Is there an explicit irrational number which is not known to be either algebraic or transcendental?

Question

There are many numbers which are not able to be classified as being rational, algebraic irrational, or transcendental. Is there an explicit number which is known to be irrational but not known to be either algebraic or transcendental?

Answer 1

Maybe the best-known example is Apery's constant, $$\zeta(3) = \sum_{n = 1}^{\infty} \frac{1}{n^3} = 1.20205\!\ldots ,$$ which Apery proved was irrational a few decades ago; this result is known as Apery's Theorem.

By contrast, $\zeta(2) = \sum_{n = 1}^{\infty} \frac{1}{n^2}$ has value $\frac{\pi^2}{6}$, which is transcendental because $\pi$ is.

Apéry, Roger (1979), Irrationalité de $\zeta(2)$ et $\zeta(3)$, Astérisque (61), 11–13.

Answer 2

The most famous have been answered. Let us be a little less constructive. At least one of $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$ is irrational, a result due to V. V. Zudilin, Communications of Moscow Mathematical Society (2001), and their true nature (algebraic and transcendental) seems unknown at the present time. This result improves the irrationality of one of the nine numbers $\zeta(5)$, $\zeta(7)$, $\ldots$ $\zeta(21)$.

Answer 3

Example 1

$$ \begin{aligned} \frac{\ln(\pi)}{\pi} &= \frac{1.14472988584940017414342735135305871164729481291531 \ldots}{3.14159265358979323846264338327950288419716939937510 \ldots} \end{aligned} $$

This example is irrational because $$\frac{\ln\pi}{\pi}=\frac{p}{q} \implies \pi^{q}=\left(e^{\pi}\right)^{p}$$

But this is impossible because Nesterenko demonstrated that $e^{\pi}$ and $\pi$ are algebraically independent.

As far as I know, there is not a demonstration that $\ln(\pi)/\pi$ is transcendental which doesn't invoke the far from proven Schanuel's Conjecture.

Example 2

The real number solution $2^t+3^t=6$. $$t \approx 1.193911477211585957200358342236350773651$$ The solution to an exponential integer equation should be demonstrably irrational but not necessarily demonstrably transcendental (But again assuming Schanuel's Conjecture one can make progress).

Example 3

\begin{aligned} \frac{L}{C_{10}} &= \frac{\sum_{n=1}^{\infty} 10^{-n!}}{ \sum_{n=1}^{\infty} n \cdot 10^{ -\left( \sum_{k=1}^{n} \left\lceil \log_{10}(k+1) \right\rceil \right) }} &= \frac{0.11000100000000000000000100000000000000100\ldots}{0.12345678910111213141516171819202122232425\ldots} \end{aligned} Any two numbers with distinct irrationality measure must have an irrational quotient (and product). For example, Liouville's Constant and the Champernowne Constant are well known constants with distinct irrationality measures.

For any real $x$ and rational $r$ we have $\mu(rx)=\mu(r/x) =\mu(x)$. So if $\mu(x) \neq \mu(y)$ then $y \neq rx$ and $y \neq r/x$ for any rational $r$ and therefore $x/y$ and $xy$ are not rational.

We have $\mu(L)=\infty \neq \mu(C_{10})=10$. And we conclude the quotient is irrational.

Example 4

$$\sum_{n=0}^\infty 1/(n^2)! \approx 2.04167 $$ is not known (as far as I know) to be demonstrably transcendental. It can be shown irrational by modifying the 'shortest' proof that $e$ is irrational.. The general proof is here. The ambient question of $\sum_{n\in S} 1/n!$ for some subset $S$ of the naturals is discussed here. In contrast, we can demonstrate $\sum_{n=0}^\infty 1/(n!)^2$ transcendental. This is written about here on MSE.

Example 5

$$\sum_{n=1}^\infty 1/n^{2^n}\approx 1.06265241602 $$ This summation has a denominator which grows fast enough to show irrationality but not transcendence.

$\sum_{n=0}^\infty 1/a(n)$ is demonstrably irrational when $a(n)$ grows fast enough: If there exists $N$ such that for $n>N$, $$a_{n+1}-a_{n}^{2}+a_{n}-1>0 \implies \sum a_n^{-1} \notin \mathbb{Q}$$ One can read this here.

And the sum is demonstrably transcendental when $a(n)$ grows even faster. For fixed $\lambda>2$ we have $\lim_{n \to \infty} a_{n+1}/{a_n}^{\lambda+1}>1 \implies \sum_{n=1}^\infty 1/a(n) $ is transcendental. This can be read in Nyblom. For example $\sum_{n=1}^\infty 1/n^{n^n}$ is transcendental by this criteria.

Example 6

$$\prod_{n=1}^ \infty \bigg( 1+1/n^{2^n} \bigg) \approx 2.1253238840491767570143378256139$$

The product $\prod \bigg( 1+1/a(n) \bigg)$ is demonstrably irrational when $a(n)^{2^{-n}} \to \infty$ as $n$ grows large. This is from Erdös.

Example 7

Let $\mathcal{P}$ denote the primes (but any subset of that naturals with an aperiodic characteristic function will do) then $$\sum_{p\in \mathcal{P}} 2^{-p} \approx 0.414683$$ is irrational: It doesn't have an eventually periodic binary expansion. This type of argument shouldn't translate into any claim about algebraic/transcendental which both have aperiodic expansions. This constant is discussed here on MSE.