Thoughts from the Outfield

So there I was in outer center field, trying to figure out what twenty-six times twenty-six times twenty-six is. (As one does.)

You may not have heard of “outer center field”; come to think of it, I haven’t either. It just seems like the right name to give to the position I always played in kickball (called “soccer baseball” in my town) back in elementary school. The captain wanted me far from the action where I couldn’t do any harm, and I was happy with the arrangement; it left me free to daydream about whatever I wanted — such as the product 26 × 26 × 26.

My interest stemmed from initials. I was JGP, my best friend was SBC, my siblings were WHP, STP, and DJP, and everyone I knew had three-letter initials. Even my dad, who had no middle name, had been assigned a middle initial on his birth certificate, perhaps because the blank field on the form had cowed his immigrant parents into improvising one. I wondered how many different possibilities there were. Of course some were unfortunate and a few were even unprintable, but leaving that issue aside, how many first initial / middle initial / last initial combinations could be constructed?

An arithmogram, a bit creased and ripped but still recognizable, courtesy of Sidney B. Cahn, my first lab-partner, now a physicist at Yale University.

To get started, I needed to figure out 26 × 26 in my head. As it happened I already knew that 25 × 25 was 625, thanks to my all-time favorite teacher, Mr. Stern. Mr. Stern had invented math activity puzzles he called arithmograms, each of which required the student to solve a few dozen math problems that resulted in decimal numbers which then were used as distances in carrying out ruler and compass constructions that would create a picture of a car, or a face, or a rocket, or something else a sixth-grader might view as an adequate reward. It’s hard to find arithmograms these days; Mr. Stern never turned arithmograms into a commercial product, and the pictures weren’t the sort of thing many students saved after completing them (for one thing, they were huge). Anyway, our rulers divided inches into sixteenths, so the arithmetic problems that appeared in arithmograms had answers that were multiples of .0625, or one sixteenth. I understood decimals well enough to know that the number fact .25 × .25 = .0625 is pretty much the same as the number fact 25 × 25 = 625. But how to get up to 26 × 26? 

It occurred to me that I could get from 25 × 25 to 25 × 26 by adding 25 to it, and then get from 25 × 26 up to 26 × 26 by adding 26 to that. So, I got 26 × 26 = 625+25+26 = 650+26 = 676.

There I got stuck, because I couldn’t use that same trick (or any other trick I knew) to multiply 676 by 26 in my head. So, I did something I’ve had occasion to do many times in my career when a question was too hard for me to think about: I thought about a different question. In this case, the question was, is there a general rule for figuring out (n+1) × (n+1) if I already knew n × n?

Following the same reasoning as before, I determined that I could get from n × n to (n+1) × (n+1) by first adding n and then adding n+1.

PROVING THINGS WITH PICTURES

Later that school year, off the field, I either discovered on my own or was shown by someone else a pictorial way to understand the relationship between n × n and (n+1) × (n+1). My formula could be seen as a special case of the more general formula involving (a+b) × (a+b). Consider an (a+b)-by-(a+b) array of dots (a 4-by-3 array for instance):

The (a+b)-by-(a+b) array can be divided into an a-by-a array, an a-by-b array, a b-by-a array, and a b-by-b array, with total number of dots equaling a×a + a×b + b×a + b×b. Replacing a by n and b by 1 gives my formula (n+1) × (n+1) = n×n + n + n + 1.

This led me to wonder what the corresponding formula for a+b times a+b times a+b would be (harking back to my original question about 26 × 26 × 26). You’d need some sort of three-dimensional picture for this, wouldn’t you?

PROVING THINGS WITH ALGEBRA

I was in the library of my elementary school, struggling to draw a picture of an a+b by a+b by a+b cube, when a teacher named Mr. Schmidt saw my scribblings and asked what I was up to. I told him about my problem, and he said “Algebra can help you with that.“

He wrote this:

I already knew some algebraic notation (for instance, I knew that 3a²b meant 3 times a times a times b); I just didn’t know that you could use algebra to prove things.

I’m not sure I really understood the proof, but since it looked a lot like the procedure for multiplying a three-digit number by a two-digit number, I was willing to accept it as valid.

“So there’s your formula: a+b squared, times a+b again, equals a³ + 3 a² b + 3 a b² + b³ .”

At that moment, I developed a sense of the power of algebra. And later that week, when I applied the method on my own to compute (a+b)⁴, (a+b)⁵, etc., my crush on algebra turned into a full-blown love affair — because when I computed the expressions

and extracted the numbers

I recognized the table of numbers as something I’d seen before, in my Giant Golden Book of Mathematics: Pascal’s triangle. (See Endnote #1.)

The Giant Golden Book had talked about Pascal’s triangle solely in terms of probability and gambling. Seeing the same numbers turn up in algebra gave me a glimpse of the deep unity of mathematics.

IT, PART TWO

A couple of weeks ago, planning this essay, I had planned to end the piece there and call it something like “How I fell in love with algebra“, perhaps padding out the story with more details about arithmograms or the Giant Golden Book of Mathematics, but then it occurred to me that the story lacked closure: I had never worked out 26 × 26 × 26 back in elementary school.

So I thought to myself: I should work it out now.

Of course I could’ve figured it out on paper, or punched the problem into a calculator, but that would’ve been cheating. Isn’t my blog supposed to be all about reveling in the marvelous things you can build in your head? Didn’t I just write an essay about that very topic, celebrating the pleasure of surveying a long chain of reasoning in the privacy of one’s own skull? So how could I do anything as crude as work out the answer on paper or with a calculator? My sense of honor required me to work out the answer in my head, using pure thought.

Here I should stress, for the nonmathematicians among you, that most mathematicians are not especially good mental calculators, and I’m probably below average for a mathematician. (See Endnote #2.) I knew that it would take me at least a minute, and probably a good deal longer, to work out the answer. The goal was not to get the answer quickly, but rather to get it right, taking as much time as I needed, using nothing outside my brain.

I should also stress that these kinds of calculations have very little to do with actual mathematical research. An isolated calculation of some particular five-digit number (see Endnote #3) tells the mathematician nothing worth knowing. Mathematicians doing research, working at the boundary between the known and the unknown, need background knowledge, persistence, imagination, and time free from distraction; by and large they do not need inner computational resources. Nor do the deductive chains of a mathematical argument resemble anything as straightforward as arithmetical computations.

But the most important difference is that, whereas my math research is guided by curiosity, in no way was I curious about the decimal expansion of the cube of 26. I wanted to figure it out for the sheer challenge of it. Also, calculating 26 × 26 × 26 in my head would give a satisfying ending to the story that began in the outfield almost fifty years ago. And maybe on a subconscious level I was influenced by the story-arc of Stephen King’s “It”, in which some middle-aged folks confront their childhood nemesis to finish it off properly.

THE WORK

I’m not going to give all the details of my mental calculation; that would be boring for me to write and even more boring for you to read. But some of you might want to pause here for a few minutes, hours, or days to see if you can find your own path to the answer.

In a way, calculating 26³ in one’s head is more of a programming challenge than a mathematical challenge. Getting human wetware (one’s own brain) to perform multidigit calculations amounts to programming a central processing unit that has very little memory and a shockingingly high error rate. But each human CPU is bad in its own way, so what worked for me may not work for you. You’ll probably choose a different route to the answer than I did.

My path to the answer used the formula (a+b)³ = a³ + 3 a² b + 3 a b² + b³ that Mr. Schmidt had shown me. I knew that formula by heart, but replacing a by 20 and b by 6 gives a formula with too many pieces for me to hold onto all at once in my head: (20 + 6)³ = 20³ + 3 × 20² × 6 + 3 × 20 × 6² + 6³. I would need to compute each of the four terms on the right hand side separately, and then find a way to hold onto them long enough to combine them somehow, without exceeding the limited capacity of my short-term memory.

The first of the four terms was easy: since 2³ = 8, 20³ = 2³ × 10³ = 8 × 1000 = 8000. That was a simple enough number to hold onto in my head. I was able to compute the second term 3 × 20² × 6 = 7200 with a little more work, and I still remembered the 8000 from before, and I was able to the 8000 to the next 7200 to get 15,200.

But there I ran into trouble. Holding the number 15,200 in my head took about all the neurons I had in that part of my brain; there was no way I could move on to calculating 3×20×6² (the third term in the expansion of (20+6)³) without the risk that I’d misremember 15,200. So I decided to put the problem aside and try again some other time.

When I returned to the task the next day, I decided to try to get around the limitations of my CPU by using supplementary memory in my brain’s auditory and motor processing units. I’m not good at memorizing numbers, but I am decent at memorizing tunes (which are both auditory and kinesthetic for me since I play piano), so I often convert strings of digits into strings of musical notes. See Endnote #4.

With that trick, I was able to separately compute, rehearse, and store 20³ + 3 × 20² × 6 (the sum of the first and second terms in the expansion) and 3 × 20 × 6² + 6³ (the sum of the third and fourth terms). It helped a lot that I already knew that 6³ = 216. With these two large numbers stored, I was able to add them — mercifully no carrying was required! — to arrive at a final answer after about ten minutes of purely mental work. (I probably could’ve done it in five minutes, but I re-did some steps several times to convince myself I hadn’t messed up.)

Did I stop there? No.

CHECKING

I realized that I had one and only one chance to solve the problem honestly. If I consulted a calculator and learned that my answer was wrong, I would also learn what the correct answer was, and it might be hard to set aside that knowledge when I redid the calculation. It’s very hard to look at something and say “OK, I’m going to try to forget this”; the very effort of trying to forget will probably have the reverse of the intended effect. (For instance, I find that trying to forget a movie spoiler only spoils the movie more.) So I had only one chance to get the answer right in a satisfyingly honest way.

One way to catch mistakes is to go through one’s work again. Unfortunately, brains are all too prone to repeat mistakes. The grooves one wears in one’s synapses the first time one makes a mistake predispose those synapses to go the same wrong way the next time.

Fortunately, in the half-century that has elapsed between my time in the outfield and the present day, I have learned a few tricks for catching arithmetic mistakes that are well-suited to the cramped confines of a human brain. The most well-known is the method of casting out nines, which usually catches single-digit errors, though it only indicates that an error has been made without revealing its location. (See Endnote #5.) A related method called casting out elevens catches errors that were caused by switching two adjacent digits (again, only indicating the occurrence of an error without telling you where it is). My five-digit proposed answer passed both tests, as well as a casting out sevens check; this gave me a sense of relief but not full confidence. I wanted a procedure, parallel to the first, that arrived at same answer by a different route. So for good measure I applied the formula (a+b)³ = a³ + 3 a² b + 3 a b² + b³ to compute (10+3)³ and then doubled the result three times. I got the same five digit answer as I’d gotten from (20+6)³.

Only then was I confident enough to whip out a calculator (or rather to whip out my phone and open the calculator app) and check that the answer was correct. No trumpets sounded; no tumblers clicked into position opening some wondrous cosmic safe. All that happened is that a mathematician on the far side of age fifty reassured himself that he can still put things into his brain and move them around, though not as easily as in his younger days.

The painstaking procedure I followed relates to one of the main themes of my current teaching, or at least a theme I’m trying to develop as I teach discrete mathematics for computer science majors: the importance of validating answers by deriving them along divergent-and-then-reconvergent solution paths. It’s a given that you’re going to make mistakes; the real question is, will you catch them in time? And, can you adopt procedures that make it easier for you to catch your own mistakes?

WHERE THAT LEAVES ME

That’s pretty much all I’ve got to say, except for one nagging, weird thought I’d like to share. Maybe you can blame the weird thought on all the Twilight Zone episodes I watched as a kid in the 1970s; perhaps thinking back to that time in my life has put me in the same spooky frame of mind. Here’s the nagging thought: What if I’m still in outer center field, and everything that I think has happened since — college, grad school, career, marriage, kids — is just an intense hallucinatory reverie, as in the great but horrible Ambrose Bierce story “Incident at Owl Creek Bridge”?

Perhaps a better literary analogue can be found in the Borges story “The Secret Miracle“, in which a writer facing a firing squad finds that external time has magically stopped, allowing him to mentally complete his magnum opus. Had I, when setting myself the problem of computing 26³ in my head, asked some Power for more time? Had that Power replied “The time for your labor has been granted”, and then delivered the illusion of a life that would equip me with the tools to carry out my self-imposed task?

What a silly idea! For one thing, if that had been the case, wouldn’t the act of mentally computing 26³ have broken the spell, waking me from my daydream of adult life and returning me to 1972? Or, in a variant scenario, wouldn’t I have suddenly found myself in a hospital bed, waking from a coma caused by getting hit in the head by a flying soccer ball I didn’t see coming because I was too busy thinking about 26³?

Since neither of these things happened, I’m inclined to think that my memories are factual, that my current experiences are real, that I did actually come in from the outfield that day and go on to lead a rich personal and mathematical life for half a century, using pictures and algebra to solve counting problems and probability problems.

… Or maybe I’m still in the outfield, and there’s something else I’m supposed to do.

Next month: Dividing by Zero.

Thanks to Art Benjamin, Sidney Cahn, Sandi Gubin, and Evan Romer.

ENDNOTES

 

#1. Although the phrase “Pascal’s triangle” is common in English-speaking countries, Pascal didn’t invent “his” triangle; centuries before he was born, the array of numbers that bears his name was studied in China, India, and elsewhere. The Eurocentrism that some attribute to mathematics has little to do with the ideas of mathematics itself and much to do with Eurocentric attributions. The culture of mathematics is truly a world culture, and math education at every level needs to catch up with what the history of math already teaches.

#2. The acclaimed performance-mathematician Art Benjamin, who can cube any two-digit number in seconds, is the exception, not the rule. The same goes for John Conway, who trained himself to do calendrical calculations with great rapidity. I suspect that Conway was partly motivated by a desire to demystify the talents of calendar savants, and more broadly to explode the “savant box” that, like the “genius box” I have written about elsewhere, divides humanity into categories instead of stressing the untapped potentials all of us have. Remember, Conway was the guy who taught himself to do tongue-tricks to debunk assertions that the ability to do various tongue-tricks is genetic, innate, and unalterable. But it should also be remembered that Conway was a huge show-off.

 

#3. Since 26 is roughly halfway between 20 and 30, 26³ should be roughly halfway between 20³ = 8000 and 30³ = 27,000. So one expects a five-digit answer even before one knows what the five digits are.

 

#4. I represent the digits 1 through 8 by the degrees of a major scale. 9 is encoded by an extra scale degree at the top; 0 is encoded by a rest. I still remember statistician David Donoho’s four-digit phone extension at Harvard back in 1981 because the tune corresponding to 7591 had a Copland-esque catchiness to it.

 

#5. In my mental working I actually combined casting out nines with other forms of reasoning. One principle I used is that if two numbers a and b differ by a multiple of nine, so must a³ and b³. This principle is easy to derive if you know the formula a³ − b³ = (a − b) (a² + ab + b²): if a − b is divisible by 9, then so is the product (a − b) (a² + ab + b²), which equals a³ − b³. Applying the principle to my outfield problem, I reasoned as follows: 26 is one less than 27, which is a multiple of nine. So 26 and −1 differ by multiple of nine, implying that 26³ and (−1)³ differ by a multiple of nine as well. But (−1)³ is just −1, so 26³ must differ from −1 by a multiple of nine. Putting that differently, 26³ + 1 must be a multiple of nine. So I took my proposed five digit value for 26³, added 1 to it, and then checked to see if the sum was a multiple of nine, using the famous digit-sum test (one of the pillars of the method of casting out nines): a counting number is a multiple of nine if and only if the sum of its digits is a multiple of nine.