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

$$ \frac{\ln\pi}{\pi}, $$

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.

Another nice example might be $t$ such that $2^t+3^t=6$. This is perfectly explicit despite us not being able to write down $t$ in terms of other constants. It's also nice and constructive in a way that should allow for making more examples. 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).