieces have the same temperature. Divide again, and again, and a few more times. You eventually get containers with so little gas in them they don’t have a temperature. Where did it go? When did it disappear?

The counterpart to an intensive property is an extensive one. This is stuff like the mass or the volume or the energy of a thing. Cut the gas’s container in two, and each has half the volume. Cut it in half again, and each of the four containers has one-quarter the volume. Keep this up and you stay in uncontroversial territory, because I am not discussing Zeno’s Paradoxes here.

And like Zeno’s Paradoxes, the Sorites Paradox can seem at first trivial. We can distinguish a heap from a non-heap; who cares where the dividing line is? Or whether the division is a gradual change? It seems easy. To show why it is easy is hard. Each potential answer is interesting, and plausible, and when you think hard enough of it, not quite satisfying. Good material to think about.

I hope to find some material think about the letter ‘T’ and have it published Friday. It’ll be available at this link, as are the rest of these glossary posts.

## Someone Else’s Homework: Was It Hard? An Umbrella Search

I wanted to follow up, at last, on this homework problem a friend had.

The question: suppose you have a function f. Its domain is the integers Z. Its rage range is also the integers Z. You know two things about the function. First, for any two integers ‘a’ and ‘b’, you know that f(a + b) equals f(a) + f(b). Second, you know there is some odd number ‘c’ for which f(c) is even. The challenge: prove that f is even for all the integers.

My friend asked, as we were working out the question, “Is this hard?” And I wasn’t sure what to say. I didn’t think it was hard, but I understand why someone would. If you’re used to mathematics problems like showing that all the roots of this polynomial are positive, then this stuff about f being even is weird. It’s a different way of thinking about problems. I’ve got experience in that thinking that my friend hasn’t.

All right, but then, what thinking? What did I see that my friend didn’t? And I’m not sure I can answer that perfectly. Part of gaining mastery of a subject is pattern recognition. Spotting how some things fit a form, while other stuff doesn’t, and some other bits yet are irrelevant. But also part of gaining that mastery is that it becomes hard to notice that’s what you’re doing.

But I can try to look with fresh eyes. There is a custom in writing this sort of problem, and that drove much of my thinking. The custom is that a mathematics problem, at this level, works by the rules of a Minute Mystery Puzzle. You are given in the setup everything that you need to solve the problem, yes. But you’re also not given stuff that you don’t need. If the detective mentions to the butler how dreary the rain is on arriving, you’re getting the tip to suspect the houseguest whose umbrella is unaccounted for.

(This format is almost unavoidable for teaching mathematics. At least it seems unavoidable given the number of problems that don’t avoid it. This can be treacherous. One of the hardest parts in stepping out to research on one’s own is that there’s nobody to tell you what the essential pieces are. Telling apart the necessary, the convenient, and the irrelevant requires expertise and I’m not sure that I know how to teach it.)

The first unaccounted-for umbrella in this problem is the function’s domain and range. They’re integers. Why wouldn’t the range, particularly, be all the real numbers? What things are true about the integers that aren’t true about the real numbers? There’s a bunch of things. The highest-level things are rooted in topology. There’s gaps between one integer and its nearest neighbor. Oh, and an integer has a nearest neighbor. A real number doesn’t. That matters for approximations and for sequences and series. Not likely to matter here. Look to more basic, obvious stuff: there’s even and odd numbers. And the problem talks about knowing something for an odd number in the domain. This is a signal to look at odds and evens for the answer.

The second unaccounted-for umbrella is the most specific thing we learn about the function. There is some odd number ‘c’, and the function matches that integer ‘c’ in the domain to some even number f(c) in the range. This makes me think: what do I know about ‘c’? Most basic thing about any odd number is it’s some even number plus one. And that made me think: can I conclude anything about f(1)? Can I conclude anything about f at the sum of two numbers?

Third unaccounted-for umbrella. The less-specific thing we learn about the function. That is that for any integers ‘a’ and ‘b’, f(a + b) is f(a) + f(b). So see how this interacts with the second umbrella. f(c) is f(some-even-number) + f(1). Do I know anything about f(some-even-number)?

Sure. If I know anything about even numbers, it’s that any even number equals two times some integer. Let me call that some-integer ‘k’. Since some-even-number equals 2*k, then, f(some-even-number) is f(2*k), which is f(k + k). And by the third umbrella, that’s f(k) + f(k). By the first umbrella, f(k) has to be an integer. So f(k) + f(k) has to be even.

So, f(c) is an even number. And it has to equal f(2*k) + f(1). f(2*k) is even; so, f(1) has to be even. These are the things that leapt out to me about the problem. This is why the problem looked, to me, easy.

Because I knew that f(1) was even, I knew that f(1 + 1), or f(2), was even. And so would be f(2 + 1), that is, f(3). And so on, for at least all the positive integers.

Now, after that, in my first version of this proof, I got hung up on what seems like a very fussy technical point. And that was, what about f(0)? What about the negative integers? f(0) is easy enough to show. It follows from one of those tricks mathematics majors are told about early. Somewhere in grad school they start to believe it. And that is: adding zero doesn’t change a number’s value, but it can give you a more useful way to express that number. Here’s how adding zero helps: we know c = c + 0. So f(c) = f(c) + f(0) and whether f(c) is even or odd, f(0) has to be even. Evens and odds don’t work any other way.

After that my proof got hung up on what may seem like a pretty fussy technical point. That amounted to whether f(-1) was even or odd. I discussed this with a couple people who could not see what my issue with this was. I admit I wasn’t sure myself. I think I’ve narrowed it down to this: my questioning whether it’s known that the number “negative one” is the same thing as what we get from the operation “zero minus one”. I mean, in general, this isn’t much questioned. Not for the last couple centuries.

You might be having trouble even figuring out why I might worry there could be a difference. In “0 – 1” the – sign there is a binary operation, meaning, “subtract the number on the right from the number on the left”. In “-1” the – sign there is a unary operation, meaning, “take the additive inverse of the number on the right”. These are two different – signs that look alike. One of them interacts with two numbers. One of them interacts with a single number. How can they mean the same thing?

With some ordinary assumptions about what we mean by “addition” and “subtraction” and “equals” and “zero” and “numbers” and stuff, the difference doesn’t matter much. We can swap between “-1” and “0 – 1” effortlessly. If we couldn’t, we probably wouldn’t use the same symbol for the two ideas. But in the context of this particular question, could we count on that?

My friend wasn’t confident in understanding what the heck I was getting on about. Fair enough. But some part of me felt like that needed to be shown. If it hadn’t been recently shown, or used, in class, then it had to go into this proof. And that’s why I went, in the first essay, into the bit about additive inverses.

This was me over-thinking the problem. I got to looking at umbrellas that likely were accounted for.

My second proof, the one thought up in the shower, uses the same unaccounted-for umbrellas. In the first proof, the second unaccounted-for umbrella seemed particularly important. Knowing that f(c) was odd, what else could I learn? In the second proof, it’s the third unaccounted-for umbrella that seemed key. Knowing that f(a + b) is f(a) + f(b), what could I learn? That right away tells me that for any even number ‘d’, f(d) must be even.

Call this the fourth unaccounted-for umbrella. Every integer is either even or odd. So right away I could prove this for what I really want to say is half of the integers. Don’t call it that. There’s not a coherent way to say the even integers are any fraction of all the integers. There’s exactly as many even integers as there are integers. But you know what I mean. (What I mean is, in any finite interval of consecutive integers, half are going to be even. Well, there’ll be at most two more odd integers than there are even integers. That’ll be close enough to half if the interval is long enough. And if we pretend we can make bigger and bigger intervals until all the integers are covered … yeah. Don’t poke at that and do not use it at your thesis defense because it doesn’t work. That’s what it feels like ought to work.)

But that I could cover the even integers in the domain with one quick sentence was a hint. The hint was, look for some thing similar that would cover the odd integers in the domain. And hey, that second unaccounted-for umbrella said something about one odd integer in the domain. Add to that one of those boring little things that a mathematician knows about odd numbers: the difference between any two odd numbers is an even number. ‘c’ is an odd number. So any odd number in the domain, let’s call it ‘d’, is equal to ‘c’ plus some even number. And f(some-even-number) has to be even and there we go.

So all this is what I see when I look at the question. And why I see those things, and why I say this is not a hard problem. It’s all in spotting these umbrellas.

## Someone Else’s Homework: Some More Thoughts

I wanted to get back to my friend’s homework problem. And a question my friend had about the question. It’s a question I figure is good for another essay.

But I also had second thoughts about the answer I gave. Not that it’s wrong, but that it could be better. Also that I’m not doing as well in spelling “range” as I had always assumed I would. This is what happens when I don’t run an essay through Hemmingway App to check whether my sentences are too convoluted. I also catch smaller word glitches.

Let me re-state the problem: Suppose you have a function f, with domain of the integers Z and rage of the integers Z. And also you know that f has the property that for any two integers ‘a’ and ‘b’, f(a + b) equals f(a) + f(b). And finally, suppose that for some odd number ‘c’, you know that f(c) is even. The challenge: prove that f is even for all the integers.

Like I say, the answer I gave on Tuesday is right. That’s fine. I just thought of a better answer. This often happens. There are very few interesting mathematical truths that only have a single proof. The ones that have only a single proof are on the cutting edge, new mathematics in a context we don’t understand well enough yet. (Yes, I am overlooking the obvious exception of ______ .) But a question so well-chewed-over that it’s fit for undergraduate homework? There’s probably dozens of ways to attack that problem.

And yes, you might only see one proof of something. Sometimes there’s an approach that works so well it’s silly to consider alternatives. Or the problem isn’t big enough to need several different proofs. There’s something to regret in that. Re-thinking an argument can make it better. As instructors we might recommend rewriting an assignment before turning it in. But I’m not sure that encourages re-thinking the assignment. It’s too easy to just copy-edit and catch obvious mistakes. Which is valuable, yes. But it’s good for communication, not for the mathematics itself.

So here’s my revised argument. It’s much cleaner, as I realized it while showering Wednesday morning.

Give me an integer. Let’s call it m. Well, m has to be either an even or an odd number. I’m supposing nothing about whether it’s positive or negative, by the way. This means what I show will work whether m is greater than, less than, or equal to zero.

Suppose that m is an even number. Then m has to equal 2*k for some integer k. (And yeah, k might be positive, might be negative, might be zero. Don’t know. Don’t care.) That is, m has to equal k + k. So f(m) = f(k) + f(k). That’s one of the two things we know about the function f. And f(k) + f(k) is is 2 * f(k). And f(k) is an integer: the integers are the function’s rage range). So 2 * f(k) is an even integer. So if m is an even number then f(m) has to be even.

All right. Suppose that m isn’t an even integer. Then it’s got to be an odd integer. So this means m has to be equal to c plus some even number, which I’m going ahead and calling 2*k. Remember c? We were given information about f for that element c in the domain. And again, k might be positive. Might be negative. Might be zero. Don’t know, and don’t need to know. So since m = c + 2*k, we know that f(m) = f(c) + f(2*k). And the other thing we know about f is that f(c) is even. f(2*k) is also even. f(c), which is even, plus f(2*k), which is even, has to be even. So if m is an odd number, then f(m) has to be even.

And so, as long as m is an integer, f(m) is even.

You see why I like that argument better. It’s shorter. It breaks things up into fewer cases. None of those cases have to worry about whether m is positive or negative or zero. Each of the cases is short, and moves straight to its goal. This is the proof I’d be happy submitting. Today, anyway. No telling what tomorrow will make me think.

## Someone Else’s Homework: A Solution

I have a friend who’s been taking mathematical logic. While talking over the past week’s work they mentioned a problem that had stumped them. But they’d figured it out — at least the critical part — about a half-hour after turning it in. And I had fun going over it. Since the assignment’s already turned in and I don’t even know which class it was, I’d like to share it with you.

So here’s the problem. Suppose you have a function f, with domain of the integers Z and rage of the integers Z. And also you know that f has the property that for any two integers ‘a’ and ‘b’, f(a + b) equals f(a) + f(b). And finally, suppose that for some odd number ‘c’, you know that f(c) is even. The challenge: prove that f is even for all the integers.

If you want to take a moment to think about that, please do.

First thing I want to do is show that f(1) is an even number. How? Well, if ‘c’ is an odd number, then ‘c’ has to equal ‘2*k + 1’ for some integer ‘k’. So f(c) = f(2*k + 1). And therefore f(c) = f(2*k) + f(1). And, since 2*k is equal to k + k, then f(2*k) has to equal f(k) + f(k). Therefore f(c) = 2*f(k) + f(1). Whatever f(k) is, 2*f(k) has to be an even number. And we’re given f(c) is even. Therefore f(1) has to be even.

Now I can prove that if ‘k’ is any positive integer, then f(k) has to be even. Why? Because ‘k’ is equal to 1 + 1 + 1 + … + 1. And so f(k) has to equal f(1) + f(1) + f(1) + … + f(1). That is, it’s k * f(1). And if f(1) is even then so is k * f(1). So that covers the positive integers.

How about zero? Can I show that f(0) is even? Oh, sure, easy. Start with ‘c’. ‘c’ equals ‘c + 0’. So f(c) = f(c) + f(0). The only way that’s going to be true is if f(0) is equal to zero, which is an even number.

By the way, here’s an alternate way of arguing this: 0 = 0 + 0. So f(0) = f(0) + f(0). And therefore f(0) = 2 * f(0) and that’s an even number. Incidentally also zero. Submit the proof you like.

What’s not covered yet? Negative integers. It’s hard not to figure, well, we know f(1) is even, we know f(a + b) if f(a) + f(b). Shouldn’t, like, f(-2) just be -2 * f(1)? Oh, it so should. I don’t feel like we have that already proven, though. So let me nail that down. I’m going to use what we know about f(k) for positive ‘k’, and the fact that f(0) is 0.

So give me any negative integer; I’m going call it ‘-k’. Its additive inverse is ‘k’, which is a positive number. -k + k = 0. And so f(-k + k) = f(-k) + f(k) = f(0). So, f(-k) + f(k) = 0, and f(-k) = -f(k). If f(k) is even — and it is — then f(-k) is also even.

So there we go: whether ‘k’ is a positive, zero, or negative integer, f(k) is even. All the integers are either positive, zero, or negative. So f is even for any integer.

## There’s Technically Still Time To Buy A Theorem For Valentine’s Day

While I have not used the service myself, it does appear that Theory Mine is still going. It’s an automated theorem-creating software. For a price, they’ll create a theorem and name it whatever you choose within reason. It will almost certainly not be an interesting theorem, nor one that anyone will ever care about. But it’ll be far more legitimate than, like, naming a star after someone, which has always been an outright scam.

I discovered this last year, and wrote a bit about how this sort of thing can work. (I’m not certain this is precisely how Theory Mine works, but I am confident that it’s something along these lines.) Also a bit about the history of this sort of system and how it’s come about. And as I say, I haven’t used the service myself. It may sound like bragging, but I’ve created my own theorems. They’re not the monumental triumph of intellect and explanatory power that attaches to theories in science. They can be much more petty things, like “when can we expect the sum of the roots of a quadratic polynomial to be greater than zero”. A theorem you make for your own project will be a little more interesting than a completely auto-generated one like this. After all, your own theorem will at least answer something you wanted to know. A computer-generated one doesn’t even promise that. But it does take less effort to send a bit of money off and get a proof mailed back to you.

## Reading the Comics, June 17, 2017: Icons Of Mathematics Edition

Comic Strip Master Command just barely missed being busy enough for me to split the week’s edition. Fine for them, I suppose, although it means I’m going to have to scramble together something for the Tuesday or the Thursday posting slot. Ah well. As befits the comics, there’s a fair bit of mathematics as an icon in the past week’s selections. So let’s discuss.

Mark Anderson’s Andertoons for the 11th is our Mark Anderson’s Andertoons for this essay. Kind of a relief to have that in right away. And while the cartoon shows a real disaster of a student at the chalkboard, there is some truth to the caption. Ruling out plausible-looking wrong answers is progress, usually. So is coming up with plausible-looking answers to work out whether they’re right or wrong. The troubling part here, I’d say, is that the kid came up with pretty poor guesses about what the answer might be. He ought to be able to guess that it’s got to be an odd number, and has to be less than 10, and really ought to be less than 7. If you spot that then you can’t make more than two wrong guesses.

Patrick J Marrin’s Francis for the 12th starts with what sounds like a logical paradox, about whether the Pope could make an infallibly true statement that he was not infallible. Really it sounds like a bit of nonsense. But the limits of what we can know about a logical system will often involve questions of this form. We ask whether something can prove whether it is provable, for example, and come up with a rigorous answer. So that’s the mathematical content which justifies my including this strip here.

Niklas Eriksson’s Carpe Diem for the 13th is a traditional use of the blackboard full of mathematics as symbolic of intelligence. Of course ‘E = mc2‘ gets in there. I’m surprised that both π and 3.14 do, too, for as little as we see on the board.

Mark Anderson’s Andertoons for the 14th is a nice bit of reassurance. Maybe the cartoonist was worried this would be a split-week edition. The kid seems to be the same one as the 11th, but the teacher looks different. Anyway there’s a lot you can tell about shapes from their perimeter alone. The one which most startles me comes up in calculus: by doing the right calculation about the lengths and directions of the edge of a shape you can tell how much area is inside the shape. There’s a lot of stuff in this field — multivariable calculus — that’s about swapping between “stuff you know about the boundary of a shape” and “stuff you know about the interior of the shape”. And finding area from tracing the boundary is one of them. It’s still glorious.

Samson’s Dark Side Of The Horse for the 14th is a counting-sheep joke and a Pi Day joke. I suspect the digits of π would be horrible for lulling one to sleep, though. They lack the just-enough-order that something needs for a semiconscious mind to drift off. Horace would probably be better off working out Collatz sequences.

Dana Simpson’s Phoebe and her Unicorn for the 14th mentions mathematics as iconic of what you do at school. Book reports also make the cut.

Dan Barry’s Flash Gordon for the 31st of July, 1962 and rerun the 16th I’m including just because I love the old-fashioned image of a mathematician in Professor Quita here. At this point in the comic strip’s run it was set in the far-distant future year of 1972, and the action here is on one of the busy multinational giant space stations. Flash himself is just back from Venus where he’d set up some dolphins as assistants to a fish-farming operation helping to feed that world and ours. And for all that early-60s futurism look at that gorgeous old adding machine he’s still got. (Professor Quinta’s discovery is a way to peer into alternate universes, according to the next day’s strip. I’m kind of hoping this means they’re going to spend a week reading Buck Rogers.)

## Reading the Comics, June 3, 2017: Feast Week Conclusion Edition

And now finally I can close out last week’s many mathematically-themed comic strips. I had hoped to post this Thursday, but the Why Stuff Can Orbit supplemental took up my writing energies and eventually timeslot. This also ends up being the first time I’ve had one of Joe Martin’s comic strips since the Houston Chronicle ended its comics pages and I admit I’m not sure how I’m going to work this. I’m also not perfectly sure what the comic strip means.

So Joe Martin’s Mister Boffo for the 1st of June seems to be about a disastrous mathematics exam with a kid bad enough he hasn’t even got numbers exactly to express the score. Also I’m not sure there is a way to link to the strip I mean exactly; the archives for Martin’s strips are not … organized the way I would have done. Well, they’re his business.

Greg Evans’s Luann Againn for the 1st reruns the strip from the 1st of June, 1989. It’s your standard resisting-the-word-problem joke. On first reading the strip I didn’t get what the problem was asking for, and supposed that the text had garbled the problem, if there were an original problem. That was my sloppiness is all; it’s a perfectly solvable question once you actually read it.

J C Duffy’s Lug Nuts for the 1st — another day that threatened to be a Reading the Comics post all on its own — is a straggler Pi Day joke. It’s just some Dadaist clowning about.

Doug Bratton’s Pop Culture Shock Therapy for the 1st is a wordplay joke that uses word problems as emblematic of mathematics. I’m okay with that; much of the mathematics that people actually want to do amounts to extracting from a situation the things that are relevant and forming an equation based on that. This is what a word problem is supposed to teach us to do.

Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 1st — maybe I should have done a Reading the Comics for that day alone — riffs on the idle speculation that God would be a mathematician. It does this by showing a God uninterested in two logical problems. The first is the question of whether there’s an odd perfect number. Perfect numbers are these things that haunt number theory. (Everything haunts number theory.) It starts with idly noticing what happens if you pick a number, find the numbers that divide into it, and add those up. For example, 4 can be divided by 1 and 2; those add to 3. 5 can only be divided by 1; that adds to 1. 6 can be divided by 1, 2, and 3; those add to 6. For a perfect number the divisors add up to the original number. Perfect numbers look rare; for a thousand years or so only four of them (6, 28, 496, and 8128) were known to exist.

All the perfect numbers we know of are even. More, they’re all numbers that can be written as the product $2^{p - 1} \cdot \left(2^p - 1\right)$ for certain prime numbers ‘p’. (They’re the ones for which $2^p - 1$ is itself a prime number.) What we don’t know, and haven’t got a hint about proving, is whether there are any odd prime numbers. We know some things about odd perfect numbers, if they exist, the most notable of them being that they’ve got to be incredibly huge numbers, much larger than a googol, the standard idea of an incredibly huge number. Presumably an omniscient God would be able to tell whether there were an odd perfect number, or at least would be able to care whether there were. (It’s also not known if there are infinitely many perfect numbers, by the way. This reminds us that number theory is pretty much nothing but a bunch of easy-to-state problems that we can’t solve.)

Some miscellaneous other things we know about an odd perfect number, other than whether any exist: if there are odd perfect numbers, they’re not divisible by 105. They’re equal to one more than a whole multiple of 12. They’re also 117 more than a whole multiple of 468, and they’re 81 more than a whole multiple of 324. They’ve got to have at least 101 prime factors, and there have to be at least ten distinct prime factors. There have to be at least twelve distinct prime factors if 3 isn’t a factor of the odd perfect number. If this seems like a screwy list of things to know about a thing we don’t even know exists, then welcome to number theory.

The beard question I believe is a reference to the logician’s paradox. This is the one postulating a village in which the village barber shaves all, but only, the people who do not shave themselves. Given that, who shaves the barber? It’s an old joke, but if you take it seriously you learn something about the limits of what a system of logic can tell you about itself.

Bud Blake’s Tiger rerun for the 2nd has Tiger’s arithmetic homework spill out into real life. This happens sometimes.

George Herriman’s Krazy Kat for the 10th of July, 1939 was rerun the 2nd of June. I’m not sure that it properly fits here, but the talk about Officer Pupp running at 60 miles per hour and Ignatz Mouse running forty and whether Pupp will catch Mouse sure reads like a word problem. Later strips in the sequence, including the ways that a tossed brick could hit someone who’d be running faster than it, did not change my mind about this. Plus I like Krazy Kat so I’ll take a flimsy excuse to feature it.

## Reading the Comics, April 22, 2017: Thought There’d Be Some More Last Week Edition

Allison Barrows’s PreTeena rerun for the 18th is a classic syllogism put into the comic strip’s terms. The thing about these sorts of deductive-logic syllogisms is that whether the argument is valid depends only on the shape of the argument. It has nothing to do with whether the thing being discussed makes any sense. This can be disorienting. It’s hard to ignore the everyday meaning of words when you hear a string of sentences. But it’s also hard to parse a string of sentences if the words don’t make sense in them. This is probably part of why on the mathematics side of things logic courses will skimp on syllogisms, using them to give an antique flavor and sense of style to the introduction of courses. It’s easier to use symbolic representations for logic instead.

Randy Glasbergen’s Glasbergen Cartoons rerun for the 20th is the old joke about arithmetic being different between school, government, and corporate work. I haven’t looked at the comments — the GoComics redesign, whatever else it does, makes it very easy to skip the comments — but I’m guessing by the second one someone’s said the Common Core method means getting the most wrong answer.

Bil Keane and Jeff Keane’s Family Circus for the 21st I don’t know is a rerun. But a lot of them are these days. Anyway, it looks like a silly joke about how nice mathematics would be without numbers; Dolly has no idea. I can sympathize with being intimidated by numerals. At the risk of being all New Math-y, I wonder if she wouldn’t like arithmetic more if it were presented as a game. Like, here’s a couple symbols — let’s say * and | for a start, and then some rules. * and * makes *, but * and | makes |. Also | and * makes |. But | and | makes |*. And so on. This is binary arithmetic, disguised, but I wonder if making it look like something inconsequential would make it more pleasant to learn, and if that would transfer over to arithmetic with 1’s and 0’s. Normal, useful arithmetic would be harder to play like this. You’d need ten symbols that are easy to write that aren’t already numbers, letters, or common symbols. But I wonder if it’d be worth it.

Tom Thaves’s Frank and Ernest for the 22nd is provided for mathematics teachers who need something to tape to their door. You’re welcome.

## Words About A Wordless Induction Proof

This pair of tweets came across my feed. And who doesn’t like a good visual proof of a mathematical fact? I hope you enjoy.

So here’s the proposition.

This is the sort of identity we normally try proving by induction. Induction is a great scheme for proving identities like this. It works by finding some index on the formula. Then show that if the formula is true for one value of the index, then it’s true for the next-higher value of the index. Finally, find some value of the index for which it’s easy to check that the formula’s true. And that proves it’s true for all the values of that index above that base.

In this case the index is ‘n’. It’s really easy to prove the base case, since 13 is equal to 12 what with ‘1’ being the number everybody likes to raise to powers. Going from proving that if it’s true in one case — $1^3 + 2^3 + 3^3 + \cdots + n^3$ — then it’s true for the next — $1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3$ — is work. But you can get it done.

And then there’s this, done visually:

It took me a bit to read fully until I was confident in what it was showing. But it is all there.

As often happens with these wordless proofs you can ask whether it is properly speaking a proof. A proof is an argument and to be complete it has to contain every step needed to deduce the conclusion from the premises, following one of the rules of inference each step. Thing is basically no proof is complete that way, because it takes forever. We elide stuff that seems obvious, confident that if we had to we could fill in the intermediate steps. A wordless proof like trusts that if we try to describe what is in the picture then we are constructing the argument.

That’s surely enough of my words.

## One Way To Get Your Own Theorem

While doing some research to better grouse about Ken Keeler’s Futurama theorem I ran across an amusing site I hadn’t known about. It is Theory Mine, a site that allows you to hire — and name — a genuine, mathematically sound theorem. The spirit of the thing is akin to that scam in which you “name” a star. But this is more legitimate in that, you know, it’s got any legitimacy. For this, you’re buying naming rights from someone who has any rights to sell. By convention the discoverer of a theorem can name it whatever she wishes, and there’s one chance in ten that anyone else will use the name.

I haven’t used it. I’ve made my own theorems, thanks, and could put them on a coffee mug or t-shirt if I wished to make a particularly boring t-shirt. But I’m delighted by the scheme. They don’t have a team of freelance mathematicians whipping up stuff and hoping it isn’t already known. Not for the kinds of prices they charge. This should inspire the question: well, where do the theorems come from?

The scheme uses an automated reasoning system. I don’t know the details of how it works, but I can think of a system by which this might work. It goes back to the Crisis of Foundations, the time in the late 19th/early 20th century when logicians got very worried that we were still letting physical intuitions and unstated assumptions stay in our mathematics. One solution: turn everything into symbols, icons with no connotations. The axioms of mathematics become a couple basic symbols. The laws of logical deduction become things we can do with the symbols, converting one line of symbols into a related other line. Every line we get is a theorem. And we know it’s correct. To write out the theorem in this scheme is to write out its proof, and to feel like you’re touching some deep magic. And there’s no human frailties in the system, besides the thrill of reeling off True Names like that.

You may not be sure what this works like. It may help to compare it to a slightly-fun number coding scheme. I mean the one where you start with a number, like, ‘1’. Then you write down how many times and which digit appears. There’s a single ‘1’ in that string, so you would write down ’11’. And repeat: In ’11’ there’s a sequence of two ‘1’s, so you would write down ’21’. And repeat: there’s a single ‘2’ and a single ‘1’, so you then write down ‘1211’. And again: there’s a single ‘1’, a single ‘2’, and then a double ‘1’, so you next write ‘111221’. And so on until you get bored or die.

When we do this for mathematics we start with a couple different basic units. And we also start with several things we may do at most symbols. So there’s rarely a single line that follows from the previous. There’s an ever-expanding tree of known truths. This may stave off boredom but I make no promises about death.

The result of this is pages and pages that look like Ancient High Martian. I don’t feel the thrill of doing this. Some people do, though. And as recreational mathematics goes I suppose it’s at least as good as sudoku. Anyway, this kind of project, rewarding indefatigability and thoroughness, is perfect for automation anyway. Let the computer work out all the things we can prove are true.

If I’m reading Theory Mine’s description correctly they seem to be doing something roughly like this. If they’re not, well, you go ahead and make your own rival service using my paragraphs as your system. All I ask is one penny for every use of L’Hôpital’s Rule, a theorem named for Guillaume de l’Hôpital and discovered by Johann Bernoulli. (I have heard that Bernoulli was paid for his work, but I do not know that’s true. I have now explained why, if we suppose that to be true, my prior sentence is a very funny joke and you should at minimum chuckle.)

This should inspire the question: what do we need mathematicians for, then? It’s for the same reason we need writers, when it would be possible to automate the composing of sentences that satisfy the rules of English grammar. I mean if there were rules to English grammar. That we can identify a theorem that’s true does not mean it has even the slightest interest to anyone, ever. There’s much more that could be known than that we could ever care about.

You can see this in Theory Mine’s example of Quentin’s Theorem. Quentin’s Theorem is about an operation you can do on a set whose elements consist of the non-negative whole numbers with a separate value, which they call color, attached. You can add these colored-numbers together according to some particular rules about how the values and the colors add. The order of this addition normally matters: blue two plus green three isn’t the same as green three plus blue two. Quentin’s Theorem finds cases where, if you add enough colored-numbers together, the order doesn’t matter. I know. I am also staggered by how useful this fact promises to be.

Yeah, maybe there is some use. I don’t know what it is. If anyone’s going to find the use it’ll be a mathematician. Or a physicist who’s found some bizarre quark properties she wants to codify. Anyway, if what you’re interested in is “what can you do to make a vertical column stable?” then the automatic proof generator isn’t helping you at all. Not without a lot of work put in to guiding it. So we can skip the hard work of finding and proving theorems, if we can do the hard work of figuring out where to look for these theorems instead. Always the way.

You also may wonder how we know the computer is doing its work right. It’s possible to write software that is logically proven to be correct. That is, the software can’t produce anything but the designed behavior. We don’t usually write software this way. It’s harder to write, because you have to actually design your software’s behavior. And we can get away without doing it. Usually there’s some human overseeing the results who can say what to do if the software seems to be going wrong. Advocates of logically-proven software point out that we’re getting more software, often passing results on to other programs. This can turn a bug in one program into a bug in the whole world faster than a responsible human can say, “I dunno. Did you try turning it off and on again?” I’d like to think we could get more logically-proven software. But I also fear I couldn’t write software that sound and, you know, mathematics blogging isn’t earning me enough to eat on.

Also, yes, even proven software will malfunction if the hardware the computer’s on malfunctions. That’s rare, but does happen. Fortunately, it’s possible to automate the checking of a proof, and that’s easier to do than creating a proof in the first place. We just have to prove we have the proof-checker working. Certainty would be a nice thing if we ever got it, I suppose.

## Reading the Comics, January 7, 2016: Just Before GoComics Breaks Everything Edition

Most of the comics I review here are printed on GoComics.com. Well, most of the comics I read online are from there. But even so I think they have more comic strips that mention mathematical themes. Anyway, they’re unleashing a complete web site redesign on Monday. I don’t know just what the final version will look like. I know that the beta versions included the incredibly useful, that is to say dumb, feature where if a particular comic you do read doesn’t have an update for the day — and many of them don’t, as they’re weekly or three-times-a-week or so — then it’ll show some other comic in its place. I mean, the idea of encouraging people to find new comics is a good one. To some extent that’s what I do here. But the beta made no distinction between “comic you don’t read because you never heard of Microcosm” and “comic you don’t read because glancing at it makes your eyes bleed”. And on an idiosyncratic note, I read a lot of comics. I don’t need to see Dude and Dude reruns in fourteen spots on my daily comics page, even if I didn’t mind it to start.

Anyway. I am hoping, desperately hoping, that with the new site all my old links to comics are going to keep working. If they don’t then I suppose I’m just ruined. We’ll see. My suggestion is if you’re at all curious about the comics you read them today (Sunday) just to be safe.

Ashleigh Brilliant’s Pot-Shots is a curious little strip I never knew of until GoComics picked it up a few years ago. Its format is compellingly simple: a little illustration alongside a wry, often despairing, caption. I love it, but I also understand why was the subject of endless queries to the Detroit Free Press (Or Whatever) about why was this thing taking up newspaper space. The strip rerun the 31st of December is a typical example of the strip and amuses me at least. And it uses arithmetic as the way to communicate reasoning, both good and bad. Brilliant’s joke does address something that logicians have to face, too. Whether an argument is logically valid depends entirely on its structure. If the form is correct the reasoning may be excellent. But to be sound an argument has to be correct and must also have its assumptions be true. We can separate whether an argument is right from whether it could ever possibly be right. If you don’t see the value in that, you have never participated in an online debate about where James T Kirk was born and whether Spock was the first Vulcan in Star Fleet.

Thom Bluemel’s Birdbrains for the 2nd of January, 2017, is a loaded-dice joke. Is this truly mathematics? Statistics, at least? Close enough for the start of the year, I suppose. Working out whether a die is loaded is one of the things any gambler would like to know, and that mathematicians might be called upon to identify or exploit. (I had a grandmother unshakably convinced that I would have some natural ability to beat the Atlantic City casinos if she could only sneak the underaged me in. I doubt I could do anything of value there besides see the stage magic show.)

Jack Pullan’s Boomerangs rerun for the 2nd is built on the one bit of statistical mechanics that everybody knows, that something or other about entropy always increasing. It’s not a quantum mechanics rule, but it’s a natural confusion. Quantum mechanics has the reputation as the source of all the most solid, irrefutable laws of the universe’s working. Statistical mechanics and thermodynamics have this musty odor of 19th-century steam engines, no matter how much there is to learn from there. Anyway, the collapse of systems into disorder is not an irrevocable thing. It takes only energy or luck to overcome disorderliness. And in many cases we can substitute time for luck.

Scott Hilburn’s The Argyle Sweater for the 3rd is the anthropomorphic-geometry-figure joke that’s I’ve been waiting for. I had thought Hilburn did this all the time, although a quick review of Reading the Comics posts suggests he’s been more about anthropomorphic numerals the past year. This is why I log even the boring strips: you never know when I’ll need to check the last time Scott Hilburn used “acute” to mean “cute” in reference to triangles.

Mike Thompson’s Grand Avenue uses some arithmetic as the visual cue for “any old kind of schoolwork, really”. Steve Breen’s name seems to have gone entirely from the comic strip. On Usenet group rec.arts.comics.strips Brian Henke found that Breen’s name hasn’t actually been on the comic strip since May, and D D Degg found a July 2014 interview indicating Thompson had mostly taken the strip over from originator Breen.

Mark Anderson’s Andertoons for the 5th is another name-drop that doesn’t have any real mathematics content. But come on, we’re talking Andertoons here. If I skipped it the world might end or something untoward like that.

Ted Shearer’s Quincy for the 14th of November, 1977, doesn’t have any mathematical content really. Just a mention. But I need some kind of visual appeal for this essay and Shearer is usually good for that.

Corey Pandolph, Phil Frank, and Joe Troise’s The Elderberries rerun for the 7th is also a very marginal mention. But, what the heck, it’s got some of your standard wordplay about angles and it’ll get this week’s essay that much closer to 800 words.

## The End 2016 Mathematics A To Z: Zermelo-Fraenkel Axioms

gaurish gave me a choice for the Z-term to finish off the End 2016 A To Z. I appreciate it. I’m picking the more abstract thing because I’m not sure that I can explain zero briefly. The foundations of mathematics are a lot easier.

## Zermelo-Fraenkel Axioms

I remember the look on my father’s face when I asked if he’d tell me what he knew about sets. He misheard what I was asking about. When we had that straightened out my father admitted that he didn’t know anything particular. I thanked him and went off disappointed. In hindsight, I kind of understand why everyone treated me like that in middle school.

My father’s always quick to dismiss how much mathematics he knows, or could understand. It’s a common habit. But in this case he was probably right. I knew a bit about set theory as a kid because I came to mathematics late in the “New Math” wave. Sets were seen as fundamental to why mathematics worked without being so exotic that kids couldn’t understand them. Perhaps so; both my love and I delighted in what we got of set theory as kids. But if you grew up before that stuff was popular you probably had a vague, intuitive, and imprecise idea of what sets were. Mathematicians had only a vague, intuitive, and imprecise idea of what sets were through to the late 19th century.

And then came what mathematics majors hear of as the Crisis of Foundations. (Or a similar name, like Foundational Crisis. I suspect there are dialect differences here.) It reflected mathematics taking seriously one of its ideals: that everything in it could be deduced from clearly stated axioms and definitions using logically rigorous arguments. As often happens, taking one’s ideals seriously produces great turmoil and strife.

Before about 1900 we could get away with saying that a set was a bunch of things which all satisfied some description. That’s how I would describe it to a new acquaintance if I didn’t want to be treated like I was in middle school. The definition is fine if we don’t look at it too hard. “The set of all roots of this polynomial”. “The set of all rectangles with area 2”. “The set of all animals with four-fingered front paws”. “The set of all houses in Central New Jersey that are yellow”. That’s all fine.

And then if we try to be logically rigorous we get problems. We always did, though. They’re embodied by ancient jokes like the person from Crete who declared that all Cretans always lie; is the statement true? Or the slightly less ancient joke about the barber who shaves only the men who do not shave themselves; does he shave himself? If not jokes these should at least be puzzles faced in fairy-tale quests. Logicians dressed this up some. Bertrand Russell gave us the quite respectable “The set consisting of all sets which are not members of themselves”, and asked us to stare hard into that set. To this we have only one logical response, which is to shout, “Look at that big, distracting thing!” and run away. This satisfies the problem only for a while.

The while ended in — well, that took a while too. But between 1908 and the early 1920s Ernst Zermelo, Abraham Fraenkel, and Thoralf Skolem paused from arguing whose name would also be the best indie rock band name long enough to put set theory right. Their structure is known as Zermelo-Fraenkel Set Theory, or ZF. It gives us a reliable base for set theory that avoids any contradictions or catastrophic pitfalls. Or does so far as we have found in a century of work.

It’s built on a set of axioms, of course. Most of them are uncontroversial, things like declaring two sets are equivalent if they have the same elements. Declaring that the union of sets is itself a set. Obvious, sure, but it’s the obvious things that we have to make axioms. Maybe you could start an argument about whether we should just assume there exists some infinitely large set. But if we’re aware sets probably have something to teach us about numbers, and that numbers can get infinitely large, then it seems fair to suppose that there must be some infinitely large set. The axioms that aren’t simple obvious things like that are too useful to do without. They assume stuff like that no set is an element of itself. Or that every set has a “power set”, a new set comprised of all the subsets of the original set. Good stuff to know.

There is one axiom that’s controversial. Not controversial the way Euclid’s Parallel Postulate was. That’s ugly one about lines crossing another line meeting on the same side they make angles smaller than something something or other. That axiom was controversial because it read so weird, so needlessly complicated. (It isn’t; it’s exactly as complicated as it must be. Or better, it’s as simple as it could possibly be and still be useful.) The controversial axiom of Zermelo-Fraenkel Set Theory is known as the Axiom of Choice. It says if we have a collection of mutually disjoint sets, each with at least one thing in them, then it’s possible to pick exactly one item from each of the sets.

It’s impossible to dispute this is what we have axioms for. It’s about something that feels like it should be obvious: we can always pick something from a set. How could this not be true?

If it is true, though, we get some unsavory conclusions. For example, it becomes possible to take a ball the size of an orange and slice it up. We slice using mathematical blades. They’re not halted by something as petty as the desire not to slice atoms down the middle. We can reassemble the pieces. Into two balls. And worse, it doesn’t require we do something like cut the orange into infinitely many pieces. We expect crazy things to happen when we let infinities get involved. No, though, we can do this cut-and-duplicate thing by cutting the orange into five pieces. When you hear that it’s hard to know whether to point to the big, distracting thing and run away. If we dump the Axiom of Choice we don’t have that problem. But can we do anything useful without the ability to make a choice like that?

And we’ve learned that we can. If we want to use the Zermelo-Fraenkel Set Theory with the Axiom of Choice we say we were working in “ZFC”, Zermelo-Fraenkel-with-Choice. We don’t have to. If we don’t want to make any assumption about choices we say we’re working in “ZF”. Which to use depends on what one wants to use.

Either way Zermelo and Fraenkel and Skolem established set theory on the foundation we use to this day. We’re not required to use them, no; there’s a construction called von Neumann-Bernays-Gödel Set Theory that’s supposed to be more elegant. They didn’t mention it in my logic classes that I remember, though.

And still there’s important stuff we would like to know which even ZFC can’t answer. The most famous of these is the continuum hypothesis. Everyone knows — excuse me. That’s wrong. Everyone who would be reading a pop mathematics blog knows there are different-sized infinitely-large sets. And knows that the set of integers is smaller than the set of real numbers. The question is: is there a set bigger than the integers yet smaller than the real numbers? The Continuum Hypothesis says there is not.

Zermelo-Fraenkel Set Theory, even though it’s all about the properties of sets, can’t tell us if the Continuum Hypothesis is true. But that’s all right; it can’t tell us if it’s false, either. Whether the Continuum Hypothesis is true or false stands independent of the rest of the theory. We can assume whichever state is more useful for our work.

Back to the ideals of mathematics. One question that produced the Crisis of Foundations was consistency. How do we know our axioms don’t contain a contradiction? It’s hard to say. Typically a set of axioms we can prove consistent are also a set too boring to do anything useful in. Zermelo-Fraenkel Set Theory, with or without the Axiom of Choice, has a lot of interesting results. Do we know the axioms are consistent?

No, not yet. We know some of the axioms are mutually consistent, at least. And we have some results which, if true, would prove the axioms to be consistent. We don’t know if they’re true. Mathematicians are generally confident that these axioms are consistent. Mostly on the grounds that if there were a problem something would have turned up by now. It’s withstood all the obvious faults. But the universe is vaster than we imagine. We could be wrong.

It’s hard to live up to our ideals. After a generation of valiant struggling we settle into hoping we’re doing good enough. And waiting for some brilliant mind that can get us a bit closer to what we ought to be.

## The End 2016 Mathematics A To Z: Cantor’s Middle Third

Today’s term is a request, the first of this series. It comes from HowardAt58, head of the Saving School Math blog. There are many letters not yet claimed; if you have a term you’d like to see my write about please head over to the “Any Requests?” page and pick a letter. Please not one I figure to get to in the next day or two.

## Cantor’s Middle Third.

I think one could make a defensible history of mathematics by describing it as a series of ridiculous things that get discovered. And then, by thinking about these ridiculous things long enough, mathematicians come to accept them. Even rely on them. Sometime later the public even comes to accept them. I don’t mean to say getting people to accept ridiculous things is the point of mathematics. But there is a pattern which happens.

Consider. People doing mathematics came to see how a number could be detached from a count or a measure of things. That we can do work on, say, “three” whether it’s three people, three kilograms, or three square meters. We’re so used to this it’s only when we try teaching mathematics to the young we realize it isn’t obvious.

Or consider that we can have, rather than a whole number of things, a fraction. Some part of a thing, as if you could have one-half pieces of chalk or two-thirds a fruit. Counting is relatively obvious; fractions are something novel but important.

We have “zero”; somehow, the lack of something is still a number, the way two or five or one-half might be. For that matter, “one” is a number. How can something that isn’t numerous be a number? We’re used to it anyway. We can have not just fraction and one and zero but irrational numbers, ones that can’t be represented as a fraction. We have negative numbers, somehow a lack of whatever we were counting so great that we might add some of what we were counting to the pile and still have nothing.

That takes us up to about eight hundred years ago or something like that. The public’s gotten to accept all this as recently as maybe three hundred years ago. They’ve still got doubts. I don’t blame folks. Complex numbers mathematicians like; the public’s still getting used to the idea, but at least they’ve heard of them.

Cantor’s Middle Third is part of the current edge. It’s something mathematicians are aware of and that defies sense at least. But we’ve come to accept it. The public, well, they don’t know about it. Maybe some do; it turns up in pop mathematics books that like sharing the strangeness of infinities. Few people read them. Sometimes it feels like all those who do go online to tell mathematicians they’re crazy. It comes to us, as you might guess from the name, from Georg Cantor. Cantor established the modern mathematical concept of how to study infinitely large sets in the late 19th century. And he was repeatedly hospitalized for depression. It’s cruel to write all that off as “and he was crazy”. His work’s withstood a hundred and thirty-five years of extremely smart people looking at it skeptically.

The Middle Third starts out easily enough. Take a line segment. Then chop it into three equal pieces and throw away the middle third. You see where the name comes from. What do you have left? Some of the original line. Two-thirds of the original line length. A big gap in the middle.

Now take the two line segments. Chop each of them into three equal pieces. Throw away the middle thirds of the two pieces. Now we’re left with four chunks of line and four-ninths of the original length. One big and two little gaps in the middle.

Now take the four little line segments. Chop each of them into three equal pieces. Throw away the middle thirds of the four pieces. We’re left with eight chunks of line, about eight-twenty-sevenths of the original length. Lots of little gaps. Keep doing this, chopping up line segments and throwing away middle pieces. Never stop. Well, pretend you never stop and imagine what’s left.

What’s left is deeply weird. What’s left has no length, no measure. That’s easy enough to prove. But we haven’t thrown everything away. There are bits of the original line segment left over. The left endpoint of the original line is left behind. So is the right endpoint of the original line. The endpoints of the line segments after the first time we chopped out a third? Those are left behind. The endpoints of the line segments after chopping out a third the second time, the third time? Those have to be in the set. We have a dust, isolated little spots of the original line, none of them combining together to cover any length. And there are infinitely many of these isolated dots.

We’ve seen that before. At least we have if we’ve read anything about the Cantor Diagonal Argument. You can find that among the first ten posts of every mathematics blog. (Not this one. I was saving the subject until I had something good to say about it. Then I realized many bloggers have covered it better than I could.) Part of it is pondering how there can be a set of infinitely many things that don’t cover any length. The whole numbers are such a set and it seems reasonable they don’t cover any length. The rational numbers, though, are also an infinitely-large set that doesn’t cover any length. And there’s exactly as many rational numbers as there are whole numbers. This is unsettling but if you’re the sort of person who reads about infinities you come to accept it. Or you get into arguments with mathematicians online and never know you’ve lost.

Here’s where things get weird. How many bits of dust are there in this middle third set? It seems like it should be countable, the same size as the whole numbers. After all, we pick up some of these points every time we throw away a middle third. So we double the number of points left behind every time we throw away a middle third. That’s countable, right?

It’s not. We can prove it. The proof looks uncannily like that of the Cantor Diagonal Argument. That’s the one that proves there are more real numbers than there are whole numbers. There are points in this leftover set that were not endpoints of any of these middle-third excerpts. This dust has more points in it than there are rational numbers, but it covers no length.

(I don’t know if the dust has the same size as the real numbers. I suspect it’s unproved whether it has or hasn’t, because otherwise I’d surely be able to find the answer easily.)

It’s got other neat properties. It’s a fractal, which is why someone might have heard of it, back in the Great Fractal Land Rush of the 80s and 90s. Look closely at part of this set and it looks like the original set, with bits of dust edging gaps of bigger and smaller sizes. It’s got a fractal dimension, or “Hausdorff dimension” in the lingo, that’s the logarithm of two divided by the logarithm of three. That’s a number actually known to be transcendental, which is reassuring. Nearly all numbers are transcendental, but we only know a few examples of them.

HowardAt58 asked me about the Middle Third set, and that’s how I’ve referred to it here. It’s more often called the “Cantor set” or “Cantor comb”. The “comb” makes sense because if you draw successive middle-thirds-thrown-away, one after the other, you get something that looks kind of like a hair comb, if you squint.

You can build sets like this that aren’t based around thirds. You can, for example, develop one by cutting lines into five chunks and throw away the second and fourth. You get results that are similar, and similarly heady, but different. They’re all astounding. They’re all hard to believe in yet. They may get to be stuff we just accept as part of how mathematics works.

## Reading the Comics, October 8, 2016: Split Week Edition Part 2

And now I can finish off last week’s comics. It was a busy week. The first few days of this week have been pretty busy too. Meanwhile, Dave Kingsbury has recently read a biography of Lewis Carroll, and been inspired to form a haiku/tanka project. You might enjoy.

Susan Camilleri Konar is a new cartoonist for the Six Chix collective. Her first strip to get mentioned around these parts is from the 5th. It’s a casual mention of the Fibonacci sequence, which is one of the few sequences that a normal audience would recognize as something going on forever. And yes, I noticed the spiral in the background. That’s one of the common visual representations of the Fibonacci sequence: it starts from the center. The rectangles inside have dimensions 1 by 2, then 2 by 3, then 3 by 5, then 5 by 8, and so on; the spiral connects vertices of these rectangles. It’s an attractive spiral and you can derive the overrated Golden Ratio from the dimensions of larger rectangles. This doesn’t make the Golden Ratio important or anything, but it is there.

Ryan North’s Dinosaur Comics for the 6th is part of a story about T-Rex looking for certain truth. Mathematics could hardly avoid coming up. And it does offer what look like universal truths: given the way deductive logic works, and some starting axioms, various things must follow. “1 + 1 = 2” is among them. But there are limits to how much that tells us. If we accept the rules of Monopoly, then owning four railroads means the rent for landing on one is a game-useful \$200. But if nobody around you cares about Monopoly, so what? And so it is with mathematics. Utahraptor and Dromiceiomimus point out that the mathematics we know is built on premises we have selected because we find them interesting or useful. We can’t know that the mathematics we’ve deduced has any particular relevance to reality. Indeed, it’s worse than North points out: How do we know whether an argument is valid? Because we believe that its conclusions follow from its premises according to our rules of deduction. We rely on our possibly deceptive senses to tell us what the argument even was. We rely on a mind possibly upset by an undigested bit of beef, a crumb of cheese, or a fragment of an underdone potato to tell us the rules are satisfied. Mathematics seems to offer us absolute truths, but it’s hard to see how we can get there.

Rick Stromoskis Soup to Nutz for the 6th has a mathematics cameo in a student-resisting-class-questions problem. But the teacher’s question is related to the figure that made my first fame around these parts.

Mark Anderson’s Andertoons for the 7th is the long-awaited Andertoon for last week. It is hard getting education in through all the overhead.

Bill Watterson’s Calvin and Hobbes rerun for the 7th is a basic joke about Calvin’s lousy student work. Fun enough. Calvin does show off one of those important skills mathematicians learn, though. He does do a sanity check. He may not know what 12 + 7 and 3 + 4 are, but he does notice that 12 + 7 has to be something larger than 3 + 4. That’s a starting point. It’s often helpful before starting work on a problem to have some idea of what you think the answer should be.

## Theorem Thursday: The Jordan Curve Theorem

There are many theorems that you have to get fairly far into mathematics to even hear of. Often they involve things that are so abstract and abstruse that it’s hard to parse just what we’re studying. This week’s entry is not one of them.

# The Jordan Curve Theorem.

There are a couple of ways to write this. I’m going to fall back on the version that Richard Courant and Herbert Robbins put in the great book What Is Mathematics?. It’s a theorem in the field of topology, the study of how shapes interact. In particular it’s about simple, closed curves on a plane. A curve is just what you figure it should be. It’s closed if it … uh … closes, makes a complete loop. It’s simple if it doesn’t cross itself or have any disconnected bits. So, something you could draw without lifting pencil from paper and without crossing back over yourself. Have all that? Good. Here’s the theorem:

A simple closed curve in the plane divides that plane into exactly two domains, an inside and an outside.

It’s named for Camille Jordan, a French mathematician who lived from 1838 to 1922, and who’s renowned for work in group theory and topology. It’s a different Jordan from the one named in Gauss-Jordan Elimination, which is a matrix thing that’s important but tedious. It’s also a different Jordan from Jordan Algebras, which I remember hearing about somewhere.

The Jordan Curve Theorem is proved by reading its proposition and then saying, “Duh”. This is compelling, although it lacks rigor. It’s obvious if your curve is a circle, or a slightly squished circle, or a rectangle or something like that. It’s less obvious if your curve is a complicated labyrinth-type shape.

It gets downright hard if the curve has a lot of corners. This is why a completely satisfying rigorous proof took decades to find. There are curves that are nowhere differentiable, that are nothing but corners, and those are hard to deal with. If you think there’s no such thing, then remember the Koch Snowflake. That’s that triangle sticking up from the middle of a straight line, that itself has triangles sticking up in the middle of its straight lines, and littler triangles still sticking up from the straight lines. Carry that on forever and you have a shape that’s continuous but always changing direction, and this is hard to deal with.

Still, you can have a good bit of fun drawing a complicated figure, then picking a point and trying to work out whether it’s inside or outside the curve. The challenging way to do that is to view your figure as a maze and look for a path leading outside. The easy way is to draw a new line. I recommend doing that in a different color.

In particular, draw a line from your target point to the outside. Some definitely outside point. You need the line to not be parallel to any of the curve’s line segments. And it’s easier if you don’t happen to intersect any vertices, but if you must, we’ll deal with that two paragraphs down.

So draw your testing line here from the point to something definitely outside. And count how many times your testing line crosses the original curve. If the testing line crosses the original curve an even number of times then the original point was outside the curve. If the testing line crosses the original an odd number of times then the original point was inside of the curve. Done.

If your testing line touches a vertex, well, then it gets fussy. It depends whether the two edges of the curve that go into that vertex stay on the same side as your testing line. If the original curve’s edges stay on the same side of your testing line, then don’t count that as a crossing. If the edges go on opposite sides of the testing line, then that does count as one crossing. With that in mind, carry on like you did before. An even number of crossings means your point was outside. An odd number of crossings means your point was inside.

So go ahead and do this a couple times with a few labyrinths and sample points. It’s fun and elevates your doodling to the heights of 19th-century mathematics. Also once you’ve done that a couple times you’ve proved the Jordan curve theorem.

Well, no, not quite. But you are most of the way to proving it for a special case. If the curve is a polygon, a shape made up of a finite number of line segments, then you’ve got almost all the proof done. You have to finish it off by choosing a ray, a direction, that isn’t parallel to any of the polygon’s line segments. (This is one reason this method only works for polygons, and fails for stuff like the Koch Snowflake. It also doesn’t work well with space-filling curves, which are things that exist. Yes, those are what they sound like: lines that squiggle around so much they fill up area. Some can fill volume. I swear. It’s fractal stuff.) Imagine all the lines that are parallel to that ray. There’s definitely some point along that line that’s outside the curve. You’ll need that for reference. Classify all the points on that line by whether there’s an even or an odd number of crossings between a starting point and your reference definitely-outside point. Keep doing that for all these many parallel lines.

And that’s it. The mess of points that have an odd number of intersections are the inside. The mess of points that have an even number of intersections are the outside.

You won’t be surprised to know there’s versions of the Jordan curve theorem for solid objects in three-dimensional space. And for hyperdimensional spaces too. You can always work out an inside and an outside, as long as space isn’t being all weird. But it might sound like it’s not much of a theorem. So you can work out an inside and an outside; so what?

But it’s one of those great utility theorems. It pops in to places, the perfect tool for a problem you were just starting to notice existed. If I can get my rhetoric organized I hope to show that off next week, when I figure to do the Five-Color Map Theorem.

## Theorem Thursday: Liouville’s Approximation Theorem And How To Make Your Own Transcendental Number

As I get into the second month of Theorem Thursdays I have, I think, the whole roster of weeks sketched out. Today, I want to dive into some real analysis, and the study of numbers. It’s the sort of thing you normally get only if you’re willing to be a mathematics major. I’ll try to be readable by people who aren’t. If you carry through to the end and follow directions you’ll have your very own mathematical construct, too, so enjoy.

# Liouville’s Approximation Theorem

It all comes back to polynomials. Of course it does. Polynomials aren’t literally everything in mathematics. They just come close. Among the things we can do with polynomials is divide up the real numbers into different sets. The tool we use is polynomials with integer coefficients. Integers are the positive and the negative whole numbers, stuff like ‘4’ and ‘5’ and ‘-12’ and ‘0’.

A polynomial is the sum of a bunch of products of coefficients multiplied by a variable raised to a power. We can use anything for the variable’s name. So we use ‘x’. Sometimes ‘t’. If we want complex-valued polynomials we use ‘z’. Some people trying to make a point will use ‘y’ or ‘s’ but they’re just showing off. Coefficients are just numbers. If we know the numbers, great. If we don’t know the numbers, or we want to write something that doesn’t commit us to any particular numbers, we use letters from the start of the alphabet. So we use ‘a’, maybe ‘b’ if we must. If we need a lot of numbers, we use subscripts: a0, a1, a2, and so on, up to some an for some big whole number n. To talk about one of these without committing ourselves to a specific example we use a subscript of i or j or k: aj, ak. It’s possible that aj and ak equal each other, but they don’t have to, unless j and k are the same whole number. They might also be zero, but they don’t have to be. They can be any numbers. Or, for this essay, they can be any integers. So we’d write a generic polynomial f(x) as:

$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_{n - 1}x^{n - 1} + a_n x^n$

(Some people put the coefficients in the other order, that is, $a_n + a_{n - 1}x + a_{n - 2}x^2$ and so on. That’s not wrong. The name we give a number doesn’t matter. But it makes it harder to remember what coefficient matches up with, say, x14.)

A zero, or root, is a value for the variable (‘x’, or ‘t’, or what have you) which makes the polynomial equal to zero. It’s possible that ‘0’ is a zero, but don’t count on it. A polynomial of degree n — meaning the highest power to which x is raised is n — can have up to n different real-valued roots. All we’re going to care about is one.

Rational numbers are what we get by dividing one whole number by another. They’re numbers like 1/2 and 5/3 and 6. They’re numbers like -2.5 and 1.0625 and negative a billion. Almost none of the real numbers are rational numbers; they’re exceptional freaks. But they are all the numbers we actually compute with, once we start working out digits. Thus we remember that to live is to live paradoxically.

And every rational number is a root of a first-degree polynomial. That is, there’s some polynomial f(x) = a_0 + a_1 x that’s made zero for your polynomial. It’s easy to tell you what it is, too. Pick your rational number. You can write that as the integer p divided by the integer q. Now look at the polynomial f(x) = p – q x. Astounded yet?

That trick will work for any rational number. It won’t work for any irrational number. There’s no first-degree polynomial with integer coefficients that has the square root of two as a root. There are polynomials that do, though. There’s f(x) = 2 – x2. You can find the square root of two as the zero of a second-degree polynomial. You can’t find it as the zero of any lower-degree polynomials. So we say that this is an algebraic number of the second degree.

This goes on higher. Look at the cube root of 2. That’s another irrational number, so no first-degree polynomials have it as a root. And there’s no second-degree polynomials that have it as a root, not if we stick to integer coefficients. Ah, but f(x) = 2 – x3? That’s got it. So the cube root of two is an algebraic number of degree three.

We can go on like this, although I admit examples for higher-order algebraic numbers start getting hard to justify. Most of the numbers people have heard of are either rational or are order-two algebraic numbers. I can tell you truly that the eighth root of two is an eighth-degree algebraic number. But I bet you don’t feel enlightened. At best you feel like I’m setting up for something. The number r(5), the smallest radius a disc can have so that five of them will completely cover a disc of radius 1, is eighth-degree and that’s interesting. But you never imagined the number before and don’t have any idea how big that is, other than “I guess that has to be smaller than 1”. (It’s just a touch less than 0.61.) I sound like I’m wasting your time, although you might start doing little puzzles trying to make smaller coins cover larger ones. Do have fun.

Liouville’s Approximation Theorem is about approximating algebraic numbers with rational ones. Almost everything we ever do is with rational numbers. That’s all right because we can make the difference between the number we want, even if it’s r(5), and the numbers we can compute with, rational numbers, as tiny as we need. We trust that the errors we make from this approximation will stay small. And then we discover chaos science. Nothing is perfect.

For example, suppose we need to estimate π. Everyone knows we can approximate this with the rational number 22/7. That’s about 3.142857, which is all right but nothing great. Some people know we can approximate it as 333/106. (I didn’t until I started writing this paragraph and did some research.) That’s about 3.141509, which is better. Then there’s 355/113, which is not as famous as 22/7 but is a celebrity compared to 333/106. That’s about 3.141529. Then we get into some numbers only mathematics hipsters know: 103993/33102 and 104348/33215 and so on. Fine.

The Liouville Approximation Theorem is about sequences that converge on an irrational number. So we have our first approximation x1, that’s the integer p1 divided by the integer q1. So, 22 and 7. Then there’s the next approximation x2, that’s the integer p2 divided by the integer q2. So, 333 and 106. Then there’s the next approximation yet, x3, that’s the integer p3 divided by the integer q3. As we look at more and more approximations, xj‘s, we get closer and closer to the actual irrational number we want, in this case π. Also, the denominators, the qj‘s, keep getting bigger.

The theorem speaks of having an algebraic number, call it x, of some degree n greater than 1. Then we have this limit on how good an approximation can be. The difference between the number x that we want, and our best approximation p / q, has to be larger than the number (1/q)n + 1. The approximation might be higher than x. It might be lower than x. But it will be off by at least the n-plus-first power of 1/q.

Polynomials let us separate the real numbers into infinitely many tiers of numbers. They also let us say how well the most accessible tier of numbers, rational numbers, can approximate these more exotic things.

One of the things we learn by looking at numbers through this polynomial screen is that there are transcendental numbers. These are numbers that can’t be the root of any polynomial with integer coefficients. π is one of them. e is another. Nearly all numbers are transcendental. But the proof that any particular number is one is hard. Joseph Liouville showed that transcendental numbers must exist by using continued fractions. But this approximation theorem tells us how to make our own transcendental numbers. This won’t be any number you or anyone else has ever heard of, unless you pick a special case. But it will be yours.

You will need:

1. a1, an integer from 1 to 9, such as ‘1’, ‘9’, or ‘5’.
2. a2, another integer from 1 to 9. It may be the same as a1 if you like, but it doesn’t have to be.
3. a3, yet another integer from 1 to 9. It may be the same as a1 or a2 or, if it so happens, both.
4. a4, one more integer from 1 to 9 and you know what? Let’s summarize things a bit.
5. A whopping great big gob of integers aj, every one of them from 1 to 9, for every possible integer ‘j’ so technically this is infinitely many of them.
6. Comfort with the notation n!, which is the factorial of n. For whole numbers that’s the product of every whole number from 1 to n, so, 2! is 1 times 2, or 2. 3! is 1 times 2 times 3, or 6. 4! is 1 times 2 times 3 times 4, or 24. And so on.
7. Not to be thrown by me writing -n!. By that I mean work out n! and then multiply that by -1. So -2! is -2. -3! is -6. -4! is -24. And so on.

Now, assemble them into your very own transcendental number z, by this formula:

$z = a_1 \cdot 10^{-1} + a_2 \cdot 10^{-2!} + a_3 \cdot 10^{-3!} + a_4 \cdot 10^{-4!} + a_5 \cdot 10^{-5!} + a_6 \cdot 10^{-6!} \cdots$

If you’ve done it right, this will look something like:

$z = 0.a_{1}a_{2}000a_{3}00000000000000000a_{4}0000000 \cdots$

Ah, but, how do you know this is transcendental? We can prove it is. The proof is by contradiction, which is how a lot of great proofs are done. We show nonsense follows if the thing isn’t true, so the thing must be true. (There are mathematicians that don’t care for proof-by-contradiction. They insist on proof by charging straight ahead and showing a thing is true directly. That’s a matter of taste. I think every mathematician feels that way sometimes, to some extent or on some issues. The proof-by-contradiction is easier, at least in this case.)

Suppose that your z here is not transcendental. Then it’s got to be an algebraic number of degree n, for some finite number n. That’s what it means not to be transcendental. I don’t know what n is; I don’t care. There is some n and that’s enough.

Now, let’s let zm be a rational number approximating z. We find this approximation by taking the first m! digits after the decimal point. So, z1 would be just the number 0.a1. z2 is the number 0.a1a2. z3 is the number 0.a1a2000a3. I don’t know what m you like, but that’s all right. We’ll pick a nice big m.

So what’s the difference between z and zm? Well, it can’t be larger than 10 times 10-(m + 1)!. This is for the same reason that π minus 3.14 can’t be any bigger than 0.01.

Now suppose we have the best possible rational approximation, p/q, of your number z. Its first m! digits are going to be p / 10m!. This will be zm And by the Liouville Approximation Theorem, then, the difference between z and zm has to be at least as big as (1/10m!)(n + 1).

So we know the difference between z and zm has to be larger than one number. And it has to be smaller than another. Let me write those out.

$\frac{1}{10^{m! (n + 1)}} < |z - z_m | < \frac{10}{10^{(m + 1)!}}$

We don’t need the z – zm anymore. That thing on the rightmost side we can write what I’ll swear is a little easier to use. What we have left is:

$\frac{1}{10^{m! (n + 1)}} < \frac{1}{10^{(m + 1)! - 1}}$

And this will be true whenever the number m! (n + 1) is greater than (m + 1)! – 1 for big enough numbers m.

But there’s the thing. This isn’t true whenever m is greater than n. So the difference between your alleged transcendental number and its best-possible rational approximation has to be simultaneously bigger than a number and smaller than that same number without being equal to it. Supposing your number is anything but transcendental produces nonsense. Therefore, congratulations! You have a transcendental number.

If you chose all 1’s for your aj‘s, then you have what is sometimes called the Liouville Constant. If you didn’t, you may have a transcendental number nobody’s ever noticed before. You can name it after someone if you like. That’s as meaningful as naming a star for someone and cheaper. But you can style it as weaving someone’s name into the universal truth of mathematics. Enjoy!

I’m glad to finally give you a mathematics essay that lets you make something you can keep.

## What’s The Shortest Proof I’ve Done?

I didn’t figure to have a bookend for last week’s “What’s The Longest Proof I’ve Done? question. I don’t keep track of these things, after all. And the length of a proof must be a fluid concept. If I show something is a direct consequence of a previous theorem, is the proof’s length the two lines of new material? Or is it all the proof of the previous theorem plus two new lines?

I would think the shortest proof I’d done was showing that the logarithm of 1 is zero. This would be starting from the definition of the natural logarithm of a number x as the definite integral of 1/t on the interval from 1 to x. But that requires a bunch of analysis to support the proof. And the Intermediate Value Theorem. Does that stuff count? Why or why not?

But this happened to cross my desk: The Shortest-Known Paper Published in a Serious Math Journal: Two Succinct Sentences, an essay by Dan Colman. It reprints a paper by L J Lander and T R Parkin which appeared in the Bulletin of the American Mathematical Society in 1966.

It’s about Euler’s Sums of Powers Conjecture. This is a spinoff of Fermat’s Last Theorem. Leonhard Euler observed that you need at least two whole numbers so that their squares add up to a square. And you need three cubes of whole numbers to add up to the cube of a whole number. Euler speculated you needed four whole numbers so that their fourth powers add up to a fourth power, five whole numbers so that their fifth powers add up to a fifth power, and so on.

And it’s not so. Lander and Parkin found that this conjecture is false. They did it the new old-fashioned way: they set a computer to test cases. And they found four whole numbers whose fifth powers add up to a fifth power. So the quite short paper answers a long-standing question, and would be hard to beat for accessibility.

There is another famous short proof sometimes credited as the most wordless mathematical presentation. Frank Nelson Cole gave it on the 31st of October, 1903. It was about the Mersenne number 267-1, or in human notation, 147,573,952,589,676,412,927. It was already known the number wasn’t prime. (People wondered because numbers of the form 2n-1 often lead us to perfect numbers. And those are interesting.) But nobody knew which factors it was. Cole gave his talk by going up to the board, working out 267-1, and then moving to the other side of the board. There he wrote out 193,707,721 × 761,838,257,287, and showed what that was. Then, per legend, he sat down without ever saying a word, and took in the standing ovation.

I don’t want to cast aspersions on a great story like that. But mathematics is full of great stories that aren’t quite so. And I notice that one of Cole’s doctoral students was Eric Temple Bell. Bell gave us a great many tales of mathematics history that are grand and great stories that just weren’t so. So I want it noted that I don’t know where we get this story from, or how it may have changed in the retellings. But Cole’s proof is correct, at least according to Octave.

So not every proof is too long to fit in the universe. But then I notice that Mathworld’s page regarding the Euler Sum of Powers Conjecture doesn’t cite the 1966 paper. It cites instead Lander and Parkin’s “A Counterexample to Euler’s Sum of Powers Conjecture” from Mathematics of Computation volume 21, number 97, of 1967. There the paper has grown to three pages, although it’s only a couple paragraphs of one page and three lines of citation on the third. It’s not so easy to read either, but it does explain how they set about searching for counterexamples. But it may give you some better idea of how numerical mathematicians find things.

## What’s The Longest Proof I’ve Done?

You know what’s a question I’m surprised I don’t get asked? I mean in the context of being a person with an advanced mathematics degree. I don’t get asked what’s the longest proof I’ve ever done. Either just reading to understand, or proving for myself. Maybe people are too intimidated by the idea of advanced mathematics to try asking such things. Maybe they’re afraid I’d bury them under a mountain of technical details. But I’d imagine musicians get asked what the hardest or the longest piece they’ve memorized is. I’m sure artists get asked what’s the painting (or sculpture, or whatnot) they’ve worked on the longest was.

It’s just as well nobody’s asked. I’m not sure what the longest proof I’ve done, or gone through, would even be. Some of it is because there’s an inherent arbitrariness to the concept of “a proof”. Proofs are arguments, and they’re almost always made up of many smaller pieces. The advantage of making these small pieces is that small proofs are usually easier to understand. We can then assemble the conclusions of many small proofs to make one large proof. But then how long was the large proof? Does it contain all the little proofs that go into it?

And, truth be told, I didn’t think to pay attention to how long any given proof was. If I had to guess I would think the longest proof I’d done, just learned, would be from a grad school course in ordinary differential equations. This is the way we study systems in which how things are changing depends on what things are now. These often match physical, dynamic, systems very well. I remember in the class spending several two-hour sessions trying to get through a major statement in a field called Kolmogorov-Arnold-Moser Theory. This is a major statement about dynamical systems being perturbed, given a little shove. And it describes what conditions make the little shove really change the way the whole system behaves.

What I’m getting to is that there appears to be a new world’s record-holder for the Longest Actually Completed Proof. It’s about a problem I never heard of before but that’s apparently been open since the 1980s. It’s known as the Boolean Pythagorean Triples problem. The MathsByAGirl blog has an essay about it, and gives some idea of its awesome size. It’s about 200 terabytes of text. As you might imagine, it’s a proof by exhaustion. That is, it divides up a problem into many separate cases, and tries out all the cases. That’s a legitimate approach. It tends to produce proofs that are long and easy to verify, at least at each particular case. They might not be insightful, that is, they might not suggest new stuff to do, but they work. (And I don’t know that this proof doesn’t suggest new stuff to do. I haven’t read it, for good reason. It’s well outside my specialty.)

But proofs can be even bigger. John Carlos Baez published a while back an essay, “Insanely Long Proofs”. And that’s awe-inspiring. Baez is able to provide theorems which we know to be true. You’ll be able to understand what they conclude, too. And in the logic system applicable to them, their proofs would be so long that the entire universe isn’t big enough just to write down the number of symbols needed to complete the proof. Let me say that again. It’s not that writing out the proof would take more than all the space in the universe. It’s that writing out how long the proof would be, written out would take more than all the space in the universe.

So you should ask, then how do we know it’s true? Baez explains.

## Reading the Comics, June 3, 2016: Word Problems Without Pictures Edition

I haven’t got Sunday’s comics under review yet. But the past seven days were slow ones for mathematically-themed comics. Maybe Comic Strip Master Command is under the impression that it’s the (United States) summer break already. It’s not, although Funky Winkerbean did a goofy sequence graduating its non-player-character students. And Zits has been doing a summer reading storyline that only makes sense if Jeremy Duncan is well into summer. Maybe Comic Strip Master Command thinks it’s a month later than it actually is?

Tony Cochrane’s Agnes for the 29th of May looks at first like a bit of nonsense wordplay. But whether a book with the subject “All About Books” would discuss itself, and how it would discuss itself, is a logic problem. And not just a logic problem. Start from pondering how the book All About Books would describe the content of itself. You can go from that to an argument that it’s impossible to compress every possible message. Imagine an All About Books which contained shorthand descriptions of every book. And the descriptions have enough detail to exactly reconstruct each original book. But then what would the book list for the description of All About Books?

And self-referential things can lead to logic paradoxes swiftly. You’d have some fine ones if Agnes were to describe a book All About Not-Described Books. Is the book described in itself? The question again sounds silly. But thinking seriously about it leads us to the decidability problem. Any interesting-enough logical system will always have statements that are meaningful and true that no one can prove.

Furthermore, the suggestion of an “All About `All About Books’ Book” suggests to me power sets. That’s the set of all the ways you can collect the elements of a set. Power sets are always bigger than the original set. They lead to the staggering idea that there are many sizes of infinitely large sets, a never-ending stack of bigness.

Robb Armstrong’s Jump Start for the 31st of May is part of a sequence about getting a tutor for a struggling kid. That it’s mathematics is incidental to the storyline, must be said. (It’s an interesting storyline, partly about the Jojo’s father, a police officer, coming to trust Ray, an ex-convict. Jump Start tells many interesting and often deeply weird storylines. And it never loses its camouflage of being an ordinary family comic strip.) It uses the familiar gimmick of motivating a word problem by making it about something tangible.

Ken Cursoe’s Tiny Sepuku for the 2nd of June uses the motif of non-Euclidean geometry as some supernatural magic. It’s a small reference, you might miss it. I suppose it is true that a high-dimensional analogue to conic sections would focus things from many dimensions. If those dimensions match time and space, maybe it would focus something from all humanity into the brain. I would try studying instead, though.

Russell Myers’s Broom Hilda for the 3rd is a resisting-the-word-problems joke. It’s funny to figure on missing big if you have to be wrong at all. But something you learn in numerical mathematics, particularly, is that it’s all right to start from a guess. Often you can take a wrong answer and improve it. If you can’t get the exact right answer, you can usually get a better answer. And often you can get as good as you need. So in practice, sorry to say, I can’t recommend going for the ridiculous answer. You can do better.

Wait for it.

## Wlog.

I’d like to say a good word for boredom. It needs the good words. The emotional state has an appalling reputation. We think it’s the sad state someone’s in when they can’t find anything interesting. It’s not. It’s the state in which we are so desperate for engagement that anything is interesting enough.

And that isn’t a bad thing! Finding something interesting enough is a precursor to noticing something curious. And curiosity is a precursor to discovery. And discovery is a precursor to seeing a fuller richness of the world.

Think of being stuck in a waiting room, deprived of reading materials or a phone to play with or much of anything to do. But there is a clock. Your classic analog-face clock. Its long minute hand sweeps out the full 360 degrees of the circle once every hour, 24 times a day. Its short hour hand sweeps out that same arc every twelve hours, only twice a day. Why is the big unit of time marked with the short hand? Good question, I don’t know. Probably, ultimately, because it changes so much less than the minute hand that it doesn’t need the attention of length drawn to it.

But let our waiting mathematician get a little more bored, and think more about the clock. The hour and minute hand must sometimes point in the same direction. They do at 12:00 by the clock, for example. And they will at … a little bit past 1:00, and a little more past 2:00, and a good while after 9:00, and so on. How many times during the day will they point the same direction?

Well, one easy way to do this is to work out how long it takes the hands, once they’ve met, to meet up again. Presumably we don’t want to wait the whole hour-and-some-more-time for it. But how long is that? Well, we know the hands start out pointing the same direction at 12:00. The first time after that will be after 1:00. At exactly 1:00 the hour hand is 30 degrees clockwise of the minute hand. The minute hand will need five minutes to catch up to that. In those five minutes the hour hand will have moved another 2.5 degrees clockwise. The minute hand needs about four-tenths of a minute to catch up to that. In that time the hour hand moves — OK, we’re starting to see why Zeno was not an idiot. He never was.

But we have this roughly worked out. It’s about one hour, five and a half minutes between one time the hands meet and the next. In the course of twelve hours there’ll be time for them to meet up … oh, of course, eleven times. Over the course of the day they’ll meet up 22 times and we can get into a fight over whether midnight counts as part of today, tomorrow, or both days, or neither. (The answer: pretend the day starts at 12:01.)

Hold on, though. How do we know that the time between the hands meeting up at 12:00 and the one at about 1:05 is the same as the time between the hands meeting up near 1:05 and the next one, sometime a little after 2:10? Or between that one and the one at a little past 3:15? What grounds do we have for saying this one interval is a fair representation of them all?

We can argue that it should be fairly enough. Imagine that all the markings were washed off the clock. It’s just two hands sweeping around in circles, one relatively fast, one relatively slow, forever. Give the clockface a spin. When the hands come together again rotate the clock so those two hands are vertical, the “12:00” position. Is this actually 12:00? … Well, we’ve got a one-in-eleven chance it is. It might be a little past 1:05; it might be that time something past 6:30. The movement of the clock hands gives no hint what time it really is.

And that is why we’re justified taking this one interval as representative of them all. The rate at which the hands move, relative to each other, doesn’t depend on what the clock face behind it says. The rate is, if the clock isn’t broken, always the same. So we can use information about one special case that happens to be easy to work out to handle all the cases.

That’s the mathematics term for this essay. We can study the one specific case without loss of generality, or as it’s inevitably abbreviated, wlog. This is the trick of studying something possibly complicated, possibly abstract, by looking for a representative case. That representative case may tell us everything we need to know, at least about this particular problem. Generality means what you might figure from the ordinary English meaning of it: it means this answer holds in general, as opposed to in this specific instance.

Some thought has to go in to choosing the representative case. We have to pick something that doesn’t, somehow, miss out on a class of problems we would want to solve. We mustn’t lose the generality. And it’s an easy mistake to make, especially as a mathematics student first venturing into more abstract waters. I remember coming up against that often when trying to prove properties of infinitely long series. It’s so hard to reason something about a bunch of numbers whose identities I have no idea about; why can’t I just use the sequence, oh, 1/1, 1/2, 1/3, 1/4, et cetera and let that be good enough? Maybe 1/1, 1/4, 1/9, 1/16, et cetera for a second test, just in case? It’s because it takes time to learn how to safely handle infinities.

It’s still worth doing. Few of us are good at manipulating things in the abstract. We have to spend more mental energy imagining the thing rather than asking the questions we want of it. Reducing that abstraction — even if it’s just a little bit, changing, say, from “an infinitely-differentiable function” to “a polynomial of high enough degree” — can rescue us. We can try out things we’re confident we understand, and derive from it things we don’t know.

I can’t say that a bored person observing a clock would deduce all this. Parts of it, certainly. Maybe all, if she thought long enough. I believe it’s worth noticing and thinking of these kinds of things. And it’s why I believe it’s fine to be bored sometimes.