I am, believe it or not, working ahead of deadline on the Little Mathematics A-to-Z for this year. I feel so happy about that. But that’s eating up time to write fresh stuff here. So please let me share some older material, this from my prolific year 2016.
However, it’s not hard to create a number that you know is transcendental. Here’s how to do it, with an easy step-by-step guide. If you want to create this and declare it’s named after you, enjoy! Nobody but you will ever care about this number, I’m afraid. Its only interesting traits will be that it’s transcendental and that you crafted it. Still, isn’t that nice anyway? I think it’s nice anyway.
With qualifiers, of course. Compute! and Compute!’s Gazette had two generations of Automatic Proofreader for Commodore computers. The magazines also had Automatic Proofreaders for the other eight-bit computers that they covered. I trust that those worked the same way, but — with one exception — don’t know. I haven’t deciphered most of those other proofreaders.
Let me introduce how it was used, though. Compute! and Compute!’s Gazette offered computer programs to type in. Many of them were in BASIC, which uses many familiar words of English as instructions. But you can still make typos entering commands, and this causes bugs or crashes in programs. The Automatic Proofreader, for the Commodore (and the Atari), put in a little extra step after you typed in a line of code. It calculated a checksum. It showed that on-screen after every line you entered. And you could check whether that matched the checksum the magazine printed. So the listing in the magazine would be something like:
You would type in all those lines up to the :rem part. ‘rem’ here stands for ‘Remark’ and means the rest of the line is a comment to the programmer, not the computer. So they’d do no harm if you did enter them. But why type text you didn’t need?
So after typing, say, 100 POKE 56,50:CLR:DIM IN$,I,J,A,B,A$,B$,A(7),N$ you’d hit return and with luck get the number 34 up on screen. The Automatic Proofreader did not force you to re-type the line. You were on your honor to do that. (Nor were you forced to type lines in order. If you wished to type line 100, then 200, then 300, then 190, then 250, then 330, you could. The checksum would calculate the same.) And it didn’t only work for entering programs, these commands starting with line numbers. It would return a result for any command you entered. But since you wouldn’t know what the checksum should be for a freeform command, that didn’t tell you much.
The first-generation Automatic Proofreader, which is what I’m talking about here, returned a number between 0 and 255. And it was a simple checksum. It could not detect transposed characters: the checksum for PIRNT was the same as PRINT and PRITN. And, it turns out, errors could offset: the checksum for PEEK(46) would be the same as that for PEEK(55).
And there was one bit of deliberate insensitivity built in. Spaces would not be counted. The checksum for FA=PEEK(45)+Z6*PEEK(46) would be the same as FA = PEEK( 45 ) + Z6 * PEEK( 46 ). So you could organize text in whatever way was most convenient.
Given this, and given the example of the first MLX, you may have a suspicion how the Automatic Proofreader calculated things. So did I and it turned out to be right. The checksum for the first-generation Automatic Proofreader, at least for the Commodore 64 and the Vic-20, was a simple sum. Take the line that’s been entered. Ignore spaces. But otherwise, take the ASCII code value for each character, and add that up, modulo 256. That is, if the sum is (say) 300, subtract 256 from that, that is, 44.
I’m fibbing a little when I say it’s the ASCII code values. The Commodore computers used a variation on ASCII, called PETSCII (Commodore’s first line of computers was the PET). For ordinary text the differences between ASCII and PETSCII don’t matter. The differences come into play for various characters Commodores had. These would be symbols like the suits of cards, or little circles, or checkerboard patterns. Symbols that, these days, we’d see as emojis, or at least part of an extended character set.
But translating all those symbols is … tedious, but not hard. If you want to do a simulated Automatic Proofreader in Octave, it’s almost no code at all. It turns out Octave and Matlab need no special command to get the ASCII code equivalent of text. So here’s a working simulation
function retval = automatic_proofreader (oneLine)
trimmedLine = strrep(oneLine, " ", "");
# In Matlab this should be replace(oneLine, " ", "");
retval = mod(sum(trimmedLine), 256);
Capitalization matters! The ASCII code for capital-P is different from that for lowercase-p. Spaces won’t matter, though. More exotic characters, though, such as the color-setting commands, are trouble and let’s not deal with that right now. Also you can enclose your line in single-quotes, in case for example you want the checksum of a line that had double-quotes. Let’s agree that lines with single- and double-quotes don’t exist.
I understand the way Commodore 64’s work well enough that I can explain the Automatic Proofreader’s code. I plan to do that soon. I don’t know how the Atari version of the Automatic Proofreader worked, but since it had the same weaknesses I assume it used the same algorithm.
There is a first-generation Automatic Proofreader with a difference, though, and I’ll come to that.
Iva Sallay, creator of the Find The Factors recreational mathematics puzzle and a kind friend to my blog, posted Yes, YOU Can Host a Playful Math Education Blog Carnival. It explains in quite good form how to join in Denise Gaskins’s roaming blog event. It tries to gather educational or recreational or fun or just delightful mathematics links.
Hosting the blog carnival is a great experience I recommend for mathematics bloggers at least once. I seem to be up to hosting it about once a year, most recently in September 2020. Most important in putting one together is looking at your mathematics reading with different eyes. Sallay, though, goes into specifics about what to look for, and how to find that.
We have goldfish, normally kept in an outdoor pond. It’s not a deep enough pond that it would be safe to leave them out for a very harsh winter. So we keep as many as we can catch in a couple 150-gallon tanks in the basement.
Recently, and irritatingly close to when we’d set them outside, the nitrate level in the tanks grew too high. Fish excrete ammonia. Microorganisms then turn the ammonia into nitrates and then nitrates. In the wild, the nitrates then get used by … I dunno, plants? Which don’t thrive enough hin our basement to clean them out. To get the nitrate out of the water all there is to do is replace the water.
We have six buckets, each holding five gallons, of water that we can use for replacement. So there’s up to 30 gallons of water that we could change out in a day. Can’t change more because tap water contains chloramines, which kill bacteria (good news for humans) but hurt fish (bad news for goldfish). We can treat the tap water to neutralize the chloramines, but want to give that time to finish. I have never found a good reference for how long this takes. I’ve adopted “about a day” because we don’t have a water tap in the basement and I don’t want to haul more than 30 gallons of water downstairs any given day.
So I got thinking, what’s the fastest way to get the nitrate level down for both tanks? Change 15 gallons in each of them once a day, or change 30 gallons in one tank one day and the other tank the next?
And, happy to say, I realized this was the tea-making problem I’d done a couple months ago. The tea-making problem had a different goal, that of keeping as much milk in the tea as possible. But the thing being studied was how partial replacements of a solution with one component affects the amount of the other component. The major difference is that the fish produce (ultimately) more nitrates in time. There’s no tea that spontaneously produces milk. But if nitrate-generation is low enough, the same conclusions follow. So, a couple days of 30-gallon changes, in alternating tanks, and we had the nitrates back to a decent level.
We’d have put the fish outside this past week if I hadn’t broken, again, the tool used for cleaning the outside pond.
These carnivals often feature recreational mathematics. Sallay’s collection this month has even more than usual, and (to my tastes) more delightful ones than usual. Even if you aren’t an educator or parent it’s worth reading, as there’s surely something you haven’t thought about before.
And if you have a blog, and would like to host the carnival some month? Denise Gaskins, who organizes the project, is taking volunteers. The 147th carnival needs a host yet, and there’s all of fall and winter available too. Hosting is an exciting and challenging thing to do, and I do recommend anyone with pop-mathematics inclinations trying it at least once.
Have a special one today. I’ve been reading a compilation of Crockett Johnson’s 1940s comic Barnaby. The title character, an almost too gentle child, follows his fairy godfather Mr O’Malley into various shenanigans. Many (the best ones, I’d say) involve the magical world. The steady complication is that Mr O’Malley boasts abilities beyond his demonstrated competence. (Although most of the magic characters are shown to be not all that good at their business.) It’s a gentle strip and everything works out all right, if farcically.
This particular strip comes from a late 1948 storyline. Mr O’Malley’s gone missing, coincidentally to a fairy cop come to arrest the pixie, who is a con artist at heart. So this sees the entry of Atlas, the Mental Giant, who’s got some pleasant gimmicks. One of them is his requiring mnemonics built on mathematical formulas to work out names. And this is a charming one, with a great little puzzle: how do you get A-T-L-A-S out of the formula Atlas has remembered?
I’m sorry the solution requires a bit of abusing notation, so please forgive it. But it’s a fun puzzle, especially as the joke would not be funnier if the formula didn’t work. I’m always impressed when a comic strip goes to that extra effort.
I am made aware that a section of Twitter argues about how to evaluate an expression. There may be more than one of these going around, but the expression I’ve seen is:
Many people feel that the challenge is knowing the order of operations. This is reasonable. That is, that to evaluate arithmetic, you evaluate terms inside parentheses first. Then terms within exponentials. Then multiplication and division. Then addition and subtraction. This is often abbreviated as PEMDAS, and made into a mnemonic like “Please Excuse My Dear Aunt Sally”.
That is fine as far as it goes. Many people likely start by adding the 1 and 2 within the parentheses, and that’s fair. Then they get:
Putting two quantities next to one another, as the 2 and the (3) are, means to multiply them. And then comes the disagreement: does this mean take and multiply that by 3, in which case the answer is 9? Or does it mean take 6 divided by , in which case the answer is 1?
And there is the trick. Depending on which way you choose to parse these instructions you get different answers. But you don’t get to do that, not and have arithmetic. So the answer is that this expression has no answer. The phrasing is ambiguous and can’t be resolved.
I’m aware there are people who reject this answer. They picked up along the line somewhere a rule like “do multiplication and division from left to right”. And a similar rule for addition and subtraction. This is wrong, but understandable. The left-to-right “rule” is a decent heuristic, a guide to how to attack a problem too big to do at once. The rule works because multiplication-and-division associates. The quantity a-times-b, multiplied by c, has to be the same number as the quantity a multiplied by the quantity b-times-c. The rule also works for addition-and-subtraction because addition associates too. The quantity a-plus-b, plus the quantity c, has to be the same as the quantity a plus the quantity b-plus-c.
This left-to-right “rule”, though, just helps you evaluate a meaningful expression. It would be just as valid to do all the multiplications-and-divisions from right-to-left. If you get different values working left-to-right from right-to-left, you have a meaningless expression.
But you also start to see why mathematicians tend to avoid the symbol. We understand, for example, to mean . Carry that out and then there’s no ambiguity about
I understand the desire to fix an ambiguity. Believe me. I’m a know-it-all; I only like ambiguities that enable logic-based jokes. (“Would you like ice cream or cake?” “Yes.”) But the rules that could remove the ambiguity in also remove associativity from multiplication. Once you do that, you’re not doing arithmetic anymore. Resist the urge.
(And the mnemonic is a bit dangerous. We can say division has the same priority as multiplication, but we also say “multiplication” first. I bet you can construct an ambiguous expression which would mislead someone who learned Please Excuse Dear Miss Sally Andrews.)
And now a qualifier: computer languages will often impose doing a calculation in some order. Usually left-to-right. The microchips doing the work need to have some instructions. Spotting all possible ambiguous phrasings ahead of time is a challenge. But we accept our computers doing not-quite-actual-arithmetic. They’re able to do not-quite-actual-arithmetic much faster and more reliably than we can. This makes the compromise worthwhile. We need to remember the difference between what the computer does and the calculation we intend.
And another qualifier: it is possible to do interesting mathematics with operations that aren’t associative. But if you are it’s in your research as a person with a postgraduate degree in mathematics. It’s possible it might fit in social media, but I would be surprised. It won’t draw great public attention, anyway.
John Golden, MathHombre, was host this month for the Playful Math Education Blog Carnival. And this month’s collection of puzzles, essays, and creative mathematics projects. Among them are some quilts and pattern-block tiles, which manifest all that talk about the structure of mathematical objects and their symmetries in easy-to-see form. There’s likely to be something of interest there.
Among the wonderful things I discovered there is Math Zine Fest 2021. It’s as the name suggests, a bunch of zines — short printable magazines on a niche topic — put together for the end of February. I had missed this organizing, but hope to get to see later installments. I don’t know what zine I might make, but I must have something I could do.
The problem I’d set out last week: I have a teapot good for about three cups of tea. I want to put milk in the once, before the first cup. How much should I drink before topping up the cup, to have the most milk at the end?
I have expectations. Some of this I know from experience, doing other problems where things get replaced at random. Here, tea or milk particles get swallowed at random, and replaced with tea particles. Yes, ‘particle’ is a strange word to apply to “a small bit of tea”. But it’s not like I can call them tea molecules. “Particle” will do and stop seeming weird someday.
Random replacement problems tend to be exponential decays. That I know from experience doing problems like this. So if I get an answer that doesn’t look like an exponential decay I’ll doubt it. I might be right, but I’ll need more convincing.
I also get some insight from extreme cases. We can call them reductios. Here “reductio” as in the word we usually follow with “ad absurdum”. Make the case ridiculous and see if that offers insight. The first reductio is to suppose I drink the entire first cup down to the last particle, then pour new tea in. By the second cup, there’s no milk left. The second reductio is to suppose I drink not a bit of the first cup of milk-with-tea. Then I have the most milk preserved. It’s not a satisfying break. But it leads me to suppose the most milk makes it through to the end if I have a lot of small sips and replacements of tea. And to look skeptically if my work suggests otherwise.
So that’s what I expect. What actually happens? Here, I do a bit of reasoning. Suppose that I have a mug. It can hold up to 1 unit of tea-and-milk. And the teapot, which holds up to 2 more units of tea-and-milk. What units? For the mathematics, I don’t care.
I’m going to suppose that I start with some amount — call it — of milk. is some number between 0 and 1. I fill the cup up to full, that is, 1 unit of tea-and-milk. And I drink some amount of the mixture. Call the amount I drink . It, too, is between 0 and 1. After this, I refill the mug up to full, so, putting in units of tea. And I repeat this until I empty the teapot. So I can do this times.
I know you noticed that I’m short on tea here. The teapot should hold 3 units of tea. I’m only pouring out . I could be more precise by refilling the mug times. I’m also going to suppose that I refill the mug with amount of tea a whole number of times. This sounds necessarily true. But consider: what if I drank and re-filled three-quarters of a cup of tea each time? How much tea is poured that third time?
I make these simplifications for good reasons. They reduce the complexity of the calculations I do without, I trust, making the result misleading. I can justify it too. I don’t drink tea from a graduated cylinder. It’s a false precision to pretend I do. I drink (say) about half my cup and refill it. How much tea I get in the teapot is variable too. Also, I don’t want to do that much work for this problem.
In fact, I’m going to do most of the work of this problem with a single drawing of a square. Here it is.
So! I start out with units of tea in the mixture. After drinking units of milk-and-tea, what’s left is units of milk in the mixture.
How about the second refill? The process is the same as the first refill. But where, before, there had been units of milk in the tea, now there are only units in. So that horizontal strip is a little narrower is all. The same reasoning applies and so, after the second refill, there’s milk in the mixture.
If you nodded to that, you’d agree that after the third refill there’s . And are pretty sure what happens at the fourth and fifth and so on. If you didn’t nod to that, it’s all right. If you’re willing to take me on faith we can continue. If you’re not, that’s good too. Try doing a couple drawings yourself and you may convince yourself. If not, I don’t know. Maybe try, like, getting six white and 24 brown beads, stir them up, take out four at random. Replace all four with brown beads and count, and do that several times over. If you’re short on beads, cut up some paper into squares and write ‘B’ and ‘W’ on each square.
But anyone comfortable with algebra can see how to reduce this. The amount of milk remaining after j refills is going to be
How many refills does it take to run out of tea? That we knew from above: it’s refills. So my last full mug of tea will have left in it
units of milk.
Anyone who does differential equations recognizes this. It’s the discrete approximation of the exponential decay curve. Discrete, here, because we take out some finite but nonzero amount of milk-and-tea, , and replace it with the same amount of pure tea.
Now, again, I’ve seen this before so I know its conclusions. The most milk will make it to the end of is as small as possible. The best possible case would be if I drink and replace an infinitesimal bit of milk-and-tea each time. Then the last mug would end with of milk. That’s as in the base of the natural logarithm. Every mathematics problem has an somewhere in it and I’m not exaggerating much. All told this would be about 13 and a half percent of the original milk.
Drinking more realistic amounts, like, half the mug before refilling, makes the milk situation more dire. Replacing half the mug at a time means the last full mug has only one-sixteenth what I started with. Drinking a quarter of the mug and replacing it lets about one-tenth the original milk survive.
But all told the lesson is clear. If I want milk in the last mug, I should put some in each refill. Putting all the milk in at the start and letting it dissolve doesn’t work.
I’ve been taking milk in my tea lately. I have a teapot good for about three cups of tea. So that’s got me thinking about how to keep the most milk in the last of my tea. You may ask why I don’t just get some more milk when I refill the cup. I answer that if I were willing to work that hard I wouldn’t be a mathematician.
It’s easy to spot the lowest amount of milk I could have. If I drank the whole of the first cup, there’d be only whatever milk was stuck by surface tension to the cup for the second. And so even less than that for the third. But if I drank half a cup, poured more tea in, drank half again, poured more in … without doing the calculation, that’s surely more milk for the last full cup.
So what’s the strategy for the most milk I could get in the final cup? And how much is in there?
My friend ChefMongoose pointed out this probability question. As with many probability questions, it comes from a dice game. Here, Yahtzee, based on rolling five dice to make combinations. I’m not sure whether my Twitter problems will get in the way of this embedding working; we’ll see.
Probability help please! You are playing Yahtzee against your insanely competitive spouse. You have two rolls left. You’re trying to get three of a kind. Is it better to commit and roll three dice here? Or split it and roll one die? pic.twitter.com/fi85UYUTUv
Probability help please! You are playing Yahtzee against your insanely competitive spouse. You have two rolls left. You’re trying to get three of a kind. Is it better to commit and roll three dice here? Or split it and roll one die? — Christopher Yost.
Of the five dice, two are showing 1’s; two are showing 2’s; and there’s one last die that’s a 3.
As with many dice questions you can in principle work this out by listing all the possible combinations of every possible outcome. A bit of reasoning takes much less work, but you have to think through the reasons.
And for the last of this year’s (planned) exhumations from my archives? It’s a piece from summer 2017: Zeta Function. As will happen in mathematics, there are many zeta functions. But there’s also one special one that people find endlessly interesting, and that’s what we mean if we say “the zeta function”. It, of course, goes back to Bernhard Riemann.
Also a cute note I saw going around. If you cut off the century years then the date today — the 16th day of the 12th month of the 20th year of the century — you get a rare Pythagorean triplet. and after a moment we notice that’s the famous 3-4-5 Pythagorean triplet all over again. If you miss it, well, that’s all right. There’ll be another along in July of 2025, and one after that in October of 2026.
I mentioned yesterday Iva Sallay’s hosting of the 140th Playful Math Education Blog Carnival. This is a collection of pieces of educational, recreational, or otherwise just delightful mathematics posts. I’d said I hoped I might have the energy to host one again this year and, you know? Denise Gaskins, who organizes this monthly event, took me up on the offer.
So, if you write, or read, or are just aware of a good mathematics or mathematics-related blog, please, leave me a comment! I’ll need all the help I can get finding things worth sharing. Anything that you’ve learned from, or that’s delighted you, is worth it. It’ll teach and delight other people too.
Greetings, friends, and thank you for visiting the 136th installment of Denise Gaskins’s Playful Math Education Blog Carnival. I apologize ahead of time that this will not be the merriest of carnivals. It has not been the merriest of months, even with it hosting Pi Day at the center.
In consideration of that, let me lead with Art in the Time of Transformation by Paula Beardell Krieg. This is from the blog Playful Bookbinding and Paper Works. The post particularly reflects on the importance of creating a thing in a time of trouble. There is great beauty to find, and make, in symmetries, and rotations, and translations. Simple polygons patterned by simple rules can be accessible to anyone. Studying just how these symmetries and other traits work leads to important mathematics. Thus how Kreig’s page has recent posts with names like “Frieze Symmetry Group F7” but also to how symmetry is for five-year-olds. I am grateful to Goldenoj for the reference.
That link was brought to my attention by Iva Sallay, another longtime friend of my little writings here. She writes fun pieces about every counting number, along with recreational puzzles. And asked to share 1458 Tangrams Can Be A Pot of Gold, as an example of what fascinating things can be found in any number. This includes a tangram. Tangrams we see in recreational-mathematics puzzles based on ways that you can recombine shapes. It’s always exciting to be able to shift between arithmetic and shapes. And that leads to a video and related thread again pointed to me by goldenoj …
This video, by Mathologer on YouTube, explains a bit of number theory. Number theory is the field of asking easy questions about whole numbers, and then learning that the answers are almost impossible to find. I exaggerate, but it does often involve questions that just suppose you understand what a prime number should be. And then, as the title asks, take centuries to prove.
Fermat’s Two-Squares Theorem, discussed here, is not the famous one about . Pierre de Fermat had a lot of theorems, some of which he proved. This one is about prime numbers, though, and particularly prime numbers that are one more than a multiple of four. This means it’s sometimes called Fermat’s 4k+1 Theorem, which is the name I remember learning it under. (k is so often a shorthand for “some counting number” that people don’t bother specifying it, the way we don’t bother to say “x is an unknown number”.) The normal proofs of this we do in the courses that convince people they’re actually not mathematics majors.
What the video offers is a wonderful alternate approach. It turns key parts of the proof into geometry, into visual statements. Into sliding tiles around and noticing patterns. It’s also a great demonstration of one standard problem-solving tool. This is to look at a related, different problem that’s easier to say things about. This leads to what seems like a long path from the original question. But it’s worth it because the path involves thinking out things like “is the count of this thing odd or even”? And that’s mathematics that you can do as soon as you can understand the question.
I again thank Iva Sallay for that link, as well as this essay. Dan Meyer’s But Artichokes Aren’t Pinecones: What Do You Do With Wrong Answers? looks at the problem of students giving wrong answers. There is no avoiding giving wrong answers. A parent’s or teacher’s response to wrong answers will vary, though, and Meyer asks why that is. Meyer has some hypotheses. His example notes that he doesn’t mind a child misidentifying an artichoke as a pinecone. Not in the same way identifying the sum of 1 and 9 as 30 would. What is different about those mistakes?
Jessannwa’s Soft Start In The Intermediate Classroom looks to the teaching of older students. No muffins and cookies here. That the students might be more advanced doesn’t change the need to think of what they have energy for, and interest in. She discusses a class setup that’s meant to provide structure in ways that don’t feel so authority-driven. And ways to turn practicing mathematics problems into optimizing game play. I will admit this is a translation of the problem which would have worked well for me. But I also know that not everybody sees a game as, in part, something to play at maximum efficiency. It depends on the game, though. They’re on Twitter as @jesannwa.
These are thoughts about how anyone can start learning mathematics. What does it look like to have learned a great deal, though, to the point of becoming renowned for it? Life Through A Mathematician’s Eyes posted Australian Mathematicians in late January. It’s a dozen biographical sketches of Australian mathematicians. It also matches each to charities or other public-works organizations. They were trying to help the continent through the troubles it had even before the pandemic struck. They’re in no less need for all that we’re exhausted. The page’s author is on Twitter as @lthmath.
I have since the start of this post avoided mentioning the big mathematical holiday of March. Pi Day had the bad luck to fall on a weekend this year, and then was further hit by the Covid-19 pandemic forcing the shutdown of many schools. Iva Sallay again helped me by noting YummyMath’s activities page It’s Time To Gear Up For Pi Day. This hosts several worksheets, about the history of π and ways to calculate it, and several formulas for π. This even gets into interesting techniques like how to use continued fractions in finding a numerical value.
Rolands Rag Bag shared A Pi-Ku for Pi-Day featuring a poem written in a form I wasn’t aware anyone did. The “Pi-Ku” as named here has 3 syllables for the first time, 1 syllable in the second line, 4 syllables in the third line, 1 syllable the next line, 5 syllables after that … you see the pattern. (One of Avery’s older poems also keeps this form.) The form could, I suppose, go on to as many lines as one likes. Or at least to the 40th line, when we would need a line of zero syllables. Probably one would make up a rule to cover that.
So the first bit of news: I’m hosting the Playful Math Education Blog Carnival later this month. This is a roaming blog link party, sharing blogs that delight or educate, or ideally both, about mathematics. As mentioned the other day Iva Sallay of Find the Factors hosted the 135th of these. My entry, the 136th, I plan to post sometime the last week of March.
And I’ll need help! If you’ve run across a web site, YouTube video, blog post, or essay that discusses something mathematical in a way that makes you grin, please let me know, and let me share it with the carnival audience.
This Saturday is March 14th, which we’ve been celebrating as Pi Day. I remain skeptical that it makes a big difference in people’s view of mathematics or in their education. But an afternoon spent talking about mathematics with everyone agreeing that, for today, we won’t complain about how hard it always was or how impossible we always found it, is pleasant. And that’s a good thing. I don’t know how much activity there’ll be for it, since the 14th is a weekend day this year. And the Covid-19 problem has got all the schools in my state closed through to April, so any calendar relevance is shattered.
But I have some things in the archive anyway. Last year I gathered Six Or Arguably Four Things For Pi Day, a collection of short essays about ways to calculate π well or poorly, and about some of the properties we’re pretty sure that π has, even if we can’t prove it. Also this fascinating physics problem that yields the digits of π.
And the middle of March often brings out Comic Strip Master Command. It looks like I’ve had at least five straight Pi Day editions of Reading the Comics, although most of them cover strips from more than just the 14th of March. From the past:
I apologize that obligations have kept me from writing some things that I mean to. So let me just point you to Iva Sallay, whose Find the Factor recreational math puzzle page hosted the 135th Playful Math Education Blog Carnival this past month. The Blog Carnival is a fun roaming thing that I’ve hosted once, and do hope to host again. It’s a curated collection of other mathematics sites that are fun or interesting or hopefully both together.
I ran across something neat. It’s something I’ve seen before, but the new element is that I have a name for it. This is the Golomb Ruler. It’s a ruler made with as few marks as possible. The marks are supposed to be arranged so that the greatest possible number of different distances can be made, by measuring between selected pairs of points.
So, like, in a regularly spaced ruler, you have a lot of ways to measure a distance of 1 unit of length. Only one fewer way to measure a distance of 2 units. One fewer still ways to measure a distance of 3 units and so on. Convenient but wasteful of marks. A Golomb ruler might, say, put marks only where the regularly spaced ruler has the units 1, 2, and 4. Then by choosing the correct pairs you can measure a distance of 1, 2, 3, or 4 units.
There’s applications of the Golomb ruler, stuff in information theory and sensor design and stuff. Also logistics. Never mind those. They present a neat little puzzle: can you find, for a given number of marks, the best possible arrangement of them into a ruler? That would be the arrangement that allows the greatest number of different lengths. Or perhaps the one that allows the longest string of whole-number differences. Your definition of best-possible determines what the answer is.
And as this suggests, you aren’t going to discover an optimal arrangement for some number of marks yourself. Unless you should be the first person to figure out an algorithm to do it. It’s not even known how complex an algorithm has to be. It’s suspected that it has to be NP-hard, though. But, while you won’t discover anything new to mathematics in pondering this, you can still have the fun of working out arrangements yourself, at least for a handful of points. There are numbers of points with more than one optimal arrangement.
(Golomb here is Solomon W Golomb, a mathematician and electrical engineer with a long history in information theory and also recreational mathematics problems. There are several parties who independently invented the problem. But Golomb actually did work with rulers, so at least they aren’t incorrectly named.)
I have another subject nominated by goldenoj today. And it even lets me get into number theory, the field of mathematics questions that everybody understands and nobody can prove.
I was once a young grad student working as a teaching assistant and unaware of the principles of student privacy. Near the end of semesters I would e-mail students their grades. This so they could correct any mistakes and know what they’d have to get on the finals. I was learning Perl, which was an acceptable pastime in the 1990s. So I wrote scripts that would take my spreadsheet of grades and turn it into e-mails that were automatically sent. And then I got all fancy.
It seemed boring to send out completely identical form letters, even if any individual would see it once. Maybe twice if they got me for another class. So I started writing variants of the boilerplate sentences. My goal was that every student would get a mass-produced yet unique e-mail. To best the chances of this I had to make sure of something about all these variant sentences and paragraphs.
So you see the trick. I needed a set of relatively prime numbers. That way, it would be the greatest possible number of students before I had a completely repeated text. We know what prime numbers are. They’re the numbers that, in your field, have exactly two factors. In the counting numbers the primes are numbers like 2, 3, 5, 7 and so on. In the Gaussian integers, these are numbers like 3 and 7 and . But not 2 or 5. We can look to primes among the polynomials. Among polynomials with rational coefficients, is prime. So is . is not.
The idea of relative primes appears wherever primes appears. We can say without contradiction that 4 and 9 are relative primes, among the whole numbers. Though neither’s prime, in the whole numbers, neither has a prime factor in common. This is an obvious way to look at it. We can use that definition for any field that has a concept of primes. There are others, though. We can say two things are relatively prime if there’s a linear combination of them that adds to the identity element. You get a linear combination by multiplying each of the things by a scalar and adding these together. Multiply 4 by -2 and 9 by 1 and add them and look what you get. Or, if the least common multiple of a set of elements is equal to their product, then the elements are relatively prime. Some make sense only for the whole numbers. Imagine the first quadrant of a plane, marked in Cartesian coordinates. Draw the line segment connecting the point at (0, 0) and the point with coordinates (m, n). If that line segment touches no dots between (0, 0) and (m, n), then the whole numbers m and n are relatively prime.
We start looking at relative primes as pairs of things. We can be interested in larger sets of relative primes, though. My little e-mail generator, for example, wouldn’t work so well if any pair of sentence replacements were not relatively prime. So, like, the set of numbers 2, 6, 9 is relatively prime; all three numbers share no prime factors. But neither the pair 2, 6 and the pair 6, 9 are not relatively prime. 2, 9 is, at least there’s that. I forget how many replaceable sentences were in my form e-mails. I’m sure I did the cowardly thing, coming up with a prime number of alternate ways to phrase as many sentences as possible. As an undergraduate I covered the student government for four years’ worth of meetings. I learned a lot of ways to say the same thing.
Which is all right, but are relative primes important? Relative primes turn up all over the place in number theory, and in corners of group theory. There are some thing that are easier to calculate in modulo arithmetic if we have relatively prime numbers to work with. I know when I see modulo arithmetic I expect encryption schemes to follow close behind. Here I admit I’m ignorant whether these imply things which make encryption schemes easier or harder.
Some of the results are neat, certainly. Suppose that the function f is a polynomial. Then, if its first derivative f’ is relatively prime to f, it turns out f has no repeated roots. And vice-versa: if f has no repeated roots, then it and its first derivative are relatively prime. You remember repeated roots. They’re factors like , that foiled your attempt to test a couple points and figure roughly where a polynomial crossed the x-axis.
I mentioned that primeness depends on the field. This is true of relative primeness. Polynomials really show this off. (Here I’m using an example explained in a 2007 Ask Dr Math essay.) Is the polynomial relatively prime to ?
It is, if we are interested in polynomials with integer coefficients. There’s no linear combination of and which gets us to 1. Go ahead and try.
It is not, if we are interested in polynomials with rational coefficients. Multiply by and multiply by . Then add those up.
Tell me what polynomials you want to deal with today and I will tell you which answer is right.
This may all seem cute if, perhaps, petty. A bunch of anonymous theorems dotting the center third of an abstract algebra text will inspire that. The most important relative-primes thing I know of is the abc conjecture, posed in the mid-80s by Joseph Oesterlé and David Masser. Start with three counting numbers, a, b, and c. Require that a + b = c.
There is a product of the unique prime factors of a, b, and c. That is, let’s say a is 36. This is 2 times 2 times 3 times 3. Let’s say b is 5. This is prime. c is 41; it’s prime. Their unique prime factors are 2, 3, 5, and 41; the product of all these is 1,230.
The conjecture deals with this product of unique prime factors for this relatively prime triplet. Almost always, c is going to be smaller than this unique prime factors product. The conjecture says that there will be, for every positive real number , at most finitely many cases where c is larger than this product raised to the power . I do not know why raising this product to this power is so important. I assume it rules out some case where this product raised to the first power would be too easy a condition.
Apart from that bit, though, this is a classic sort of number theory conjecture. Like, it involves some technical terms, but nothing too involved. You could almost explain it at a party and expect to be understood, and to get some people writing down numbers, testing out specific cases. Nobody will go away solving the problem, but they’ll have some good exercise and that’s worthwhile.
And it has consequences. We do not know whether the abc conjecture is true. We do know that if it is true, then a bunch of other things follow. The one that a non-mathematician would appreciate is that Fermat’s Last Theorem would be provable by an alterante route. The abc conjecture would only prove the cases for Fermat’s Last Theorem for powers greater than 5. But that’s all right. We can separately work out the cases for the third, fourth, and fifth powers, and then cover everything else at once. (That we know Fermat’s Last Theorem is true doesn’t let us conclude the abc conjecture is true, unfortunately.)
There are other implications. Some are about problems that seem like fun to play with. If the abc conjecture is true, then for every integer A, there are finitely many values of n for which is a perfect square. Some are of specialist interest: Lang’s conjecture, about elliptic curves, would be true. This is a lower bound for the height of non-torsion rational points. I’d stick to the stuff at a party. A host of conjectures about Diophantine equations — (high school) algebra problems where only integers may be solutions — become theorems. Also coming true: the Fermat-Catalan conjecture. This is a neat problem; it claims that the equation
where a, b, and c are relatively prime, and m, n, and k are positive integers satisfying the constraint
has only finitely many solutions with distinct triplets . The inequality about reciprocals of m, n, and k is needed so we don’t have boring solutions like clogging us up. The bit about distinct triplets is so we don’t clog things up with a or b being 1 and then technically every possible m or n giving us a “different” set. To date we know something like ten solutions, one of them having a equal to 1.
Another implication is Pillai’s Conjecture. This one asks whether every positive integer occurs only finitely many times as the difference between perfect powers. Perfect powers are, like 32 (two to the fifth power) or 81 (three to the fourth power) or such.
So as often happens when we stumble into a number theory thing, the idea of relative primes is easy. And there are deep implications to them. But those in turn give us things that seem like fun arithmetic puzzles.
I’m more fluent in graph theory, and my writing will reflect that. But its critical insight involves looking at spaces and ignoring things like distance and area and angle. It is amazing that one can discard so much of geometry and still have anything to consider. What we do learn then applies to very many problems.
Königsberg Bridge Problem.
Once upon a time there was a city named Königsberg. It no longer is. It is Kaliningrad now. It’s no longer in that odd non-contiguous chunk of Prussia facing the Baltic Sea. It’s now in that odd non-contiguous chunk of Russia facing the Baltic Sea.
I put it this way because what the city evokes, to mathematicians, is a story. I do not have specific reason to think the story untrue. But it is a good story, and as I think more about history I grow more skeptical of good stories. A good story teaches, though not always the thing it means to convey.
The story is this. The city is on two sides of the Pregel river, now the Pregolya River. Two large islands are in the river. For several centuries these four land masses were connected by a total of seven bridges. And we are told that people in the city would enjoy free time with an idle puzzle. Was there a way to walk all seven bridges one and only one time? If no one did something fowl like taking a boat to cross the river, or not going the whole way across a bridge, anyway? There were enough bridges, though, and enough possible ways to cross them, that trying out every option was hopeless.
Then came Leonhard Euler. Who is himself a preposterous number of stories. Pick any major field of mathematics; there is an Euler’s Theorem at its center. Or an Euler’s Formula. Euler’s Method. Euler’s Function. Likely he brought great new light to it.
And in 1736 he solved the Königsberg Bridge Problem. The answer was to look at what would have to be true for a solution to exist. He noticed something so obvious it required genius not to dismiss it. It seems too simple to be useful. In a successful walk you enter each land mass (river bank or island) the same number of times you leave it. So if you cross each bridge exactly once, you use an even number of bridges per land mass. The exceptions are that you must start at one land mass, and end at a land mass. Maybe a different one. How you get there doesn’t count for the problem. How you leave doesn’t either. So the land mass you start from may have an odd number of bridges. So may the one you end on. So there are up to two land masses that may have an odd number of bridges.
Once this is observed, it’s easy to tell that Königsberg’s Bridges did not match that. All four land masses in Königsberg have an odd number of bridges. And so we could stop looking. It’s impossible to walk the seven bridges exactly once each in a tour, not without cheating.
Graph theoreticians, like the topologists of my prologue, now consider this foundational to their field. To look at a geographic problem and not concern oneself with areas and surfaces and shapes? To worry only about how sets connect? This guides graph theory in how to think about networks.
The city exists, as do the islands, and the bridges existed as described. So does Euler’s solution. And his reasoning is sound. The reasoning is ingenious, too. Everything hard about the problem evaporates. So what do I doubt about this fine story?
Teo Paoletti, author of that web page, says Danzig mayor Carl Leonhard Gottlieb Ehler wrote Euler, asking for a solution. This falls short of proving that the bridges were a common subject of speculation. It does show at least that Ehler thought it worth pondering. Euler apparently did not think it was even mathematics. Not that he thought it was hard; he simply thought it didn’t depend on mathematical principles. It took only reason. But he did find something interesting: why was it not mathematics? Paoletti quotes Euler as writing:
This question is so banal, but seemed to me worthy of attention in that [neither] geometry, nor algebra, nor even the art of counting was sufficient to solve it.
I am reminded of a mathematical joke. It’s about the professor who always went on at great length about any topic, however slight. I have no idea why this should stick with me. Finally one day the professor admitted of something, “This problem is not interesting.” The students barely had time to feel relief. The professor went on: “But the reasons why it is not interesting are very interesting. So let us explore that.”
The Königsberg Bridge Problem is in the first chapter of every graph theory book ever. And it is a good graph theory problem. It may not be fair to say it created graph theory, though. Euler seems to have treated this as a little side bit of business, unrelated to his real mathematics. Graph theory as we know it — as a genre — formed in the 19th century. So did topology. In hindsight we can see how studying these bridges brought us good questions to ask, and ways to solve them. But for something like a century after Euler published this, it was just the clever solution to a recreational mathematics puzzle. It was as important as finding knight’s tours of chessboards.
That we take it as the introduction to graph theory, and maybe topology, tells us something. It is an easy problem to pose. Its solution is clever, but not obscure. It takes no long chains of complex reasoning. Many people approach mathematics problems with fear. By telling this story, we promise mathematics that feels as secure as a stroll along the riverfront. This promise is good through about chapter three, section four, where there are four definitions on one page and the notation summons obscure demons of LaTeX.
Still. Look at what the story of the bridges tells us. We notice something curious about our environment. The problem seems mathematical, or at least geographic. The problem is of no consequence. But it lingers in the mind. The obvious approaches to solving it won’t work. But think of the problem differently. The problem becomes simple. And better than simple. It guides one to new insights. In a century it gives birth to two fields of mathematics. In two centuries these are significant fields. They’re things even non-mathematicians have heard of. It’s almost a mathematician’s fantasy of insight and accomplishment.
But this does happen. The world suggests no end of little mathematics problems. Sometimes they are wonderful. Richard Feynman’s memoirs tell of his imagination being captured by a plate spinning in the air. Solving that helped him resolve a problem in developing Quantum Electrodynamics. There are more mundane problems. One of my professors in grad school remembered tossing and catching a tennis racket and realizing he didn’t know why sometimes it flipped over and sometimes didn’t. His specialty was in dynamical systems, and he could work out the mechanics of what a tennis racket should do, and when. And I know that within me is the ability to work out when a pile of books becomes too tall to stand on its own. I just need to work up to it.
The story of the Königsberg Bridge Problem is about this. Even if nobody but the mayor of Danzig pondered how to cross the bridges, and he only got an answer because he infected Euler with the need to know? It is a story of an important piece of mathematics. Good stories will tell us things that are true, which are not necessarily the things that happen in them.
A friend was playing with that cute little particle-physics simulator idea I mentioned last week. And encountered a problem. With a little bit of thought, I was able to not solve the problem. But I was able to explain why it was a subtler and more difficult problem than they had realized. These are the moments that make me feel justified calling myself a mathematician.
The proposed simulation was simple enough: imagine a bunch of particles that interact by rules that aren’t necessarily symmetric. Like, the attraction particle A exerts on particle B isn’t the same as what B exerts on A. Or there are multiple species of particles. So (say) red particles are attracted to blue but repelled by green. But green is attracted to red and repelled by blue twice as strongly as red is attracted to blue. Your choice.
Give a mathematician a perfectly good model of something. She’ll have the impulse to try tinkering with it. One reliable way to tinker with it is to change the domain on which it works. If your simulation supposes you have particles moving on the plane, then, what if they were in space instead? Or on the surface of a sphere? Or what if something was strange about the plane? My friend had this idea: what if the particles were moving on the surface of a cube?
And the problem was how to find the shortest distance between two particles on the surface of a cube. The distance matters since most any attraction rule depends on the distance. This may be as simple as “particles more than this distance apart don’t interact in any way”. The obvious approach, or if you prefer the naive approach, is to pretend the cube is a sphere and find distances that way. This doesn’t get it right, not if the two points are on different faces of the cube. If they’re on adjacent faces, ones which share an edge — think the floor and the wall of a room — it seems straightforward enough. My friend got into trouble with points on opposite faces. Think the floor and the ceiling.
Inside a rectangular room, measuring 30 feet in length and 12 feet in width and height, a spider is at a point on the middle of one of the end walls, 1 foot from the ceiling, as at A; and a fly is on the opposite wall, 1 foot from the floor in the centre, as shown at B. What is the shortest distance that the spider must crawl in order to reach the fly, which remains stationary? Of course the spider never drops or uses its web, but crawls fairly.
(Also I admire Dudeney’s efficient closing off of the snarky, problem-breaking answer someone was sure to give. It suggests experienced thought about how to pose problems.)
What makes this a puzzle, even a paradox, is that the obvious answer is wrong. At least, what seems like the obvious answer is to start at point A, move to one of the surfaces connecting the spider’s and the fly’s starting points, and from that move to the fly’s surface. But, no: you get a shorter answer by using more surfaces. Going on a path that seems like it wanders more gets you a shorter distance. The solution’s presented here, along with some follow-up problems. In this case, the spider’s shortest path uses five of the six surfaces of the room.
The approach to finding this is an ingenious one. Imagine the room as a box, and unfold it into something flat. Then find the shortest distance on that flat surface. Then fold the box back up. It’s a good trick. It turns out to be useful in many problems. Mathematical physicists often have reason to ponder paths of things on flattenable surfaces like this. Sometimes they’re boxes. Sometimes they’re toruses, the shape of a doughnut. This kind of unfolding often makes questions like “what’s the shortest distance between points” easier to solve.
There are wrinkles to the unfolding. Of course there are. How interesting would it be if there weren’t? The wrinkles amount to this. Imagine you start at the corner of the room, and walk up a wall at a 45 degree angle to the horizon. You’ll get to the far corner eventually, if the room has proportions that allow it. All right. But suppose you walked up at an angle of 30 degrees to the horizon? At an angle of 75 degrees? You’ll wind your way around the walls (and maybe floor and ceiling) some number of times, each path you start with. Probably different numbers of times. Some path will be shortest, and that’s fine. But … like, think about the path that goes along the walls and ceiling and floor three times over. The room, unfolded into a flat panel, has only one floor and one ceiling and each wall once. The straight line you might be walking goes right off the page.
And this is the wrinkle. You might need to tile the room. In a column of blocks (like in Dudeney’s solution) every fourth block might be the floor, with, between any two of them, a ceiling. This is fine, and what’s needed. It can be a bit dizzying to imagine such a state of affairs. But if you’ve ever zoomed a map of the globe out far enough that you see Australia six times over then you’ve understood how this works.
I cannot attest that this has helped my friend in the slightest. I am glad that my friend wanted to think about the surface of the cube. The surface of a dodecahedron would be far, far past my ability to help with.
Next bit is an article that relates to my years-long odd interest in pasta making. Mathematicians solve age-old spaghetti mystery reports a group of researchers at MIT — the renowned “Rensselaer Polytechnic Institute of Boston” [*] — studying why dry spaghetti fractures the way it does. Like many great problems, it sounds ridiculous to study at first. Who cares why, basically, you can’t snap a dry spaghetti strand in two equal pieces by bending it at the edges? The problem has familiarity to it and seems to have little else. But then you realize this is a matter of how materials work, and how they break. And realize it’s a great question. It’s easy to understand and subtle to solve.
The article finishes with (as teased in the tweet above) a report of an electric toothbrush that should keep track of positions inside the user’s head, even as the head rotates. This is intriguing. I say as a person who’s reluctantly started using an electric toothbrush. I’m one of those who brushes, manually, too hard, to the point of damaging my gums. The electric toothbrush makes that harder to do. I’m not sure how an orientation-aware electric toothbrush will improve the situation any, but I’m open-minded.
[*] I went to graduate school at Rensselaer Polytechnic Institute, the “RPI of New York”. The school would be a rival to MIT if RPI had any self-esteem. I’m guessing, as I never went to a school that had self-esteem.
Greetings one and all! Come, gather round! Wonder and spectate and — above all else — tell your friends of the Playful Mathematics Blog Carnival! Within is a buffet of delights and treats, fortifications for the mind and fire for the imagination.
121 is a special number. When I was a mere tot, growing in the wilds of suburban central New Jersey, it stood there. It held a spot of privilege in the multiplication tables on the inside front cover of composition books. On the forward diagonal, yet insulated from the borders. It anchors the safe interior. A square number, eleventh of that set in the positive numbers.
121 is more than just a square. It is the lone square known to be the sum of the first several powers of a prime number: it is , a fantastic combination. If there is another square that is such a sum of primes, it is unknown to any human — and must be at least 35 digits long.
We look now for a moment at some astounding animals. From the renowned Dr Nic: Introducing Cat Maths cards, activities, games and lessons — a fine collection of feline companions, such toys as will enterain them. A dozen attributes each; twenty-seven value cards. These cats, and these cards, and these activity puzzles, promise games and delights, to teach counting, subtraction, statistics, and inference!
Next and no less incredible is the wooly Mathstodon. Christian Lawson-Perfect hosts this site, an instance of the open-source Twitter-like service Mastodon. Its focus: a place for people interested in mathematicians to write of what they know. To date over 1,300 users have joined, and have shared nearly 25,000 messages. You need not join to read many of these posts — your host here has yet to — but may sample its wares as you like.
The Second Tent
121 is one of only two perfect squares known to be four less than the cube of a whole number. The great Fermat conjectured that 4 and 121 are the only such numbers; no one has found a counter-example. Nor a proof.
Friends, do you know the secret to popularity? There is an astonishing truth behind it. Elias Worth of the MathSection blog explains the Friendship Paradox. This mind-warping phenomenon tells us your friends have more friends than you do. It will change forever how you look at your followers and following accounts.
And now to thoughts of learning. Stepping forward now is Monica Utsey, @Liveonpurpose47 of Chocolate Covered Boy Joy. Her declaration: “I incorporated Montessori Math materials with my right brain learner because he needed literal representations of the work we were doing. It worked and we still use it.” See now for yourself the representations, counting and comparing and all the joys of several aspects of arithmetic.
Take now a moment for your own fun. Blog Carnival patron and organizer Denise Gaskins wishes us to know: “The fun of mathematical coloring isn’t limited to one day. Enjoy these coloring resources all year ’round!” Happy National Coloring Book Day offers the title, and we may keep the spirit of National Coloring Book Day all the year round.
121 is a star number, the fifth of that select set. 121 identical items can be tiled to form a centered hexagon. You may have seen it in the German game of Chinese Checkers, as the board of that has 121 holes.
We come back again to teaching. “Many homeschoolers struggle with teaching their children math. Here are some tips to make it easier”, offers Denise Gaskins. Step forth and benefit from this FAQ: Struggling with Arithmetic, a collection of tips and thoughts and resources to help make arithmetic the more manageable.
Step now over to the arcade, and to the challenge of Pac-Man. This humble circle-inspired polygon must visit the entirety of a maze, and avoid ghosts as he does. Matthew Scroggs of Chalk Dust Magazine here seeks and shows us Optimal Pac-Man. Graph theory tells us there are thirteen billion different paths to take. Which of them is shortest? Which is fastest? Can it be known, and can it help you through the game?
121 is furthermore the sixth of the centered octagonal numbers. 121 of a thing may be set into six concentric octagons of one, then two, then three, then four, then five, and then six of them on a side.
Step now back to the amazing Mathstodon. Gaze in wonder at the account @dudeney_puzzles. Since the September of 2017 it has brought out challenges from Henry Ernest Dudeney’s Amusements in Mathematics. Puzzles given, yes, with answers that follow along. The impatient may find Dudeney’s 1917 book on Project Gutenberg among other places.
The Fifth Tent
Sum the digits of 121; you will find that you have four. Take its prime factors, 11 and 11, and sum their digits; you will find that this is four again. This makes 121 a Smith number. These marvels of the ages were named by Albert Wilansky, in honor of his brother-in-law, a man known to history as Harold Smith, and whose telephone number of 4,937,775 was one such.
And now to an astounding challenge. Imagine an assassin readies your death. Can you protect yourself? At all? Tai-Danae Bradley invites you to consider: Is the Square a Secure Polygon? This question takes you on a tour of geometries familiar and exotic. Learn how mathematicians consider how to walk between places on a torus — and the lessons this has for a square room. The fate of the universe itself may depend on the methods described herein — the techniques used to study it relate to those that study whether a physical system can return to its original state. And then J2kun turned this into code, Visualizing an Assassin Puzzle, for those who dare to program it.
Have you overcome this challenge? Then step into the world of linear algebra, and this delight from the Mathstodon account of Christian Lawson-Perfect. The puzzle is built on the wonders of eigenvectors, those marvels of matrix multiplication. They emerge from multiplication longer or shorter but unchanged in direction. Lawson-Perfect uses whole numbers, represented by Scrabble tiles, and finds a great matrix with a neat eigenvalue. Can you prove that this is true?
The Sixth Tent
Another wonder of the digits of 121. Take them apart, then put them together again. Contorted into the form 112 they represent the same number. 121 is, in the base ten commonly used in the land, a Friedman Number, second of that line. These marvels, in the Arabic, the Roman, or even the Mayan numerals schemes, are named for Erich Friedman, a figure of mystery from the Stetson University.
Denise Gaskins coordinates the Playful Mathematics Education Blog Carnival. Upcoming scheduled carnivals, including the chance to volunteer to host it yourself, or to recommend your site for mention, are listed here. And October’s 122nd Playful Mathematics Education Blog Carnival is scheduled to be hosted by Arithmophobia No More, and may this new host have the best of days!
I apologize for falling even more silent than usual, and shall get to reviewing the past week’s comic strips soon. I had a big pile of life land on me, although, not so big a pile as landed on other people.
I apologize for not having a more robust introduction here. My week’s been chopped up by concern with the health of the older of our rabbits. Today’s proved to be less alarming than we had feared, but it’s still a lot to deal with. I appreciate your kind thoughts. Thank you.
Meanwhile the comics from last week have led me to discover something really weird going on with the Mutt and Jeff reruns.
Charles Schulz’s Peanuts Classics for the 6th has the not-quite-fully-formed Lucy trying to count the vast. She’d spend a while trying to count the stars and it never went well. It does inspire the question of how to count things when doing a simple tally is too complicated. There are many mathematical approaches. Most of them are some kind of sampling. Take a small enough part that you can tally it, and estimate the whole based on what your sample is. This can require ingenuity. For example, when estimating our goldfish population, it was impossible to get a good sample at one time. When tallying the number of visible stars in the sky, we have the problem that the Galaxy has a shape, and there are more stars in some directions than in others. This is why we need statisticians.
Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 6th looks initially like it’s meant for a philosophy blog’s Reading the Comics post. It’s often fruitful in the study of ethics to ponder doing something that is initially horrible, but would likely have good consequences. Or something initially good, but that has bad effects. These questions challenge our ideas about what it is to do good or bad things, and whether transient or permanent effects are more important, and whether it is better to be responsible for something (or to allow something) by action or inaction.
It comes to mathematics in the caption, though, and with an assist from the economics department. Utilitarianism seems to offer an answer to many ethical problems. It posits that we need to select a primary good of society, and then act so as to maximize that good. This does have an appeal, I suspect even to people who don’t thrill of the idea of finding the formula that describes society. After all, if we know the primary good of society, why should we settle for anything but the greatest value of that good? It might be difficult in practice, say, to discount the joy a musician would bring over her lifetime with her performances fairly against the misery created by making her practice the flute after school when she’d rather be playing. But we can imagine working with a rough approximation, at least. Then the skilled thinkers point out even worse problems and we see why utilitarianism didn’t settle all the big ethical questions, even in principle.
The mathematics, though. As Weinersmith’s caption puts it, we can phrase moral dilemmas as problems of maximizing evil. Typically we pose them as ones of maximizing good. Or at least of minimizing evil. But if we have the mechanism in place to find where evil is maximized, don’t we have the tools to find where good is? If we can find the set of social parameters x, y, and z which make E(x, y, z) as big as possible, can’t we find where -E(x, y, z) is as big, too? And isn’t that then where E(x, y, z) has to be smallest?
And, sure. As long as the maximum exists, or the minimum exists. Maybe we can tell whether or not there is one. But this is why when you look at the mathematics of finding maximums you realize you’re also doing minimums, or vice-versa. Pretty soon you either start referring to what you find as extremums. Or you stop worrying about the difference between a maximum and a minimum, at least unless you need to check just what you have found. Or unless someone who isn’t mathematically expert looks at you wondering if you know the difference between positive and negative numbers.
Bud Fisher’s Mutt and Jeff for the 7th has run here before. Except that was before they redid the lettering; it was a roast beef in earlier iterations. I was thinking to drop Mutt and Jeff from my Reading the Comics routine before all these mysteries in the lettering turned up. Anyway. The strip’s joke starts with a work-rate problems. Given how long some people take to do a thing, how long does it take a different number of people to do a thing? These are problems that demand paying attention to units, to the dimensions of a thing. That seems to be out of fashion these days, which is probably why these questions get to be baffling. But if eating a ham takes 25 person-minutes to do, and you have ten persons eating, you can see almost right away how long to expect it to take. If the ham’s the same size, anyway.
Olivia Jaimes’s Nancy for the 7th is built on a spot of recreational mathematics. Also on the frustration one can have when a problem looks like it’s harmless innocent fun and turns out to take just forever and you’re never sure you have the answers just right. The commenters on GoComics.com have settled on 18. I’m content with that answer.
My friend’s finished the last of the exams and been happy with the results. And I’m stuck thinking harder about a little thing that came across my Twitter feed last night. So let me share a different problem that we had discussed over the term.
It’s a probability question. Probability’s a great subject. So much of what people actually do involves estimating probabilities and making judgements based on them. In real life, yes, but also for fun. Like a lot of probability questions, this one is abstracted into a puzzle that’s nothing like anything anybody does for fun. But that makes it practical, anyway.
So. You have a bowl with fifteen balls inside. Five of the balls are labelled ‘1’. Five of the balls are labelled ‘2’. Five of the balls are labelled ‘3’. The balls are well-mixed, which is how mathematicians say that all of the balls are equally likely to be drawn out. Three balls are picked out, without being put back in. What’s the probability that the three balls have values which, together, add up to 6?
My friend’s instincts about this were right, knowing what things to calculate. There was part of actually doing one of these calculations that went wrong. And was complicated by my making a dumb mistake in my arithmetic. Fortunately my friend wasn’t shaken by my authority, and we got to what we’re pretty sure is the right answer.
Their plan was to make more exciting the discussion of some of Deep Space Nine‘s episodes by recording their reviews while drinking a lot. The plan was, for the fifteen episodes they had in the season, there would be a one-in-fifteen chance of doing any particular episode drunk. So how many drunk episodes would you expect to get, on this basis?
It’s a well-formed expectation value problem. There could be as few as zero or as many as fifteen, but some cases are more likely than others. Each episode could be recorded drunk or not-drunk. There’s an equal chance of each episode being recorded drunk. Whether one episode is drunk or not doesn’t depend on whether the one before was, and doesn’t affect whether the next one is. (I’ll come back to this.)
The most likely case was for there to be one drunk episode. The probability of exactly one drunk episode was a little over 38%. No drunk episodes was also a likely outcome. There was a better than 35% chance it would never have turned up. The chance of exactly two drunk episodes was about 19%. There drunk episodes had a slightly less than 6% chance of happening. Four drunk episodes a slightly more than 1% chance of happening. And after that you get into the deeply unlikely cases.
As the Deep Space Nine season turned out, this one-in-fifteen chance came up twice. It turned out they sort of did three drunk episodes, though. One of the drunk episodes turned out to be the first of two they planned to record that day. I’m not sure why they didn’t just swap what episode they recorded first, but I trust they had logistical reasons. As often happens with probability questions, the independence of events — whether a success for one affects the outcome of another — changes calculations.
There’s not going to be a second-season update to this. They’ve chosen to make a more elaborate recording game of things. They’ve set up a modified Snakes and Ladders type board with a handful of spots marked for stunts. Some sound like fun, such as recording without taking any notes about the episode. Some are, yes, drinking episodes. But this is all a very different and more complicated thing to project. If I were going to tackle that it’d probably be by running a bunch of simulations and taking averages from that.
Also I trust they’ve been warned about the episode where Quark has a sex change so he can meet a top Ferengi soda magnate after accidentally giving his mother a heart attack because gads but that was a thing that happened somehow.
I’m slow about sharing them is all. It’s a simple dynamic: I want to write enough about each tweet that it’s interesting to share, and then once a little time has passed, I need to do something more impressive to be worth the wait. Eventually, nothing is ever shared. Let me try to fix that.
Just as it says: a link to Leonhard Euler’s Elements of Algebra, as rendered by Google Books. Euler you’ll remember from every field of mathematics ever. This 1770 textbook is one of the earliest that presents algebra that looks like, you know, algebra, the way we study it today. Much of that is because this book presented algebra so well that everyone wanted to imitate it.
This Theorem of the Day from back in November already is one about elliptic functions. Those came up several times in the Summer 2017 Mathematics A To Z. This day about the Goins-Maddox-Rusin Theorem on Heron Triangles, is dense reading even by the standards of the Theorem of the Day tweet (which fits each day’s theorem into a single slide). Still, it’s worth lounging about in the mathematics.
Elke Stangl, writing about one of those endlessly-to-me interesting subjects: phase space. This is a particular way of representing complicated physical systems. Set it up right and all sorts of physics problems become, if not easy, at least things there’s a standard set of tools for. Thermodynamics really encourages learning about such phase spaces, and about entropy, and here she writes about some of this.
Non-limit calculating e by hand. https://t.co/Kv80RotboJ Fun activity & easily reproducible. Anyone know the author?
So ‘e’ is an interesting number. At least, it’s a number that’s got a lot of interesting things built around it. Here, John Golden points out a neat, fun, and inefficient way to find the value of ‘e’. It’s kin to that scheme for calculating π inefficiently that I was being all curmudgeonly about a couple of Pi Days ago.
Jo Morgan comes to the rescue of everyone who tries to read old-time mathematics. There were a lot of great and surprisingly readable great minds publishing in the 19th century, but then you get partway through a paragraph and it might as well be Old High Martian with talk about diminishings and consequents and so on. So here’s some help.
For college students that will be taking partial differential equations next semester, here is a very good online book. https://t.co/txtfbMaRKc
As it says on the tin: a textbook on partial differential equations. If you find yourself adrift in the subject, maybe seeing how another author addresses the same subject will help, if nothing else for finding something familiar written in a different fashion.
Here's a cool way to paper-fold an ellipse:
1) Cut a circle and fold it so that the circumference falls on a fixed point inside 2) Repeat this procedure using random folds pic.twitter.com/TAU50pvgll
And this is just fun: creating an ellipse as the locus of points that are never on the fold line when a circle’s folded by a particular rule.
Finally, something whose tweet origin I lost. It was from one of the surprisingly many economists I follow considering I don’t do financial mathematics. But it links to a bit of economic history: Origins of the Sicilian Mafia: The Market for Lemons. It’s 31 pages plus references. And more charts about wheat production in 19th century Sicily than I would have previously expected to see.
By the way, if you’re interested in me on Twitter, that would be @Nebusj. Thanks for stopping in, should you choose to.
There were a good number of mathematically-themed comic strips in the syndicated comics last week. Those from the first part of the week gave me topics I could really sink my rhetorical teeth into, too. So I’m going to lop those off into the first essay for last week and circle around to the other comics later on.
Jef Mallett’s Frazz started a week of calendar talk on the 31st of December. I’ve usually counted that as mathematical enough to mention here. The 1st of January as we know it derives, as best I can figure, from the 1st of January as Julius Caesar established for 45 BCE. This was the first Roman calendar to run basically automatically. Its length was quite close to the solar year’s length. It had leap days added according to a rule that should have been easy enough to understand (one day every fourth year). Before then the Roman calendar year was far enough off the solar year that they had to be kept in synch by interventions. Mostly, by that time, adding a short extra month to put things more nearly right. This had gotten all confusingly messed up and Caesar took the chance to set things right, running 46 BCE to 445 days long.
But why 445 and not, say, 443 or 457? And I find on research that my recollection might not be right. That is, I recall that the plan was to set the 1st of January, Reformed, to the first new moon after the winter solstice. A choice that makes sense only for that one year, but, where to set the 1st is literally arbitrary. While that apparently passes astronomical muster (the new moon as seen from Rome then would be just after midnight the 2nd of January, but hitting the night of 1/2 January is good enough), there’s apparently dispute about whether that was the objective. It might have been to set the winter solstice to the 25th of December. Or it might have been that the extra days matched neatly the length of two intercalated months that by rights should have gone into earlier years. It’s a good reminder of the difficulty of reading motivation.
Brian Fies’s The Last Mechanical Monster for the 1st of January, 2018, continues his story about the mad scientist from the Fleischer studios’ first Superman cartoon, back in 1941. In this panel he’s describing how he realized, over the course of his long prison sentence, that his intelligence was fading with age. He uses the ability to do arithmetic in his head as proof of that. These types never try naming, like, rulers of the Byzantine Empire. Anyway, to calculate the cube root of 50,653 in his head? As he used to be able to do? … guh. It’s not the sort of mental arithmetic that I find fun.
But I could think of a couple ways to do it. The one I’d use is based on a technique called Newton-Raphson iteration that can often be used to find where a function’s value is zero. Raphson here is Joseph Raphson, a late 17th century English mathematician known for the Newton-Raphson method. Newton is that falling-apples fellow. It’s an iterative scheme because you start with a guess about what the answer would be, and do calculations to make the answer better. I don’t say this is the best method, but it’s the one that demands me remember the least stuff to re-generate the algorithm. And it’ll work for any positive number ‘A’ and any root, to the ‘n’-th power.
So you want the n-th root of ‘A’. Start with your current guess about what this root is. (If you have no idea, try ‘1’ or ‘A’.) Call that guess ‘x’. Then work out this number:
Ta-da! You have, probably, now a better guess of the n-th root of ‘A’. If you want a better guess yet, take the result you just got and call that ‘x’, and go back calculating that again. Stop when you feel like your answer is good enough. This is going to be tedious but, hey, if you’re serving a prison term of the length of US copyright you’ve got time. (It’s possible with this sort of iterator to get a worse approximation, although I don’t think that happens with n-th root process. Most of the time, a couple more iterations will get you back on track.)
But that’s work. Can we think instead? Now, most n-th roots of whole numbers aren’t going to be whole numbers. Most integers aren’t perfect powers of some other integer. If you think 50,653 is a perfect cube of something, though, you can say some things about it. For one, it’s going to have to be a two-digit number. 103 is 1,000; 1003 is 1,000,000. The second digit has to be a 7. 73 is 343. The cube of any number ending in 7 has to end in 3. There’s not another number from 1 to 9 with a cube that ends in 3. That’s one of those things you learn from playing with arithmetic. (A number ending in 1 cubes to something ending in 1. A number ending in 2 cubes to something ending in 8. And so on.)
So the cube root has to be one of 17, 27, 37, 47, 57, 67, 77, 87, or 97. Again, if 50,653 is a perfect cube. And we can do better than saying it’s merely one of those nine possibilities. 40 times 40 times 40 is 64,000. This means, first, that 47 and up are definitely too large. But it also means that 40 is just a little more than the cube root of 50,653. So, if 50,653 is a perfect cube, then it’s most likely going to be the cube of 37.
Bill Watterson’s Calvin and Hobbes rerun for the 2nd is a great sequence of Hobbes explaining arithmetic to Calvin. There is nothing which could be added to Hobbes’s explanation of 3 + 8 which would make it better. I will modify Hobbes’s explanation of what the numerator. It’s ridiculous to think it’s Latin for “number eighter”. The reality is possibly more ridiculous, as it means “a numberer”. Apparently it derives from “numeratus”, meaning, “to number”. The “denominator” comes from “de nomen”, as in “name”. So, you know, “the thing that’s named”. Which does show the terms mean something. A poet could turn “numerator over denominator” into “the number of parts of the thing we name”, or something near enough that.
Hobbes continues the next day, introducing Calvin to imaginary numbers. The term “imaginary numbers” tells us their history: they looked, when first noticed in formulas for finding roots of third- and fourth-degree polynomials, like obvious nonsense. But if you carry on, following the rules as best you can, that nonsense would often shake out and you’d get back to normal numbers again. And as generations of mathematicians grew up realizing these acted like numbers we started to ask: well, how is an imaginary number any less real than, oh, the square root of six?
Hobbes’s particular examples of imaginary numbers — “eleventenn” and “thirty-twelve” — are great-sounding compositions. They put me in mind, as many of Watterson’s best words do, of a 1960s Peanuts in which Charlie Brown is trying to help Sally practice arithmetic. (I can’t find it online, as that meme with edited text about Sally Brown and the sixty grapefruits confounds my web searches.) She offers suggestions like “eleventy-Q” and asks if she’s close, which Charlie Brown admits is hard to say.
And finally, James Allen’s Mark Trail for the 3rd just mentions mathematics as the subject that Rusty Trail is going to have to do some work on instead of allowing the experience of a family trip to Mexico to count. This is of extremely marginal relevance, but it lets me include a picture of a comic strip, and I always like getting to do that.
If anything dominated the week in mathematically-themed comic strips it was Zach Weinersmith’s Saturday Morning Breakfast Cereal. I don’t know how GoComics selects the strips to (re?)print on their site. But there were at least four that seemed on-point enough for me to mention. So, okay. He’s got my attention. What’s he do with it?
On the 3rd of December is a strip I can say is about conditional probability. The mathematician might be right that the chance someone will be murdered by a serial killer are less than one in ten million. But that is the chance of someone drawn from the whole universe of human experiences. There are people who will never be near a serial killer, for example, or who never come to his attention or who evade his interest. But if we know someone is near a serial killer, or does attract his interest? The information changes the probability. And this is where you get all those counter-intuitive and somewhat annoying logic puzzles about, like, the chance someone’s other child is a girl if the one who just walked in was, and how that changes if you’re told whether the girl who just entered was the elder.
On the 5th is a strip about sequences. And built on the famous example of exponential growth from doubling a reward enough times. Well, you know these things never work out for the wise guy. The “Fibonacci Spiral” spoken of in the next-to-last panel is a spiral, like you figure. The dimensions of the spiral are based on those of golden-ratio rectangles. It looks a great deal like a logarithmic spiral to the untrained eye. Also to the trained eye, but you knew that. I think it’s supposed to be humiliating that someone would call such a spiral “random”. But I admit I don’t get that part.
The strip for the 6th has a more implicit mathematical content. It hypothesizes that mathematicians, given the chance, will be more interested in doing recreational puzzles than even in eating and drinking. It’s amusing, but I’ll admit I’ve found very few puzzles all that compelling. This isn’t to say there aren’t problems I keep coming back to because I’m curious about them, just that they don’t overwhelm my common sense. Don’t ask me when I last received actual pay for doing something mathematical.
And then on the 9th is one more strip, about logicians. And logic puzzles, such as you might get in a Martin Gardner collection. The problem is written out on the chalkboard with some shorthand logical symbols. And they’re symbols both philosophers and mathematicians use. The letter that looks like a V with a crossbar means “for all”. (The mnemonic I got was “it’s an A-for-all, upside-down”. This paired with the other common symbol, which looks like a backwards E and means there exists: “E-for-exists, backwards”. Later I noticed upside-down A and backwards E could both be just 180-degree-rotated A and E. But try saying “180-degree-rotated” in a quick way.) The curvy E between the letters ‘x’ and ‘S’ means “belongs to the set”. So that first line says “for all x that belong to the set S this follows”. Writing out “isLiar(x)” instead of, say, “L(x)”, is more a philosopher’s thing than a mathematician’s. But it wouldn’t throw anyway. And the T just means emphasizing that this is true.
And that is as much about Saturday Morning Breakfast Cereal as I have to say this week.
Sam Hurt’s Eyebeam for the 4th tells a cute story about twins trying to explain infinity to one another. I’m not sure I can agree with the older twin’s assertion that infinity means there’s no biggest number. But that’s just because I worry there’s something imprecise going on there. I’m looking forward to the kids learning about negative numbers, though, and getting to wonder what’s the biggest negative real number.
Percy Crosby’s Skippy for the 4th starts with Skippy explaining a story problem. One about buying potatoes, in this case. I’m tickled by how cranky Skippy is about boring old story problems. Motivation is always a challenge. The strip originally ran the 7th of October, 1930.
Dave Whamond’s Reality Check for the 6th uses a panel of (gibberish) mathematics as an example of an algorithm. Algorithms are mathematical, in origin at least. The word comes to us from the 9th century Persian mathematician Al-Khwarizmi’s text about how to calculate. The modern sense of the word comes from trying to describe the methods by which a problem can be solved. So, legitimate use of mathematics to show off the idea. The symbols still don’t mean anything.
Rick Detorie’s One Big Happy for the 7th has Joe trying to get his mathematics homework done at the last minute. … And it’s caused me to reflect on how twenty multiplication problems seems like a reasonable number to do. But there’s only fifty multiplications to even do, at least if you’re doing the times tables up to the 10s. No wonder students get so bored seeing the same problems over and over. It’s a little less dire if you’re learning times tables up to the 12s, but not that much better. Yow.
Olivia Walch’s Imogen Quest for the 8th looks pretty legitimate to me. It’s going to read as gibberish to people who haven’t done parametric functions, though. Start with the plane and the familiar old idea of ‘x’ and ‘y’ representing how far one is along a horizontal and a vertical direction. Here, we’re given a dummy variable ‘t’, and functions to describe a value for ‘x’ and ‘y’ matching each value of ‘t’. The plot then shows all the points that ever match a pair of ‘x’ and ‘y’ coordinates for some ‘t’. The top drawing is a shape known as the cardioid, because it kind of looks like a Valentine-heart. The lower figure is a much more complicated parametric equation. It looks more anatomically accurate,
Still no sign of Mark Anderson’s Andertoons and the drought is worrying me, yes.
Was there an uptick in mathematics-themed comic strips in the syndicated comics this past week? It depends how tight a definition of “theme” you use. I have enough to write about that I’m splitting the week’s load. And I’ve got a follow-up to that Wronski post the other day, so I’m feeling nice and full of content right now. So here goes.
Zach Weinersmith’s Saturday Morning Breakfast Cereal posted the 5th gets my week off to an annoying start. Science and mathematics and engineering people have a tendency to be smug about their subjects. And to see aptitude or interest in their subjects as virtue, or at least intelligence. (If they see a distinction between virtue and intelligence.) To presume that an interest in the field I like is a demonstration of intelligence is a pretty nasty and arrogant move.
And yes, I also dislike the attitude that school should be about training people. Teaching should be about letting people be literate with the great thoughts people have had. Mathematics has a privileged spot here. The field, as we’ve developed it, seems to build on human aptitudes for number and space. It’s easy to find useful sides to it. Doesn’t mean it’s vocational training.
Lincoln Peirce’s Big Nate on the 6th discovered mathematics puzzles. And this gave him the desire to create a new mathematical puzzle that he would use to get rich. Good luck with that. Coming up with interesting enough recreational mathematics puzzles is hard. Presenting it in a way that people will buy is another, possibly greater, challenge. It takes luck and timing and presentation, just as getting a hit song does. Sudoku, for example, spent five years in the Dell Magazine puzzle books before getting a foothold in Japanese newspapers. And then twenty years there before being noticed in the English-speaking puzzle world. Big Nate’s teacher tries to encourage him, although that doesn’t go as Mr Staples might have hoped. (The storyline continues to the 11th. Spoiler: Nate does not invent the next great recreational mathematics puzzle.)
Jef Mallett’s Frazz for the 7th start out in a mathematics class, at least. I suppose the mathematical content doesn’t matter, though. Mallett’s making a point about questions that, I confess, I’m not sure I get. I’ll leave it for wiser heads to understand.
Mike Thompson’s Grand Avenue for the 8th is a subverted word-problem joke. And I suppose a reminder about the need for word problems to parse as things people would do, or might be interested in. I can’t go along with characterizing buying twelve candy bars “gluttonous” though. Not if they’re in a pack of twelve or something like that. I may be unfair to Grand Avenue. Mind, until a few years ago I was large enough my main method of getting around was “being rolled by Oompa-Loompas”, so I could be a poor judge.
If the sphere has radius r, then the surface area of the sphere is 4πr2. And the volume is (4/3)πr3. What’s interesting about this is that there’s a relationship between these two expressions. The first is the derivative of the second. The derivative is one of the earliest things one learns in calculus. It describes how much a quantity changes with a tiny change in something it depends on.
And this got him to thinking about the surface area of a cube. Call the length of a cube’s side s. Its surface is six squares, each of them with a side of length s. So the surface area of each of the six squares is s2, which is obvious when you remember we call raising something to the second power “squaring”. Its total surface area then is 6s2. But its volume is is s3. This is why we even call raising something to the third power “cubing”. And the derivative of s3 is 3s2. (If you don’t know calculus, but you suspect you see a pattern here, you’re learning calculus. If you’re not sure about the pattern, let me tell you that the derivative of s4 would be 4s3, and the derivative of (1/3)s2 would be (2/3)s.)
There’s an obvious flaw there, and Austin’s aware of it. But it got him pondering different ways to characterize how big a cube is. He can find one that makes the relationship between volume and surface area work out like he expects. But the question remains, why that? And what about other shapes?
I think that’s an interesting discussion to have, and mean to think about it some more myself. And I wanted to point people who’d be interested over there to join in.
(By the way, I’m @Nebusj on Twitter. I’m happy to pick up new conversational partners even if I never quite feel right starting to chat with someone.)
Schmidt does assume normal, ordinary, six-sided dice for this. You can work out the problem for four- or eight- or twenty- or whatever-sided dice, with most likely a different answer.
But given that, the problem hasn’t quite got an answer right away. Reasonable people could disagree about what it means to say “if you roll a die four times, what is the probability you create a correct proportion?” For example, do you have to put the die result in a particular order? Or can you take the four numbers you get and arrange them any way at all? This is important. If you have the numbers 1, 4, 2, and 2, then obviously 1/4 = 2/2 is false. But rearrange them to 1/2 = 2/4 and you have something true.
We can reason this out. We can work out how many ways there are to throw a die four times, and so how many different outcomes there are. Then we count the number of outcomes that give us a valid proportion. That count divided by the number of possible outcomes is the probability of a successful outcome. It’s getting a correct count of the desired outcomes that’s tricky.
I’m thinking to do a second Mathematics A-To-Z Glossary. For those who missed it, last summer I had a fun string of several weeks in which I picked a mathematical term and explained it to within an inch of its life, or 950 words, whichever came first. I’m curious if there’s anything readers out there would like to see me attempt to explain. So, please, let me know of any requests. All requests must begin with a letter, although numbers might be considered.
Meanwhile since there’s been some golden ratio talk around these parts the last few days, I thought people might like to see this neat Algebra Fact of the Day:
In a comment on my “Gilded Ratios” essay fluffy wondered about a variation on the Golden and Golden-like ratios. What’s interesting about the Golden Ratio and similar numbers is that their reciprocal — one divided by them — is a whole number less than the original number. That is, 1 divided by 1.618(etc) is 0.618(etc), which is 1 less than the original number. 1 divided by 2.414(etc) is 0.414(etc), exactly 2 less than the original 2.414(etc). 1 divided by 3.302(etc) is 0.302(etc), exactly 3 less than the original 3.302(etc).
fluffy wondered about a variation. Is there some number x that’s exactly 2 less than 2 divided by x? Or a (presumably) differently number that’s exactly 3 less than 3 divided by it? Yes, there is.
Let me call the whole number difference — the 1 or 2 or 3 or so on, referred to above — by the name b. And let me call the other number — the one that’s b less than b divided by it — by the name x. Then a number x, for which b divided by x is exactly b less than itself, makes true the equation . This is slightly different from the equation used last time, but not very different. Multiply both sides by x, which we know not to be zero, and we get a polynomial.
Yes, quadratic formula, I see you waving your hand in the back there. And you’re right. There are two x’s that will make that equation true. The positive one is . The negative one you get by changing the + sign, just before the square root, to a – sign, but who cares about that root? Here’s the first several of the (positive) silver-leaf ratios:
Some More Numbers With Cute Reciprocals
Looking over those hypnotic rows of digits past the decimal inspires thoughts. The part beyond the decimal keeps rising, closer and closer to 1. Does it ever get past 1? That is, might (say) the silver-leaf number that’s 2,038 more than its reciprocal be 2,039.11111 (or something)?
No, it never does. There are a couple of ways to prove that, if you feel like. We can take the approach that’s easiest (to my eyes) to imagine. It takes a little algebraic grinding to complete. That is to look for the smallest number b for which the silver-leaf number, , is larger than . Follow that out and you realize that it’s any value of b for which 0 is greater than 4. Logically, therefore, we need to take b into a private room and have a serious talk about its job performance, what with it not existing.
A harder proof to imagine working out, but that takes no symbol manipulation, comes from thinking about these reciprocals. Let’s imagine we had some b for which its corresponding silver-leaf number x is more than b + 1. Then, x – b has to be greater than 1. But if x is greater than 1, then its reciprocal has to be less than 1. We again have to talk with b about how its nonexistence is keeping it from doing its job.
Are there other proofs? Most likely. I was satisfied by this point, and resolved not to work on it more until the shower. Updates after breakfast, I suppose.
I may have mentioned that I regard the Golden Ratio as a lot of bunk. If I haven’t, allow me to mention: the Golden Ratio is a lot of bunk. I concede it’s a cute number. I found it compelling when I first had a calculator that let me use the last answer for a new operation. You can pretty quickly find that 1.618033 (etc, and the next digit is a 9 by the way) has a reciprocal that’s 0.618033 (etc).
There’s no denying that. And there’s no denying that’s a neat pattern. But it is not some aesthetic ideal. When people evaluate rectangles that “look best” they go to stuff that’s a fair but not too much wider in one direction than the other. But people aren’t drawn to 1.618 (etc) any more reliably than they like 1.6, or 1.8, or 1.5, or other possible ratios. And it is not any kind of law of nature that the Golden Ratio will turn up. It’s often found within the error bars of a measurement, but so are a lot of numbers.
The Golden Ratio is an irrational number, but basically all real numbers are irrational except for a few peculiar ones. Those peculiar ones happen to be the whole numbers and the rational numbers, which we find interesting, but which are the rare exception. It’s not a “transcendental number”, which is a kind of real number I don’t want to describe here. That’s a bit unusual, since basically all real numbers are transcendental numbers except for a few peculiar ones. Those peculiar ones include whole and rational numbers, and square roots and such, which we use so much we think they’re common. But not being transcendental isn’t that outstanding a feature. The Golden Ratio is one of those strange celebrities who’s famous for being a celebrity, and not for any actual accomplishment worth celebrating.
I started wondering: are there other Golden-Ratio-like numbers, though? The title of this essay gives what I suppose is the best name for this set. The Golden Ratio is interesting because its reciprocal — 1 divided by it — is equal to it minus 1. Is there another number whose reciprocal is equal to it minus 2? Another number yet whose reciprocal is equal to it minus 3?
So I looked. Is there a number between 2 and 3 whose reciprocal is it minus 2? Certainly there is. How do I know this?
Let me call this number, if it exists, x. The reciprocal of x is the number 1/x. The number x minus 2 is the number x – 2. We’ll pick up the pace in a little bit. Now imagine trying out every single number from 2 to 3, in order. The reciprocals 1/x start out at 1/2 and drop to 1/3. The subtracted numbers start out at 0 and grow to 1. There’s no gaps or sudden jumps or anything in either the reciprocals or the subtracted numbers. So there must be some x for which 1/x and x – 2 are the same number.
In the trade we call that an existence proof. It shows there’s got to be some answer. It doesn’t tell us much about what the answer is. Often it’s worth looking for an existence proof first. In this case, it’s probably overkill. But you can go from this to reasoning that there have to be Golden-Like-Ratio numbers between any two counting numbers. So, yes, there’s some number between 2,038 and 2,039 whose reciprocal is that number minus 2,038. That’s nice to know.
So what is the number that’s two more than its reciprocal? That’s whatever number or numbers make true the equation . That’s straightforward to solve. Multiply both sides by x, which won’t change whether the equation is true unless x is zero. (And x can’t be zero, or else we wouldn’t talk of 1/x except in hushed, embarrassed whispers.) This gets an equivalent equation . Subtract 1 from both sides, and we get and we’re set up to use the quadratic formula. The answer will be . The answer is about 2.414213562373095 (and on). (No, is not an answer; it’s not between 2 and 3.)
The number that’s three more than its reciprocal? We’ll call that x again, trusting that we remember this is a different number with the same name. For that we need to solve and that turns into the equation . And so and so it’s about 3.30277563773200. Yes, there’s another possible answer we rule out because it isn’t between 3 and 4.
We can do the same thing to find another number, named x, that’s four more than its reciprocal. That starts with and gets eventually to or about 4.23606797749979. We could go on like this. The number x that’s 2,038 more than its reciprocal is about 2038.00049082160.
If your eyes haven’t just slid gently past the equations you noticed the pattern. Suppose instead of saying 2 or 3 or 4 or 2038 we say the number b. b is some whole number, any that we like. The number whose reciprocal is exactly b less than it is the number x that makes true the equation . And that leads to the finding the number that makes the equation true.
And, what the heck. Here’s the first twenty or so gilded numbers. You can read this either as a list of the numbers I’ve been calling x — 1.618034, 2.414214, 3.302776 — or as an ordered list of the reciprocals of x — 0.618034, 0.414214, 0.302276 — as you like. I’ll call that the gilt; you add it to the whole number to its left to get that a number that, cutely, has a reciprocal that’s the same after the decimal.
I did think about including a graph of these numbers, but the appeal of them is that you can take the reciprocal and see digits not changing. A graph doesn’t give you that.
Some Numbers With Cute Reciprocals
None of these are important numbers. But they are pretty, and that can be enough on a quiet day.