common confusions

Devlin's Angle

October 2005

Common confusions

I often get letters and emails from people asking me to clear up a confusion with some mathematical issue. Sometimes the confusion is occasioned by something I've written in a book or article or said on the radio, other times the question is sparked by something having nothing to do with me. But regardless of the origin, three issues crop up so often that I sometimes wonder if those of us who teach mathematics for a living aren't always getting the message across quite clearly enough.

For instance, since the publication of my book The Math Gene in 2000, I've received a steady stream of messages from readers who believe there is an error in the proof I give there that there are infinitely many primes. There isn't. The confusion that arises in some readers' minds comes from my attempt to lay out the standard argument in a completely linear fashion. When I wrote the book, I formulated the proof the way I did because experience had shown me that an argument by cases can cause problems for some people. Unfortunately, eliminating the cases approach results in a different problem - rather, two different problems, as it turned out.

The argument by cases will be familiar to most readers of this column. The idea is to show that if you start to list the primes in order, P(1),P(2), P(3), etc., then the list continues for ever. To show this, you assume you have reached some stage P(N), and you prove that you can always find at least one more prime with which to continue the list. The idea - and it's a brilliant one, known to the ancient Greeks and described in Euclid's book Elements - is to consider the number

Q = [P(1) x P(2) x ... x P(N)] + 1
obtained by multiplying together all the primes listed so far and then adding 1 to the product. Now the argument splits into two cases. If Q is prime, then since Q is bigger than P(N), it is a prime not already listed. If Q is not prime, then there must be some prime P that divides into Q. But P cannot be any one of P(1), ..., P(N), since dividing any of those into Q leaves a remainder of 1. So P is a prime not already listed. Either way, there is a prime not in the list, so the list can be continued.

Now, if you are one of the many people who finds the above argument unproblematic, you may have trouble understanding why anyone would object to it. But believe me, many people do. (One problem that some people have - a problem I hoped to avoid by organizing the argument differently - is to assume that we have in some way identified the next prime to be added to the list. We have not; we've just showed that there is at least one more prime out there, and hence the list may be continued by picking the next one, whatever it may be.)

Knowing that the above argument causes some people difficulties, when I wrote The Math Gene, I formulated the proof in a slightly different way, as a proof by contradiction. To whit:

Suppose that there are only finitely many primes. List them all as P(1), P(2), ..., P(N). (Thus P(N) is the largest prime.) Now look at the number

Q = [P(1) x P(2) x ... x P(N)] + 1
Next we show that Q is prime. To do this, we show that it has no prime divisors. Well, all primes are in the list P(1), ..., P(N), and each of those numbers leaves a remainder of 1 when you divide it into Q. Hence Q can have no prime divisors. Hence Q must itself be prime.

But Q is bigger than P(N), and P(N) is supposedly the largest prime, so we have a contradiction. Hence our initial assumption is false, which is to say, there must be infinitely many primes.

Now you might suspect that the reason some people have problems with this argument is that it uses the method of contradiction, and for some of my correspondents that seems to have been the case. But I also get letters from readers who clearly understand how proof by contradiction works, but who nevertheless feel the proof is wrong. Their misconception almost always comes from believing the argument claims that the number

Q = [P(1) x P(2) x ... x P(N)] + 1
is prime for any value of N. That is easily seen to be false. For example,
[2 x 3 x 5 x 7 x 11 x 13] + 1 = 30031 = 59 x 509
But the argument I give makes no such claim. The claim that "Q is prime" is made for a specific (and decidedly hypothetical) value of N, namely the N for which P(N) is the supposed largest prime number, and the claim depends on that assumed property of P(N).

Now, I might have avoided problems if I had phrased the last part of my proof to avoid using the phrase "Q is prime". This can easily be done. I didn't because I wanted to split the entire proof into small, easily digested pieces, but maybe that was a poor choice. My choice did, however, highlight a problem that is far more fundamental. After engaging in several exchanges with readers who had trouble with my proof, I eventually realized that the problem was, at heart, confusion between variables and fixed, but unknown, constants. (Actually, in the case of my primes proof, a constant that does not exist at all outside the context of the proof!) And here is the lesson to be learned, I think. We mathematicians happily use letters to denote both variables and constants. But many people don't see the distinction. They see a formula like

Q = [P(1) x P(2) x ... x P(N)] + 1
and because N is a letter and not a specific number, such as 3, 6, or 29, they assume it denotes ANY number.

I suspect, but do not know for certain, that computer programmers would not make such an error. They, after all, have to know that once you set a variable X equal to a specific numerical value in a program, then it denotes that specific value, and nothing else, until reset to another specific value. They understand too the difference between a free variable and a variable that is bound by a context, such as being introduced in a subroutine. But I suspect that many students pass through one or more college level mathematics courses without ever coming to grips with the way mathematicians use letters to denote any one of a variable, a fixed but unknown number, or a known specific number.

The second matter that regularly lands upon my desk - or in my email inbox - is occasioned by attempts (in my own writing and that by others) to try to explain the basic ideas of the topological study of surfaces (specifically, orientability) by distinguishing between a wedding band (or cylinder) and a Moebius strip, in terms of the former "having two sides" and the latter "one side". In my book Mathematics: The New Golden Age, for example, I try to explain orientability by considering what happens when you take a closed loop with an arrowhead in the surface and sliding it around, whereupon, in the case of a Moebius band, the arrow changes from clockwise to counterclockwise when you return to the starting point - something that does not happen in the case of a cylindrical (wedding) band. I illustrate the point with the following diagram. (The loop starts just to the left of the line and moves around the band in a clockwise direction until it arrives back at the starting line.)

Several readers have written to me over the years saying that the diagram does not show what I claim it does. When you get back to the starting line, they say, the loop is actually on "the other side" of the band, and you have to go around again to get back to the starting point. But if you interpret the diagram as intended, this is not the case. The key issue is, the surfaces that mathematicians study don't have sides. They just are. That's why I was careful just now to say that the arrowed loop is in the surface, not "on" it. Strictly speaking, there is nothing for the loop to be "on"; it can only be in the surface. The two "sides" that a physical band exhibits are features of the physical model, not the mathematical surface it is intended to represent. (A more mathematical way of saying this is that the two "sides" that you see are part of the three-dimensional space surrounding the surface, not the surface itself.) To carry out the loop transportation process physically, and have it represent what is really going on in terms of the mathematical surface being represented, it's best to create the band from a strip of a clear material such as a slide for an overhead projector, and then draw the loop in successive positions as you move around the band, being careful to copy the arrow's direction faithfully at each stage. Then, although you have to draw the loop on one (physical) side of the material, since you can see it easily well from either (physical) side, it's easier to imagine the loop being in the surface and not "on" it. Do that and you will indeed see that the loop-arrow changes direction when you have transported it around the band and back to its starting point.

The third confusion that I get asked to clear up (or accused - often with great vehemence - of getting wrong in my writing) concerns probability. But I've already used up my monthly allotment of space, so I'll turn to that in a later column. Stay tuned.

Devlin's Angle is updated at the beginning of each month.
Mathematician Keith Devlin (email: is the Executive Director of the Center for the Study of Language and Information at Stanford University and The Math Guy on NPR's Weekend Edition. Devlin's newest book, THE MATH INSTINCT: Why You're a Mathematical Genius (along with Lobsters, Birds, Cats, and Dogs) was published recently by Thunder's Mouth Press.