11:28
Let $r\in \Bbb{R}$ and let $P \in \Bbb{P}(\Bbb{Z})$
where $P(x) = \sum_{k=0}^n a_kx^k$ for some $n \in \Bbb{N}$
if $r \in \Bbb{A}$, then:
$P(r) = \sum_{k=0}^n a_kr^k = 0$
else if $r \in \Bbb{T}$, then:
$P(r) = \sum_{k=0}^n a_kr^k \neq 0$ for all $P$
Let $[r]_m$ be the mth order approximation of $r$ under some protocol $\text{Proc}$. For example $\text{Proc}$ can be defined as follows:
$$\text{Proc} : [r]_m = \min_{(p,q)} |r - \frac{p}{q^m}|$$
Thus if $r$ is algebraic, then there is some $M$ such that $ P([r]_M) < \epsilon$ for some $\epsilon$ expression (to be determined), otherwise, it is transcendental
Currently figuring how to relate this to Diophantine approximations and Liuoville's theorem
The basic idea is that only approximations of transcendentals cannot be zeroed out quick enough by any fixed polynomial, whereas algebraics can reach zero after a polynomial mapping and its approximation should approach zero quick enough to do so
That is, if $r$ is transcendental, then the sequence $P_m([r]_m)$ will only have some terms zero if $P_m\neq P_s$ for some $m,s \in \Bbb{N}$
likewise if $r$ is algebraic, there exists a unique polynomial $P$ such that $\lim_{m\to \infty }P([r]_m) = 0$
$$P([r]_{m+1}) - P([r]_m)$$
$$\frac{\partial}{\partial y} P(x) = \frac {\partial a_k(y)}{\partial y}$$
$$\frac{\partial}{\partial y} P(x) = \frac {\partial a_k(y)}{\partial y} x^k$$
Hmm... given a set of vectors:
where $x \in \Bbb{R}$. Then $x$ is transcendental if:
$a_0+a_1x+a_2x^2+\cdots = 0$ implies $a_0=a_1=\cdots = 0$
Thus algebraic independence is similar to the good ol' linear independence of vector spaces
except, not really, because the vectors can add to each other...
Let $r > 1$, Then $\frac{1}{r} < 1$. $[\frac{1}{r}]_m$ should start to truncate higher order terms in any given polynomial $P$