So I came across the concept of extending the notion of irrationality to higher degree polynomials. The base case of this is standard irrationality. That is, a number is irrational if it cannot be expressed as the ratio of two integers. This is equivalent to it not being able to be expressed as a root to a linear equation with integer coefficients:
$\sqrt 2$ is an example of a number that cannot satisfy the above. Thus it is irrational. However , we can extend this condition to quadraic polynomials:
$\sqrt 2$ does satsify the above given $a=1$, $b=0$ and $c=-2$. Thus in this scheme $\sqrt 2$ would be a $1$st degree irrational.
(I think degree of irrationality is already a term in math. Is it equivalent to what I am talking about? If it's not we can use something like "level $1$ irrational".)
And in general an $n$th degree irrational number is a real number that can be expressed as a zero of an $n$-degree polynomial and not an $n+1$ degree polynomial. (This is so every number has a unique degree of irrationality.)
This brings us to transcendental numbers, which are not the solution to any finite degree polynomial with integer coefficients. This is their definition. This leads me to my question:
Is a transcendental number a real number whose degree of irrationality is infinite (or maybe countably infinite if that distinction is relevant or even valid)? So any transcendental (and any real as a result) is the root of a power series with integer coefficients (centered at 0)?
Or is a transcendental number a number that simply isn't the solution to any finite degree polynomial? This would seem to be a weaker statement then the one above.