0
votes
1answer
33 views

Let $K = $ algebraic numbers. Then is $\operatorname{Span}_K(\pi, \pi^2, \dots)$ a vector space of transcendentals?

$V = {\rm Span}_K(\pi, \pi^2, \dots)$ is clearly a $K$-vector space. If we let $K = \Bbb{Q}$ temporarily, then every element of $V$ is transcendental as it's a finite linear combination $Q(X), \ X = ...
0
votes
0answers
41 views

What are some algorithms that can be used to test if a number is transcendental or not?

Well according to the definition of transcendental numbers I find that its any number that doesn't have any polynomial equation of any degree with integer coefficients summing up to 0. So ...
4
votes
1answer
71 views

Transcendental Basis

Can you say that because $\pi$ is transcendental, that a basis of $\{\pi, \pi^2, \pi^3, \dots\}$ in the rational numbers $\mathbb{Q}$ spans the entire real numbers? It seems likely, although I can't ...
0
votes
1answer
62 views

number of the form $\frac{a_0+a_1\pi +\text{…}+a_n\pi ^n}{b_0+b_1\pi +\text{…}+b_m\pi ^m}$ [closed]

all numbers of the form $$\begin{align*}\frac{a_0+a_1\pi +\text{...}+a_n\pi ^n}{b_0+b_1\pi +\text{...}+b_m\pi ^m}\end{align*}$$ form an number field. $n,m$ are arbitrary non-negative itegers, ...
4
votes
2answers
176 views

Why is the concept of transcendental numbers linked with rational coefficients? Why not real nor complex coefficients?

I've read this: In mathematics, a transcendental number is a (possibly complex) number that is not algebraic—that is, it is not a root of a non-zero polynomial equation with rational ...
2
votes
3answers
189 views

Does the Abel-Ruffini Theorem contradict the Fundamental Theorem of Algebra?

It is my understanding that the Abel-Ruffini Theorem implies that certain polynomial equations $(x^5-x+1=0$, for instance) have transcendental roots. However, the Fundamental Theorem of Algebra states ...
0
votes
1answer
59 views

Polynomials with roots having the same module and linear dependent arguments

Is it possible for a polynomial with integer coefficients to have some of its roots: $$m_1e^{i\theta_1 \pi}, m_2e^{i\theta_2 \pi}, \ldots, m_ke^{i\theta_k \pi}$$ such that there exist nonzero integers ...
6
votes
2answers
575 views

Can $\pi$ be a root of a polynomial under special cases?

What if we consider polynomials whose coefficients are either rational or $e$, that is, a polynomial in $\mathbb{Q} \cup \{e\}$ with $\pi$ as a root. Can this happen? Does it matter if we change ...