Philosophically why should proving that $\gamma$ is irrational (let alone transcendental) be so much harder than proving $\pi$ or $e$ are irrational?

We know that for algebraic $e^{it\pi}$, $t$ can not be algebraic irrational by Baker's theorem, but his proof is analytic; is there some algebraic understanding for such fact? I …

I am curious to know if the following number is irrational or transcendental: $$\displaystyle A = \sum_p 2^{-p},$$ where the sum is over all positive primes. A similar question ca …

Quadratic irrational numbers are perhaps the most basic examples of irrational numbers that arise as basic physical ratios: think of $\sqrt{2}$ as the distance between the corners …

I'm almost finishing my PhD in applied mathematics, but I'm planning soon (after doing post-doc) to start seriously doing research on problems about proving irrationality of real …

$\sum\limits_{n=0}^{\infty}\dfrac{1}{n!}$ doesn't converge in $\mathbb{Q}_p$, however, $e^p:=\sum\limits_{n=0}^{\infty}\dfrac{p^n}{n!}$ does converge for $p\neq 2$. So my question …

Suppose $L/K$ and $M/K$ are algebraic extensions of a field $K$, such that $L\cap M=K$, and $L/K$ is a normal extension. It is well-known that, with these conditions, we have: $$ …

Define an elliptic modulus $k_n$ such that, $$\frac{K'(k_n)}{K(k_n)}=\sqrt{n}$$ This is the case $m=2$ of the more general, $$\frac{\;_2F_1(\frac{1}{m},1-\frac{1}{m},1,1-x)}{\;_ …

I recently discovered a constant that is constructed as follows: $\chi=\sum_{n=1}^\infty (\frac{\cos{n}}{2|\cos{n}|}+\frac{1}{2}) 2^{-n}$ Furthermore I can prove that it is an i …

A fundamental result in Diophantine approximation, which was largely responsible for Klaus Roth being awarded the Fields Medal in 1958, is the following simple-to-state result: If …

An entire function $f$ is said to be of exponential type if there exist constants $c$ and $k$ such that $|f(z)|\leq c e^{k |z|}$. A famous result of Polya says if $f$ is an entir …

Last year, in a talk of Michel Waldschmidt's, I remember hearing a statement along the lines of the title of this question, that is, "The Galois group of $\pi$ is $\mathbb{Z}$.". I …

For an algebraic number $\alpha$ one can define its "height" in many ways. Informally, you could use its minimal polynomial over $\mathbf{Q}$ and consider the maximum of the height …

Can anyone suggest me an ingenious proof of the transcendence of $\pi$. I have seen Lindemann's proof but it appears intricate.

Let $\mathbb{A^+}$ be the set of non-negative algebraic numbers. Consider the set of "polynomials" : $$\mathbb{P} = \lbrace a_0 + a_1x^{r_1} + a_2x^{r_2} + a_3x^{r_3} +\cdots + a_n …