49

One has $$\int_0^1 x^k e^x dx = A_ke+B_k \to 0$$ as $k\to \infty$, $A_k,B_k\in \mathbb{Z}$, by integration-by-parts and induction. If $e=a/b$, then $(A_k a+B_kb)/b \to 0$, so $A_ke+B_k=0$ for $k$ large, a contradiction.

44

I believe most such questions are still very far from being resolved.
Apparently, it is not even known if $\pi^{\pi^{\pi^\pi}}$ is an integer (let alone irrational).

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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24

There is a really easy proof that just uses the Pigeonhole principle. Let $\alpha$ be irrational.
Lemma: For any $\delta >0$, there is an $n>0$ such that $(n \alpha) \in (- \delta, \delta) \setminus \{ 0 \}$.
Proof: Choose $N$ large enough that $1/N < \delta$. Divide $[0,1]$ up into $N$ segments of length $1/N$. By pigeonhole, there are $j$ and $k$ ...

answered Oct 8 '12 at 16:32

David E Speyer

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22

Your set has measure zero by theorems of Khinchin. First a theorem of Khinchin shows that for almost all real numbers $x$ (i.e. outside a set of measure zero) one has
$$
\lim_{n\to \infty} \frac{\log q_n}{n}= C
$$
for a positive constant $C$. Here $p_n/q_n$ are the convergents of $x$. So almost surely, the denominators $q_n$ are exponential in $n$. In ...

22

This is not a solution, but a plot suggesting that the recursion has the
geometric structure of a planar quasicrystal with the fivefold symmetry of
a Penrose tiling:
(source: harvard.edu)
It's obtained by connecting each integer point $(x,y)$ to its image,
rotated by 72 degrees with respect to the quadratic form invariant
under the linearized recursion $(...

answered Jan 31 '16 at 16:50

Noam D. Elkies

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19

Another visualization, similar to Noam's, together with a modification of a comment by David, might lead to a solution of the question: Define $T:\mathbb Z^2\to\mathbb Z^2$ by $T(a_1,a_2)=(a_6,a_7)$. It is easy to see that $T(v)-v$ is in $\{-1,0,1\}^2$ (and distinct from $(1,-1)$ and $(-1,1)$) for all $v\in\mathbb Z^2$. Let $F_i$ be the $i$-th Fibonacci ...

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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15

Rationals are equidistributed in the sense that If you take any "nice" function, then if you approximate the integral of $f$ over (say) $[0, 1]$ by $\frac{1}{N_B} \sum f(r),$ where $N_B$ is the number of rationals with denominator bounded by $B,$ and the sum is over all such rationals, then this converges to the actual integral.
This equidistribution is ...

answered Mar 31 '18 at 20:01

Igor Rivin

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15

Regarding your second question, Apéry's amazing formula
$$\zeta(3) = {5\over 2} \sum_{n\ge1} {(-1)^{n-1} \over n^3 {2n \choose n}}$$
has inspired the search for analogous formulas for other zeta function values. I think that the earliest such was conjectured by Borwein and Bailey and proved by Almkvist and Granville:
$$\sum_{k\ge0} \zeta(4k+3) z^{4k} = {5\...

answered Sep 26 '18 at 21:55

Timothy Chow

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14

Yes. the fundamental theorem is that the constructible angles in the non-Euclidean plane are exactly the constructible angle in the Euclidean plane. Lengths come from the two laws of cosines and the law of sines. In particular, length 1 is not constructible. As i recall, positive length $x$ is constructible if and only if $\sinh x$ is constructible in the ...

14

Define the Fibonacci numbers by $f_0 = 0$, $f_1 = 1$, $f_{n+1} = f_n + f_{n-1}$, and the golden ratio by $\phi = \frac{\sqrt{5}+1}{2}$. Then it is easy to check that $f_n = \frac{\phi^n - (\tfrac{-1}{\phi})^n}{\sqrt{5}}$ and $(\phi-1)f_n = f_{n-1}-(\tfrac{-1}{\phi})^n$.
We will show that if five terms in a row are all less than $f_{2k}$ for some $k > 1$, ...

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 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

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 ...

12

There is a positive number $a$ such that $a^a=2$. If $a=m/n$ with $m,n\in{\mathbb N}$ coprime, then $m^m=2^nn^m$. As $n\ge1$, we conclude that $m$ is even, sayt $m=2^kl$ with $k\ge1$ and odd $l$. So, $2^{km}l^m=2^nn^m$, implying $2^{km}=2^n$ and $km=n$. A contradiction.

12

The irrationality measure of $\arctan(1/3)$ is not known. It lies between $2$ and $6.096755\dots$. The lower bound is trivial (it holds for every irrational number), while the upper bound is the main result of the paper you quote. It is reasonable to guess that the irrationality measure of $\arctan(1/3)$ is exactly $2$, because almost all irrational numbers (...

11

This is related to Noam Elkies's answer but is not exactly the same.
Rayleigh's theorem, a.k.a. Beatty's theorem, says that if $a$ and $b$ are positive irrational numbers such that $1/a + 1/b=1$, then the sets $\lbrace \lfloor na\rfloor : n\in \mathbb{N}\rbrace$ and $\lbrace \lfloor nb\rfloor : n\in \mathbb{N}\rbrace$ comprise a partition of $\mathbb N$ into ...

answered Sep 18 '13 at 15:48

Timothy Chow

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10

The condition $\lambda>1$ is sufficient and, at least almost, necessary:
To clarify, the space of irrational numbers $(0,1)-\mathbb Q$ is homeomorphic to
$\omega^\omega$ under the map that sends $\frac{1}{a_1+\frac{1}{a_2+\cdots}}$ to a function $f$ satisfying
$f(n)=a_{n+1}-1$. (As is well known.)
This way Lebesgue measure on $(0,1)$ induces a "Gaussian"...

answered Jul 16 '14 at 22:57

Bjørn Kjos-Hanssen

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10

Expanding on my comment, here is a reason why you shouldn't find any better (in your sense) approximation.
Let $p_n/q_n$ be the $n$-th convergent of the continued fraction of $\pi$, and $R_n$ its quality as you defined it in your question.
We purposefully ignore integer parts and off-by-one errors in expressing the number of decimal digits and simply ...

10

The continued fraction expansion is related to the Gauss transformation $T:(0,1)\to(0,1)$, defined by
$$ Tx:=\frac{1}{x} \mod 1. $$
(Indeed, if $x=[a_1,a_2,\ldots)$, then $Tx=[a_2,a_3,\ldots)$.)
It is well known that $T$ admits an absolutely continuous invariant probability measure $\mu$, given by
$$ \mu(A):=\frac{1}{\ln 2}\int_A \frac{dx}{1+x}, $$
and that ...

10

Answered by paul Monsky in the comments: The perimeter is transcendental!
Established by Theodor Schneider in 1937:
Roughly (Those with better knowledge of Deutsch, please correct!):
The value of an elliptic integral of the first or second kind with algebraic
coefficients and between algebraic limits is transcendental.

9

As Timothy Chow pointed out, the OP's question is most naturally
formulated as a promise problem: given a finite system of
equations and inequalities in finitely many variables and the
promise that it has a solution and furthermore that all solutions
have the same value for the particular variable $\xi$, determine
whether or not that value of $\xi$ is ...

answered Nov 17 '15 at 12:21

Joel David Hamkins

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8

A result due to W. H. Mills (1950) mentioned in Apostol's Introduction to Analytic Number Theory (in the Historical Introduction section) states: there is a positive number $A$, which is not an integer, such that $\left\lfloor{A^{3^x}}\right\rfloor$ is a prime for all $x = 1,2,3,\ldots$. (It is probably not known if $A$ has to be irrational, but I thought ...

8

This is Włodzimierz Holsztyński's translation (that he deleted) with my finish:
A circle $K$ has circumference $1$. $O$ belongs to $K$, and it is a beginning of the time axis which is wounded around $K$. On this axis the time intervals of consecutive lengths $1,\frac 12,\frac 13,\dots,\frac 1n,\ldots$ are positioned in a way that $n$-th interval $I_n$ is ...

8

Tilting Peter Mueller's image by -18° as suggested by Noam D. Elkies and coloring the interior of certain orbits shows again the 5-fold symmetry, self-similarity and the closeness to Penrose tilings.

8

As Frits Beukers write in http://www.staff.science.uu.nl/~beuke106/caen.pdf "Ironically all generalisations tried so far did not give any new interesting results. Only through a combination of miracles such generalisations seem to work, which in practice means that we fall back to $\zeta(2)$ or $\zeta(3)$ again".
Nevertheless Apéry-like numbers do appear in ...

8

As mentioned by GH from MO, the irrationality measure
for almost all real numbers is 2. However,
computing it for a particular number
is a notoriously difficult problem. For an irrational algebraic number this measure is indeed 2, but this is a pretty hard theorem by Roth for which he got a Fields medal.
Other then that, to my knowledge the only "famous ...

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