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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?

Travis Willse
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Charles
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  • 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). – lulu Apr 18 '16 at 18:07
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    $e+\pi$ or $e\cdot\pi$. – student forever Apr 18 '16 at 18:07
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    @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.) – Charles Apr 18 '16 at 18:09
  • @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. – Steven Stadnicki Apr 18 '16 at 18:20
  • @StevenStadnicki Ah...let's see. Certainly a mistake to use only $1's$...makes it too easy to subtract $\frac 19$. Is my example better if we cycle through the digits? $.192293339\dots$? Surely there's a simple way to write a non-periodic decimal to which we can't apply Liouville or the like... – lulu Apr 18 '16 at 18:26
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    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: http://math.stackexchange.com/a/266638/155629 . https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Borwein_constant – Travis Willse Apr 18 '16 at 18:26

3 Answers3

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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.

Travis Willse
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    And $\zeta(2n+1)$ in general for $n\geq 1$ is not known either, I think. – Arkady Apr 18 '16 at 18:17
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    @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.) – Travis Willse Apr 18 '16 at 18:20
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    Ah yes. I was being hasty. Thank you :) – Arkady Apr 18 '16 at 18:44
  • 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. – Mason Wheeler Apr 18 '16 at 19:44
  • @MasonWheeler: in maths undergrad terms that's "other squiggle", with "squiggle" being $\xi$. – Steve Jessop Apr 19 '16 at 07:51
  • Shouldn't the summation in Riemann's Zeta function start at n=1? – Vinícius Barros Apr 21 '16 at 22:58
  • @ViníciusBarros Yes, of course, thanks for pointing it out (I've fixed the typo). – Travis Willse Apr 22 '16 at 08:55
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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 $\zeta(5)$, $\zeta(7)$, $\ldots$ $\zeta(21)$.

Laurent Duval
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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).

Mason
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  • Great example, Mason! – Charles Aug 31 '22 at 20:06
  • ${\ln(\pi)}/{\ln\Gamma(\frac{22}{3})}$ 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. – Mason Sep 15 '22 at 05:33