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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?

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  • $\begingroup$ I would think numbers like $.121121112\dots$ would do the job, no? Liouville doesn't obviously give us transcendence (at least it isn't obvious to me). $\endgroup$
    – lulu
    Commented Apr 18, 2016 at 18:07
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    $\begingroup$ $e+\pi$ or $e\cdot\pi$. $\endgroup$ Commented Apr 18, 2016 at 18:07
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    $\begingroup$ @studentforever: I don't know that either of those numbers has the indicated property. (Of course at most one of them is rational, but I can't see how that helps here.) $\endgroup$
    – Charles
    Commented Apr 18, 2016 at 18:09
  • $\begingroup$ @lulu I believe that number is known to be transcendental. It's the sum of a rational number ($\frac19$) and a theta-value at a rational argument that I'm pretty sure is known not to be algebraic. $\endgroup$ Commented Apr 18, 2016 at 18:20
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    $\begingroup$ Erdős proved that the Erdős–Borwein constant $\sum_{n = 1}^{\infty} \frac{1}{2^n - 1} = 1.60669\!\ldots$ is irrational, and to my knowledge whether it's algebraic remains open; the comments below this old answer suggest that's the case: math.stackexchange.com/a/266638/155629 . en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Borwein_constant $\endgroup$ Commented Apr 18, 2016 at 18:26

3 Answers 3

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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.

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    $\begingroup$ And $\zeta(2n+1)$ in general for $n\geq 1$ is not known either, I think. $\endgroup$
    – Arkady
    Commented Apr 18, 2016 at 18:17
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    $\begingroup$ @Jake I think that's true, but I don't think it's known for any odd $m > 3$ whether $\zeta(m)$ is even rational. It is known, perhaps a little oddly, that at least one of the first four such numbers is irrational. (See Zudilin, Wadim (2001), One of the numbers $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$ is irrational, Russ. Math. Surv. 56 (4), 774–776.) $\endgroup$ Commented Apr 18, 2016 at 18:20
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    $\begingroup$ Ah yes. I was being hasty. Thank you :) $\endgroup$
    – Arkady
    Commented Apr 18, 2016 at 18:44
  • $\begingroup$ Wow, I thought I knew the Greek symbols by sight, but I had to look up that weird salamander-looking thing to find out it's a lower-case Zeta. $\endgroup$ Commented Apr 18, 2016 at 19:44
  • $\begingroup$ @MasonWheeler: in maths undergrad terms that's "other squiggle", with "squiggle" being $\xi$. $\endgroup$ Commented Apr 19, 2016 at 7:51
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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 $\zeta(5)$, $\zeta(7)$, $\ldots$ $\zeta(21)$.

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  • $\begingroup$ Is it known that these numbers are all irrational? $\endgroup$
    – Charles
    Commented Apr 18, 2016 at 18:22
  • $\begingroup$ @Charles not by me, but at least of them (explicitely) is irrational, and unknown to be algebraic or transcendental. This is why I called my answer non constructive $\endgroup$ Commented Apr 18, 2016 at 18:35
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    $\begingroup$ Fair enough. :) $\endgroup$
    – Charles
    Commented Apr 18, 2016 at 18:38
  • $\begingroup$ @Charles glad you liked it, pfiou... $\endgroup$ Commented Apr 18, 2016 at 21:07
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$$ \frac{\ln\pi}{\pi}, $$

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).

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  • $\begingroup$ Great example, Mason! $\endgroup$
    – Charles
    Commented Aug 31, 2022 at 20:06
  • $\begingroup$ ${\ln(\pi)}/{\ln\Gamma(\frac{22}{3})}$ has the same property of being clearly irrational but not necessarily demonstrably transcendental because $\pi$ and $\Gamma(22/3)$ are algebraically independent according to this.. $\{a,b\}$ has transcendence degree $2$ implies $\ln(a) /\ln(b)$ irrational(but not transcendental). The question becomes: Did the theory that allowed for proving $a$, $b$ algebraic also provide an an inroad for proving $\ln a / \ln b$ transcendental. $\endgroup$
    – Mason
    Commented Sep 15, 2022 at 5:33

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