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It's hard to prove a number is transcendental (non-algebraic) yet there are some wonderful examples amongst them like π,e and Liouville's number. What's so special about them?

Are most numbers transcendental?

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closed as off-topic by Lucia, Chris Godsil, Andreas Blass, Joël, Willie Wong Apr 3 '14 at 9:37

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I'm a grumpy old pedant, and I'll start closing questions with poor grammar if too many start appearing. I fixed this one ... – Scott Morrison Oct 21 '09 at 0:48
not to be mean or anything... Diab is NOT a toddler... and is this really a suitable question for MO?? – Yaakov Baruch Mar 13 '12 at 13:07
Absolutely nothing is special about transcendental numbers. Algebraic numbers are special, though. Rational numbers even more so, natural numbers are particularly special, and $0$ and $1$ are the most special numbers of them all. Except $42$, which is special because it is the answer. – Asaf Karagila Apr 2 '14 at 19:17
up vote 14 down vote accepted

The set of real numbers is uncountable, but the set of algebraic numbers is countable, so "most" real numbers are transcendental in a very strong sense of "most". This is actually a capsule description of Cantor's proof of the existence of transcendental numbers; just note that an uncountable set cannot be empty. Looking at the interval [0,1], the set of algebraic numbers there have (Lebesgue) measure zero, so a number picked "at random" (uniform distribution) from that interval is transcendental with probability 1. Transcendental numbers are a dime a dozen - but to prove that particular real numbers are transcendental is either hard or just too hard. It is known that $ e^\pi $ is transcendental, but as to $\pi^e$, nobody knows. Transcendental numbers are studied for their own sake. Important mathematicians found them interesting, so they must be important :) Euler was the first to consider the possibility that there might be real numbers that were not algebraic, Liouville constructed the first ones, Hermite proved e transcendental, and so forth. The closely connected subject of linear forms in logarithms has applications in more mainstream number theory, especially to Diophantine equations.

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The cool thing about transcendental numbers is that they set the set of real numbers, or more specifically uncountable sets, apart from every other set of numbers. The natural numbers for instance, you can get by counting, the integers by adding closure under subtraction, and the rationals by further adding closure under division, and finally reaching the point of a field. Once you get to this point, you still don't have all the numbers out there. The greeks didn't know what to do with sqrt(2) when they found that it was irrational, because everything they knew was integers and ratios (ie. rational numbers.) Let's say we add closure under the square root operation, we then get the constructable numbers (those that are reachable by Euclidean constructions), but we're still missing, for instance the cube root of 2, so let's add all the cube roots, and while we're at it, just add all the n roots, but we still don't have the real numbers.

As mentioned above, even once we've added all of the (non-complex) roots of every polynomial of finite degree, we still have a countable set. This is because the integers are countable, and the set of polynomials of degree n is countable, and countable unions of countable sets are also countable. The set of algebraic numbers is, effectively, the set of all numbers for which we can give an exact description such as (1 + sqrt(5))/2 (the golden ratio). But we know that the real numbers are uncountable (there are several proofs of this). So along comes the skeptic who says, "If you know that infinitely many transcendental numbers exist, then I dare you to show me one." At this point, you can point the skeptic to pi or e, and show that such numbers do, in fact, exist and are rather useful.

When it comes to sets of numbers, we can start with our fingers and toes and build up to some very large sets, but we can never quite construct the real numbers. Once we make the jump to the real numbers we can go on to all sorts of other wonderful sets such as R^2 and C, but you cannot construct the reals from the naturals. The transcendental numbers are form the gap between all the countable sets and the reals, and so finding examples of some do actually exist is very important.

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I like a lot of things about this answer! The one statement I disagree with is "we can never quite construct the real numbers". Dedekind cuts and equivalence classes of Cauchy sequences are two ways to construct the real numbers (more precisely, an ordered field that satisfies all the properties we want the real numbers to satisfy) from the rational numbers, which are themselves constructed from the integers. Of course this construction is pretty far from fingers and toes, I concede! – Greg Martin Apr 2 '14 at 20:28

I would turn this on its head. It's not that transcendentals are special. It's that the algebraic numbers are special. It's very unusual (in a precisely defined sense) for a real number to be expressible as the solution to an algebraic equation with rational coefficients.

As a related aside, there is an interesting class of real numbers containing the algebraic numbers: the periods. These are the numbers that can be constructed by integrating algebraic functions over an algebraically defined domain. For example you can define the unit disk in the plane with x^2+y^2<=1 and so you can construct pi by integrating 1 over this domain. It's amazing how many of the usual constants of mathematics can be obtained in this way, although a few important ones like e get left out.

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Most numbers are transcendental. In particular, the set of algebraic numbers is countable -- basically because there are countably many polynomials, each with countably many roots. (This is a generalization of the argument that the rationals are countable.) But there are uncountably many real numbers.

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Adding to the above, it is worth noting that the transcendental numbers that are commonly known/used in mathematics (e, pi, etc.) belong to the countable subset of transcendental numbers with a finite recursive function generator. The uncountable transcendental numbers, or the random numbers, are those which cannot be generated by a finite recursive function, and they can therefore also not be the solution to a finite analytic equation (unless it contains random numbers or is solved by a subset of the reals). In other words, if you pick a real number at random, the resulting number will be random.

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So, yes, most numbers are transcendental. But you have to be more inventive to define one; you cannot just say take the solution of this or that polynomial ...

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... with rational coefficients. – alekzander Nov 2 '09 at 5:57
... or more generally, algebraic coefficients; thus one can't get a transcendental number this way unless one already has a transcendental number to start with. Instead, one has to do something non-algebraic, such as sum an infinite series. – Terry Tao Nov 2 '09 at 20:27

I would like to mention a highly readable paper by Ram Murty on transcendental numbers. In a narrative fashion, he explains transcendental numbers from Euler via Lindemann, Hermite and Baker to his own work. He describes their esoteric nature by results like - $\zeta(2n) \in \pi^{2n} \mathbb Q$. For example, we know from complex or Fourier analysis that $\displaystyle \zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ but its unknown if $\zeta(2n+1)$ is transcendental even for $n=1$!

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I don't think the result on $\zeta(2n)$ is esoteric. It is a matter of taste of course. – GH from MO Mar 13 '12 at 19:37
It sure is. (now comment is large enough) – Abhishek Parab Mar 13 '12 at 19:42

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