# How hard is the proof of $\pi$ or $e$ being transcendental?

I understand that $\pi$ and $e$ are transcendental and that these are not simple facts. I mean, I have been told that these results are deep and difficult, and I am happy to believe them. I am curious what types of techniques are used and just how difficult of a problem it is. Would this result be a reasonable capstone to any course? (any course that isn't essentially "how to prove...")

Another part of this is the following observation: as time passes deep results become easier to understand or rather assimilate into ones body of knowledge and some problems are just hard. I am wondering if people feel like this result is something that a grad student could spend some leisure time and understand, or if it really is something only graspable by "experts" (meaning people in the appropriate field and not a general mathematical audience). How specialized are the techniques used for the problem at hand? Have they been used to prove different results? are the techniques drastically different for $e$ and $\pi$?

Hope this isn't too soft of a question. I was talking with my roommate, also a math grad student, and it came up. I said that it was a classically difficult result, but then wondered if that was right, so here I am.

Note: I don't want a proof or a sketch of one, but maybe a heuristic as to why new techniques were needed or explaining the troubles one has when being naive or using early methods to attack the problem.

Edit: As Matt E pointed out I should also ask: what are the old techniques? Also, did I even tag the question correctly? These seem to be the areas I would put them in, but I don't know the anything about this stuff.

In the above it also isn't clear that I am wondering if the proof of this result is gotten by some clever new trick or by lots of hard had work that people couldn't/didn't do before? essentially, is it all elbow grease, or some clever new machinery or something completely different?

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If you look at the proof, e.g. in user3971's answer below, you can see that it is easy enough to follow, but not so easy to come up with! (As for your question about "why new techniques were needed", what "old techniques" did you have in mind? I think one basic difficulty with transcendence results, at least one that I have, is figuring out what technique at all you could possibly use to check that some given number is transcendental.) – Matt E Dec 3 '10 at 6:31
user3971's answer seems to have disappeared, but it linked to a useful planetmath page: planetmath.org/… – Matt E Dec 3 '10 at 6:37
I think even an undergrad can understand the proofs, but I haven't read them myslef so I can't be sure. – Adam Dec 8 '13 at 14:16
Here is a simple proof to understand for e. – Edi Dec 17 '15 at 4:45

This is an answer elaborating on my comment above, and largely addressing the edits to the original question:

I don't think there were any "old techniques" before Hermite proved transcendence of $e$ in the early 1870s. As far as I know, the only transcendental numbers known before then were Liouville's interesting but somewhat artificial examples from early that century.

The subject of transcendence, and the related subject of Diophantine approximation (i.e. approximating irrational numbers, especially irrational algebraic numbers, by rationals), is relatively new. Liouville proved the first results showing that it is not so easy to approximate an irrational algebraic number by rational ones, and used this to construct his transcendental numbers, which he could recognize as being transcendental because they are too well approximated by rational numbers. ("Too well" and "not so easy" here refer to the following problem: if you try to approximate $\alpha$ by the rational number $p/q$, can you get within a distance of $O(q^{-n})$ for some given $n$ as you let the denominator $q$ get arbitrarily large. (The larger $n$ is, the smaller $q^{-n}$ is, and so the better the rate of approximation.) Liouville showed, using the pigeon hole principal more-or-less, that if $\alpha$ is algebraic of degree $d$ then you can't do better than $n = d$.) But this left open the problem of showing that various given numbers (like $e$ and $\pi$) are transcendental.

If you like, here is one way to think of the problem: if you want to show that $\alpha$ is transcendental, then you want to show that $f(\alpha) \neq 0$ for any non-zero polynomial $f$ with rational coefficients. The difficulty is that there will be lots of polynomials with real coefficients that have $\alpha$ as a root, and any one of them can be approximated as closely as you like by an $f$ with rational coefficients, so we can find (lots of!) $f$ with rational coefficients such that $f(\alpha)$ is as close to zero as we like.

So you have to find some way to pin down the difference between $f(\alpha)$ being zero and $f(\alpha)$ being very close to zero. This is not so easy! (For example, computationally, you can't tell the difference between $0$ and any real number that is smaller than your computational accuracy can recognize.) And now one sees why Diophantine approximation ideas of the type mentioned above are relevant. They are related to quantifying how close we can make $f(\alpha)$ to zero while bounding the denominators of the rational numbers involved.

It is not coincidence that bounding the denominators is relevant: morally this is an attempt to pass from working over $\mathbb Q$ to working over $\mathbb Z$. Why do we want to do this? Well, as I already noted, it's pretty hard to tell the difference between $\mathbb Q$ and $\mathbb R$, since the former is dense in the latter, but we can tell the difference between $\mathbb Z$ and $\mathbb R$, since the former is discrete: a non-zero integer is some definite positive distance (i.e. at least 1) away from $0$.

The preceding remarks are somewhat philosophical, and they reflect my (limited) experience of thinking about these kinds of questions. If you look at the proofs in the link above, it may not be obvious that they are relevant, but I believe that they in fact do have some relevance: e.g. you will see that the arguments reduce to considering integer rather than rational polynomials, and that growth considerations play a key role. Another thing you will see is that certain auxiliary polynomials enter the proof, and a key fact about them is that they have a high order of vanishing at their zeroes. The appearance of auxiliary polynomials, often with a high order of vanishing, is ubiquitous in this theory.

One more (somewhat cultural) remark: Roth's theorem, for which he got the Fields medal, is the ultimate strengthening of Liouville's theorem: he shows that if $\alpha$ is irrational algebraic, then one can't do better than $O(q^{-2})$ in the problem of Diophantine approximation discussed above. The proof involves (among other things) constructions with auxiliary polynomials. So my impression is that Liouville, Hermite, and Lindemann (and there are probably other names that should be here) invented a new subject, namely Diophantine approximation and transcendence theory, whose modern methods are an outgrowth of the methods that they introduced.

P.S. Reading the first few pages of Baker's book Transcendental Number Theory (Google Books link) might help.

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Matt, I cleaned the Google Books link up a wee bit; I hope you don't mind. – J. M. Dec 3 '10 at 11:22
@J.M.: Dear J.M., I don't mind at all; thanks a lot! Best wishes, – Matt E Dec 3 '10 at 13:48

As mentioned above, it can be easily proven that e is transcendental. One can then combine it with $e^{iπ}+1=0$ (Euler's equation). Lindemann (I think) proved that if $e^x+1=0$, then $x$ has to be transcendental. And since $i$π is transcendental, ($i$ is obviously algebraic), π has to be transcendental, too. So, this way the proof seems rather easy, but there are much more complicated ways to prove it (as far as I know), which I will leave for others to elaborate on.

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How recent/difficult is Lindemann's result? – Sean Tilson Dec 3 '10 at 6:33
According to the planetmath page linked in my above comment, Hermite proved transcendence of $e$ in 1873, and then Lindemann proved his more general results (including transcendence of $\pi$) later in the same decade. – Matt E Dec 3 '10 at 6:37
@Sean Tilson: Here's the Wiki page: en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem So it's from somewhere between 1875 and 1885. The difficulty you can judge for yourself. – Adrian Petrescu Dec 3 '10 at 6:38
"The proof seems rather easy": I think any difficulty has just been moved to Lindemman's result. – Matt E Dec 3 '10 at 6:40
It was proven in 1882, if I am not mistaken. Actually, you can see the proofs (both) here: mayer.dial.pipex.com/maths/docs/pi.html . Matt -- sorry, that's right. – InterestedGuest Dec 3 '10 at 6:40

This paper by Ivan Niven, The transcendence of π [Amer. Math. Monthly 46, (1939). 469–471], is short and readable. His book Irrational numbers contains this result and much more and is highly recommended.

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Here is a link to Niven's paper, and here is a link to Niven's book. – J. M. Dec 3 '10 at 11:25
This paper has been reproduced in Appendix A of Jonathan M. Borwein and Scott T. Chapman, "I Prefer Pi: A Brief History and Anthology of Articles in the American Mathematical Monthly", American Mathematical Monthly, Volume 122, Issue 03, pp. 189 - 296, March 2015 (downloadable from carma.newcastle.edu.au/jon/31415.pdf from the first author's web page at carma.newcastle.edu.au/jon/index-papers.shtml). – Marnix Klooster Oct 1 '15 at 6:00

For the proofs, you would of course need to know the necessary notions from mathematical analysis, covered in a good course, for instance the equivalent of what is in W. Rudin's principles of mathematical analysis.

Assuming you are comfortable with the analytic notions, then proving the transcendence of $e$ is quite easy. It can be done using the series expansion for $e$. This is done in the introductory algebra book of I. N. Herstein, "Topics in algebra". The transcendence of $pi$ is harder. But it is still within the grasp of a beginning graduate student, though the proof is usually not included in the curriculum. It is given for instance in the algebra book of N. Jacobson, "Basic Algebra", Vol 1..

A proof sketch of the Lindemann-Weierstrass theorem is in fact given in wikipedia.

So, the answer to your question of how difficult the theorems are, the answer is that they are accessible to beginning graduate students, and undergraduates who are willing to put in the extra labor.

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Dear George, Comparing Hernstein's proof with the one on Planetmath (which is taken from Baker's book), they are not so different. The proof of transcendence of $\pi$ on Planetmath (again taken from Baker), while harder, is not so much harder. The main new ingredient is the basics of algebraic numbers and integers. (The point of this comment is just to be a little more explicit about the difference between the $e$ and $\pi$ cases, the point being that if you are comfortable with arguments involving algebraic numbers, then the difference in difficulty is not very big.) Best wishes, – Matt E Dec 3 '10 at 14:04
@Matt E: All I wanted to convey in this answer is that the proof can be found in algebra books for beginning graduate students. Baker's book is very good; but some students are easier to convince if the proof is available in an introductory algebra textbook instead of a number theory textbook. I have met some students and more senior people who are on the one hand reasonably skilled in some topic in mathematics but on the other hand are not very eager to get into number theory. – user1119 Dec 4 '10 at 19:51

I would refer to Calculus by Spivak, a complete proof of the transcendence of e is given and is presented as a gateway proof, distinguising the mathematician from the student.

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You'll need to know one thing that you may not have learned in high school: Stirling's approximation. Find it on wiki and just keep the formula in mind. If you know basic integration well enough, it should be all okay to understand. And no, the techniques used for proving their transcendence is much the same, in fact, they are almost the same.

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