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I teach, among many other things, a class of wonderful and inquisitive 7th graders. We've recently been studying and discussing various number systems (N, Z, Q, R, C, algebraic numbers, and even quaternions and surreals). One thing that's been hanging in the air is giving a proof that there really do exist transcendental numbers (and in particular, real ones). They're willing to take my word for it, but I'd really like to show them if I can.

I've brainstormed two possible approaches:

1) Use diagonalization on a list of algebraic numbers enumerated by their heights (in the usual way) to construct a transcendental number. This seems doable to me, and would let me share some cool facts about cardinality along the way. The asterisk by it is that, while the argument is constructive, we don't start with a number in hand and then prove that it's transcendental--a feature that I think would be nice.

2) More or less use Liouville's original proof, put as simply as I can manage. The upshots of this route are that we start with a number in hand, it's a nice bit of history, and there are some cool fraction things that we could talk about (we've been discussing repeating decimals and continued fractions). The downside is that I'm not sure if I can actually make it accessible to my students.

So here is where you come in. Is there a simple, elementary proof that some particular number is transcendental? Two kinds of responses that would be helpful would be:

a) to point out some different kind of argument that has a chance of being elementary enough, and

b) to suggest how to recouch or bring to its essence a Liouville-like argument. My model for this is the proof Conway popularized of the fact that $\sqrt{2}$ is irrational. You can find it as proof 8''' on this page.

I realize that transcendence is deep waters, and I certainly don't expect something easy to arise, but I thought I'd tap this community's expertise and ingenuity. Thanks for thinking on it.

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I would be really surprised if going through 1) gave you a number with a reasonably concrete description. Have you tried it? –  Qiaochu Yuan Dec 14 '10 at 4:21
You can describe a fairly short program for a specific enumeration of integer polynomials and use fairly rapid numerical methods to find the real roots to a finite precision. So you can get a number out to a fair number of digits without great grief. But the ease does not matter because there is no real interest in doing it, only that one can (and there are many nice candidates for an order). –  Aaron Meyerowitz Dec 14 '10 at 4:50
A mind opener for students (at some level) is that since the algebraic numbers are enumerable we can list them (in principle) and put the kth one in the center of an interval of diameter $1/2^k$. Then we have a collection of intervals covering a set of which the (dense) set of rationals is a "tiny" part. But together these intervals have combined length 1 so "most" real numbers are excluded (hence transcendental). At least this shows that our intuition is far from the full story. –  Aaron Meyerowitz Dec 14 '10 at 4:55

6 Answers 6

The original Liouville's number is probably the easiest, but most of the proofs tend to invoke calculus (because why not?), so let me try to show it in a more 7th-grade friendly way. I'll call this the swaths-of-zero approach.

So we know that Liouville's number $L$ looks like this: .1100010000000000000000010... with a 1 in the $n!$ places.

When we square it, we get this: .012100220001000000000000220002...

What happens is that in the $2n!$ places we get a 1, and in the $p!+q!$ places we get a 2. (The great thing about this is that it can be explained using the elementary-school algorithm, the one they are all familiar with, for multiplication.)

If we multiply $L$ by an integer and write down the answer, the value of that integer will be "laid bare" as we go deeply enough into $L$'s decimal expansion, as eventually the 1s are far enough away to become that integer without stepping on each other.

Similarly, if we multiply $L^2$ by an integer, we will see that integer in some places, and 2 times that integer in others. For large enough $n,$ if we look between the $n!$ place and the $(n+1)!$ place, the last thing we'll see is that integer written at the $2n!$ place.

Thus the swaths of zero in the multiple of $L$ are, $n!-(n-1)!=(n-1)(n-1)!$ long (minus a constant), whereas the widest swaths of zero in the multiple of $L^2$ are $n!-2(n-1)!=(n-2)(n-1)!$ (minus a constant) long, which is shorter, so there is no way to add positive multiples of $L$ and $L^2$ together to clear everything after the decimal point, or find positive multiples of each so that everything after the decimal point is equal.

More generally:

Suppose $a_jL^j+...$ and $a_kL^k+...$ are integer polynomials in $L,$ where $j>k.$ We show that their values cannot match up fully past the decimal point. The swaths of zero in the first polynomial, moving back from the $n!$ spot, are a constant away from $(n-j)(n-1)!$ long (the constant being the length of the sum of the coefficients), whereas in the second they are a constant away from $(n-k)(n-1)!$ long, in the same place (moving back from the $n!$ spot).

I don't know if this explanation holds up to the standards of rigor you like to maintain when teaching them, but I think they will find it fascinating.

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@David Feldman: Agreed! It was, for me, a very thought-provoking question. –  Daniel Briggs Dec 14 '10 at 6:55
Daniel, thanks so much for your thoughtful answer. A few clarifying questions: 1) How do you know that a 4 doesn't pop up somewhere in L^2, or more generally that the sum of some factorials doesn't equal the sum of some others? 2) Do you mean if we look between the 2(n-1)! and the 2n! place for large enough n, we'll see the integer multiplier bare at the end of that stretch? If so, how do we know that there aren't any 2's cropping up along the way that would mess things up? 3) What with the 2's, I feel lost on how you're calculating the length of the swaths of zeros. Could you expand on this? –  Justin Lanier Dec 15 '10 at 1:08
Given mL^2 for an integer m, let's go back from the n! spot towards the 2(n-1)! spot. Near the n! spot there can be contributions of the form 2m 10^-n!10^-k! for small k, but notice that the effect of the 10^-k! is to move right, rather than left, and the smallest values k! can take on are 1, 2, 6, 24, ... . So .2m+.02m+.000002m+... will be seen here, but that makes at most one more positive digit left than .2m does, and it's the same near the n! spot for any n. Once you get past it, there's nothing going back to 2(n-1)!, and then there's something (if m has final 0s, once we get past them). –  Daniel Briggs Dec 15 '10 at 13:53
Or, increasing from 2(n-1)!, the first factorial sum to be seen is n!+1!. (And decreasing from 2(n-1)!, it's (n-1)!+(n-2)!, which is very far away, and this shows that the factorial sums can't conspire to make "magic" 0s.) Similarly, with mL^3, all the products involving at least one 1 from the n! place on make less than 3m 10^-n! L^2 (the 3 is from choosing the 1 to be in the n! place in the first, second, third L, and the "less than" from the microscopic overcounting); moving left from here, the first thing we see is m at the 3(n-1)! place, which is sooner than in a multiple of L^2. –  Daniel Briggs Dec 15 '10 at 14:20

I keep meaning to write this up nicely, but one can prove the transcendentality of Liouville's number in a very, very elementary way.

Write $L$ for the Liouville number. Suppose $p(L)=0$ for some polynomial with integer coefficients. Then $p_+(L)=p_-(L)$ where $p=p_+-p_-$ and polynomials $p_+$ and $p_-$ have only positive coefficients and no terms of the same degree. Assume WLOG that $p_+$ has the higher degree, say, $k$, so $p_+(x)=cx^k+\cdots$. Then when you calculate $p_+(L)$ (via the distributive law) you'll get contributions of $c10^{-kn!}$ for $n=1,2,3\ldots$. With $n$ large enough, nothing arising out of $p_-(L)$ can balance these contributions, so contradiction.

So no calculus! Just a little thought about how grade school arithmetic goes.

I believe that the distinction between Liouville and Cantor actually turns out artificial.

My argument above shows that one can view Liouville's construction as a diagonalization against the polynomials. Each nonzero digit "kills" a collection of polynomials of low degree and low height until one has killed them all!

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Hi Daniel Briggs...two minds with but a single thought... your answer appeared just as I posted! –  David Feldman Dec 14 '10 at 6:46
This type of idea came up in a letter from Goldbach to Daniel Bernoulli. Goldbach claimed that the number with 1s in every 2^k-th place and 0 everywhere else is irrational because the decimal expansion was not periodic. Liouville, in his article containing his number, actually refers to a letter by Goldbach. –  Franz Lemmermeyer Dec 14 '10 at 18:17
It would be interesting to see what portion of Liouville numbers can be taken care of without much trouble by using an appropriate base for expansion and discussing the number in terms of the base (it seems that (1) q must be able to be chosen in a relatively uniform way, such as powers of the base, and (2) the places with the positive digits would have to become sparse enough so that the number wouldn't get muddy: would this requirement be equivalent to the Liouville criterion? Or stronger?) –  Daniel Briggs Dec 14 '10 at 23:15
Hi, David. Thanks for your answer. Can you expand upon why "With n large enough, nothing arising out of p_(L) can balance these contributions"? I like the approach of focusing on the term of highest power, but I only see that for large n, the contribution is very small, and so I don't see the contradiction. I also like you remark about Cantor and Liouville being equivalent. But I don't see how exactly it could be that each nonzero digit "kills" a collection of polynomials, since if you removed one 1 and left the rest of L the same, wouldn't it still be transcendental? Thanks again. –  Justin Lanier Dec 15 '10 at 1:21

I favour the Liouvillian approach over the Cantorian approach, because although the diagonal argument will in principle let you write down the decimal expansion of a transcendental number, you will never see the whole expansion "all at once" (so to speak), and there will be nothing special about the truncated expansion you do see; it will just be some random decimal expansion, and any finite decimal expansion can be completed to be transcendental.

In the approach via Liouville, of course, you get to actually see the transcendental number.
And Liouville's argument is not so difficult; it boils down to the pigeonhole principle (which, by the sounds of what you've been teaching them, your 7th graders will have no trouble understanding, if they don't already know it).

I don't know an optimal reference, but if I was to pursue this path, I would fix my Liouville number first, and just focus on proving that that particular number is transcendental. (In other words, don't prove a general criterion and then check that your number satisfies; keep things more concrete by just directly proving the criterion for your chosen Liouville number.) I think that if you do this, and you just write down a putative polynomial equation with integer coeffs. that your Liouville number is supposed to satisfy, it won't be hard to argue your way to a contradiction. And because you have a concrete number, you can really work this through with your class, e.g. by beginning with a quadratic , actually plugging in your Liouville number, and staring at it and seeing why this couldn't give zero. I think this would make things quite intuitive and concrete.

Added: See Daniel Briggs's very nice answer for exactly this kind of explicit argument with a concrete Liouville number.

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The Liouville number is certainly the first concrete example to mention (it is also, not by chance, the first one historically). I love the way trascendence can be shown by means of elementary operations as shown here by David Feldman and Daniel Briggs. I don't know how much time you will devote to this fascinating topic, but I wouldn't omit to mention the origin, the quadrature of the circle, the most famous problem of the antiquity, and possibly the one who remained open the longest time in all history of mathematics. A small historical perspective renders justice to Liouville example (1851), since it then appears to be not just a mathematical curiosity, but a first step towards a possible proof of trascendence of other important constants. In a sense, Liouville number is trascendental because it has a too fast rational approximation. So it also illustrates a kind of paradoxical character of trascendental numbers: rational numbers seems to be "closer" to trascendental numbers than to non-rational algebraic numbers. Indeed, the first case of an already known constant proven to be trascendental was e (Hermite, 1873), exploiting the very good rational approximation given by the exponential series (depending on time, you may consider including in the course also Hilbert's simpler version of the proof, that only requires elementary calculus): and the case of e was certainly a starting point for Lindemann's work for the more difficult trascendence of $\pi$ (1882).

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How about Chaitin's Constant $\Omega$ (for some fixed encoding)? The proof of transcendence is doable, and is in some ways a compromise between the two approaches, with the drawback that for any given encoding, one obviously can't actually write down the constant beyond the first few digits, though the proof is in some sense "constructive."

Of course this is basically diagonalization (hidden in the proof of the insolubility of the Halting Problem) but I think it might be easier for a seventh-grader to get his hands on the first few digits of the constant than in your option (1).

Plus the language of computability is wonderful, and I think easily understood by middle-schoolers.

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Full details require a short paper rather than a long MO comment and Daniel Briggs and I have started discussing writing one jointly. But let me address some of what you ask right now.

A finite collection of polynomials has only a finite number of roots on the real line. So you can find an interval in the real that avoids all of them. Fixing the first digit of Liouville's number together with a minimum gap till the next digit does just that. Note that $1/10$ itself can't be the root of an irreducible polynomial of degree higher than $1$. Now you increase the degree and the height (maximum magnitude of the coefficients) you'll allow your polynomials and this gives you new roots to avoid, but you can still find a small "safe" interval inside the interval you had before. This is the essence of diagonalization. Each further specification of the number kills more polynomials. The trouble is that the set of roots of polynomials of bounded degree and height looks very complicated, so Liouville's trick takes extreme measures to avoid them all but without much computation (where Cantor would determined merely one more digit to avoid the fully calculated root of just one more polynomial).

Now of course you're right - changing a transcendental number by a rational leaves it transcendental. That just means that each polynomial gets killed many times (who says you can't beat a dead horse in mathematics).

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Justin - it occurs to me that there's time value on this for you, right? Email me and I'll be glad to discuss this with you in detail so you have something soon to show your students! –  David Feldman Dec 15 '10 at 2:29
Thanks! Will do! –  Justin Lanier Dec 15 '10 at 3:21

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