74

Philosophically, there is essentially only one way to prove that a number is irrational/transcendental, which is to use the fact that there is no integer between 0 and 1. That is, one assumes that the number in question is rational/algebraic, and constructs some quantity that can be shown to be bounded away from 0, less than 1, and also an integer. To get ...

answered May 3 '13 at 18:21

Timothy Chow

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53

In point of fact, K. Mahler proved in this paper that, if $p(x)$ in a non-constant polynomial such that $p(n) \in \mathbb{N}$ for every $n\in \mathbb{N}$, then the number
$$0.p(1)p(2)p(3)p(4)\ldots,$$
which is formed concatenating after the decimal point the values of $p(1), p(2), p(3), \ldots$ (in that order), is a transcendental and non-Liouville number.

45

There are number theorists who understand this subject much better than I do. However, I feel obliged to post an incomplete answer quickly before people have a chance to close this question.
There are a lot more connections known between $\pi$ and $e$ and other numbers than between $\gamma$ and other numbers. We can get proofs of their irrationality by ...

34

(Turning comments into an answer, as requested)
This follows from Schanuel's conjecture but it's probably hard to prove unconditionally.
Apply Schanuel to $2\pi i,\log n,\log m$. The last two numbers are linearly independent over $\mathbb Q$ because of your hypothesis and the fact that $\pi$ is irrational. Then all three numbers are linearly independent ...

31

Unconditionally, this can essentially happen at most once. It is a consequence of the Five Exponentials Theorem, which itself follows from a combination of the classical Six Exponentials Theorem and Baker's Theorem on linear forms in logarithms.
Five Exponentials Theorem: (See Waldschmidt, Diophantine Approximation on Linear Alebgraic Groups, 2000, Section ...

answered Jun 17 '15 at 23:18

Matt Papanikolas

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17

I gather that the idea behind $\mathrm{Gal}(\pi)=\mathbb{Z}\backslash\{0\}$ (not $\mathbb{Z}$, $0$ is not a conjugate of $\pi$!) comes from Euler's formula:
$$\prod_{n\in \mathbb{Z\backslash\{0\}}}\bigg(1-\frac{x}{n\pi}\bigg)=\frac{\mathrm{sin}(x)}{x} \in\mathbb{Q}\{x\} $$
which can make you think of those $n\pi$ as the conjugates of $\pi$.
But in order ...

17

Your first question can be taken with several senses, from weak to
strong. Let us understand your "explicit" terminology to mean
"computable".
(weak version) Is there a computable generic value?
(medium) Can we give a specific algorithm for computing a generic
value?
(strong) Can we identify a computable generic value that we can
also understand in a simple ...

answered Aug 23 '13 at 0:04

Joel David Hamkins

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17

At the present time, we do not even know how to prove that the Euler-Mascheroni constant $\gamma=\lim_{n\to\infty} \sum_{k=1}^n\frac{1}{k} - \log n$ is irrational, much less transcendental; although it is conjectured to be transcendental. The reason you won't find a lot about this topic online (or in the research literature) is because so little is known ...

answered Apr 30 '17 at 20:12

Joe Silverman

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16

The Gauss-Kuzmin Theorem says that if $x$ is chosen uniformly at random from (say) $[0,1)$ (thanks, John Bentin) then as $n\to\infty$ the probability that the $n$th partial quotient of $x$ is $k$ tends to $$\log_2{(k+1)^2\over k(k+2)}$$ It was widely believed in 1978, and is still widely believed today, that $\pi$ acts, in this regard, like a random number. ...

answered Apr 23 '14 at 6:20

Gerry Myerson

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16

Robert Israel's guess that this is true for $t$ an algebraic irrational does actually follow from Schanuel's conjecture.
Apply Schanuel's conjecture to $\log p$ and $t \log p$ for $n$ different primes $p$. These are linearly independent unless $t$ is the ratio of the logarithms of two integers, which it isn't because $t$ is an algebraic irrational.
So we ...

16

Computable, absolutely normal numbers do actually exist. See
V. Becher, S. Figueira: An example of a computable absolutely normal number, Theoretical Computer Science 270 (2002), 947-958.

nt.number-theory pr.probability stochastic-differential-equations transcendental-number-theory algorithmic-randomness

answered Aug 21 '17 at 18:13

Francesco Polizzi

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16

What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
$$\frac{p}{q} < x < \frac{p}{q} + \frac{1}{q^2}$$
so that the first two digits in the base-$q$ expansion of $x$ ...

nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution

15

Barry, this post is coming a "little bit" after your class ended, but I want to direct you to a paper on the Lindemann-Weierstrass theorem by Beukers, Bezivin, and Robba in the Amer. Math. Monthly in 1990: https://www.jstor.org/stable/pdf/2324683.pdf. They give a proof of LW theorem that does not explicitly involve a choice of an auxiliary prime number, and ...

14

Irrationality measure is a question about approximation by rationals. The continued fraction expansion gives the best approximations and controls their quality. Irrationality measure is a kind of asymptotic growth of the continued fraction expansion. Asking about the irrationality measure of a particular number is asking properties of its continued fraction ...

13

Two observations:
Not every transcendental is generic. Indeed the real solution $r$ of the equation $x^x=2$ is not generic because $r$ satisfies
$$x^{x^{x^x}}=\left( x^x\right)^x,$$
and $r$ is transcendental by the Gelfond-Schneider theorem.
Here is an argument showing that no positive real algebraic is generic (by which I mean that every positive real ...

13

According to the last paragraph in Section 3 of the paper "Transcendental numbers in the p-adic domain" by William W. Adams (Amer. J. of Math., Vol. 88, 1966):
http://www.jstor.org/discover/10.2307/2373193?uid=3738736&uid=2&uid=4&sid=21101389828747
the answer is yes. More specifically, they prove that if $a \in \mathbf{Q}_p$ is a non-zero ...

13

The conjecture has been neither refuted nor proved. The state of the art, as far as I know, is contained in the papers of Adamczewski and Bugeaud, in which they show that anything with a very low complexity decimal expansion cannot be an algebraic irrational. The complexity is the function $c_x(n)$ giving the number of blocks of length $n$ in the decimal ...

13

Algebraic Independence of Values of Exponential and Elliptic Functions, G. V. Chudnovsky (1978)
(a.i. = algebraically independent)

answered Jun 23 '15 at 6:19

Carlo Beenakker

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13

The irrationality of $\log \pi$ is an open problem (see for example this recent paper).
It is expected to be transcendental (page 34 of this slides by Michel Waldschmidt), and in fact this follows from Schanuel's conjecture (this is referenced here, beginning of section 3), which is widely believed to be true.
In particular, to answer the question in the ...

13

Here is a 2018 paper, A Note on One-dimensional Varieties Over the Complex p-adic Field, that still lists the "full" statement as a conjecture; "half" of the statement, meaning that at least $\lfloor N/2\rfloor$ of the exponents are independent, has been proven by Nesterenko.

answered Dec 12 '18 at 21:47

Carlo Beenakker

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12

They are not always independent. There are two problematic cases: the trivial case $a=b$, and the trinomial case when $c_1x^n + c_2x^m +c_3 =0$ for rational $c_1,c_2,c_3$, $a= x^n$, and $ b= x^m$, so that $ (a^a)^{c_1 /n} (b^b)^{c_2/m} x^{c_3}=1$, giving an algebraic relation between $a^a$ and $b^b$.
In every other case they are independent conditional on ...

11

The answer is probably no.
In his paper Transcendence of Periods: the State of the Art (Pure and Applied Mathematics Quarterly, Volume 2, Number 2, p. 435-463, 2006), Michel Waldschmidt conjectures that no period is a Liouville number (see questions 2 & 3 in the introduction). This expectation is supported by the main advances of transcendental number ...

11

This is most likely open, since alredy $e^{\pi^2}$ is not known to be transcendental.
As an added difficulty, I don't think that $\frac{\pi^2}{12 \log 2}$ is known to be transcendental either.
There are very few, very limited, tricks to prove this kind of result: things like taking $(-1)^{-i}$ and $i^i$ and applying Gelfond–Schneider, or building the ...

10

The expected value is $2$, simply because almost all real numbers have irrationality measure $2$. Note that this is "almost all" in the sense of measure theory. In the sense of topological category, it's the other way around: the set of reals of measure $\infty$ is residual. As I pointed out recently, it's a bit mysterious why measure theory makes better ...

nt.number-theory soft-question analytic-number-theory diophantine-approximation transcendental-number-theory

10

I think the follwing is the answer of this question. It is not depend on the above comments because I didn't understand them. This is just my approach.
(But not fully my idea because it depends on a strong proposition which is already known, and the remaining part is just a corollary.)
Proposition. Let $f$ be a modular form of weight $k$, defined over $\...

9

By the Gelfond–Schneider theorem, $2^{\sqrt2}$ is transcendental.
$2^{\sqrt 2}$ is called the Gelfond–Schneider constant.
See also https://math.stackexchange.com/questions/173804/deciding-whether-2-sqrt2-is-irrational-transcendental

9

There is also Simpson's proof that isolated points of the characteristic varieties of fundamental groups of projective manifolds are torsion.
It also relies on Gelfond-Schneider Theorem.
The moduli space of representations of those fundamental groups on $\mathbb C^*$ admit three different algebraic/analytic structures. And in each of these the ...

answered Feb 25 '16 at 1:43

Jorge Vitório Pereira

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