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I am reading A. van der Poorten's 1978 paper on Apery's constant, and it cited the Thue-Siegel-Roth Theorem (that if $\beta$ is algebraic, then for all $\epsilon > 0$ the inequality $|\beta - p/q| \leq 1/q^{2+\epsilon}$ has only finitely many solutions) as a way to test whether a given number is transcendental, but that this method is not very satisfactory since only a set of measure zero of transcendental numbers can be detected to be transcendental in this way.

So my question is, since 1978 has anyone devised a method that can (at least in theory) test whether a set of numbers with positive measure is transcendental or not?

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  • $\begingroup$ What does it mean to "test" an arbitrary a set of real numbers of positive measure? Perhaps you would be satisfied with an "explicit" description of a single set of transcendentals of positive measure? If so what counts as "explicit? $\endgroup$
    – Sidney Raffer
    Commented Feb 21, 2011 at 5:09
  • $\begingroup$ As a simple example, we can start with Liouville's constant, $\displaystyle \sum_{n=1}^\infty \frac{1}{10^{n!}}$ which can easily proved to be transcendental, and at the same time conclude that $\displaystyle \sum_{n=1}^\infty \frac{a_n}{10^{n!}}$ for any sequence $(a_n)$ such that $a_n = 0,1$ for all $n$ is also transcendental. Thus, we can conclude that an uncountable set of real numbers is transcendental, but this set will have measure 0. $\endgroup$
    – Stanley Yao Xiao
    Commented Feb 21, 2011 at 5:09

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How about checking if the number is computable?

It doesn't seem to me that the question is well posed, because you haven't given a precise notion of "test" and how a real number is given to you. My belief is that most explicit reals appearing in number theory are computable and thus inside a set of measure zero. See this previous answer by Joel David Hamkins for some of the hierarchies one can consider this way.

Notice the similarity that computability is also a property defined in terms of an approximation-by-rationals property, but even though it gives a way to recognize way more transcendental numbers, it is unlikely to have direct applications in number theory for the reason mentioned above. The results from Diophantine approximations you mention are, on the other hand, more valuable since they have applications ranging from solutions to Diophantine equations to transcendence of constants arising in various places (still within the computable world). See also this answer by Greg Kuperberg to a previous question, which is close to this point of view.

I'm sorry if this long comment isn't very enlightening, but maybe you can edit your question a bit to express what you would look for in a helpful answer.

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The question is indeed not well posed. Here is a nice example from [J.M. Borwein and P.B. Borwein, On the generating function of the integer part: $[n\alpha +\gamma]$, J. Number Theory 43 (1993), no. 3, 293--318], namely, part (b) of Theorem 0.4 there.

Consider the function $F(\alpha):=\sum_{n=1}^\infty\lfloor n\alpha\rfloor/2^n$ and the set of irrational numbers $A\subset\mathbb R$ that have unbounded partial quotients in their partial fraction expansions. (For example, $$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,\dots]=2+\frac1{1+\dfrac1{2+\dfrac1{\ddots}}} $$ belongs to $A$.) The set $A$ has full measure in $\mathbb R$ as its complement is of measure 0. The values $F(\alpha)$ at $\alpha\in A$ are all transcendental Liouville numbers. (Note the image set has... measure zero.)

A related variant of test for a given $\beta\in\mathbb R$ would be: if there exists $\alpha\in A$ such $\beta=F(\alpha)$, then $\beta$ is transcendental.

Is this a transcendence method for the full-measure set or a zero-measure set? $\ddot\smile$

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Controlling the measure doesn't help much, since the algebraic numbers have measure zero. You'd need some kind of closure property to make the question nontrivial.

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  • $\begingroup$ Are you referring to the availability of silly examples, like taking the positive-measure set to be the set of all transcendental numbers and taking the "test" to be "say yes" (which works fine on that set)? I agree that any meaningful version of the question would have to exclude this example, as well as variations of it where the silliness is more hidden. $\endgroup$
    – Andreas Blass
    Commented Apr 29, 2012 at 23:43

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