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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:

$$r=\frac{p}{q}\rightarrow qr-p=0$$

where $p,q\in\mathbb{Z}$.

$\sqrt 2$ is an example of a number that cannot satisfy the above. Thus it is irrational. However , we can extend this condition to quadraic polynomials:

$$ax^2+bx+c=0$$

$\sqrt 2$ 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.

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    $\begingroup$ (You probably want only power series centered at 0 with integer coefficients.) $\endgroup$ – Andrés E. Caicedo Sep 8 '18 at 17:03
  • $\begingroup$ You're right, changed it. $\endgroup$ – Ozaner Hansha Sep 8 '18 at 23:09
  • $\begingroup$ In your quadratic, with c=-4, the roots are 2 and -2. If you want √2 for a root, change it to c=2. $\endgroup$ – richard1941 Sep 12 '18 at 5:55
  • $\begingroup$ Whoops, you're right, changed it. Although I think you mean $c=-2$. $\endgroup$ – Ozaner Hansha Sep 12 '18 at 21:40
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What you call the "degree of irrationality" of a number is the degree of its minimal polynomial over $\mathbb Q$, sometimes called the degree of the number over $\mathbb Q$. This concept is defined and explored in field theory.

Every real number is the root of a power series with integer coefficients.

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    $\begingroup$ Only the real numbers in $(-1,1)$. Transcendental numbers outside of this range cannot be roots of power series because their powers get larger and larger, so the addends in any power series of it cannot converge to 0 and thus the sum does not converge. $\endgroup$ – alphacapture Sep 8 '18 at 23:38
  • $\begingroup$ @alphacapture can't you get a series centered around a different number $c$ (i.e. $\sum_{n=1}^\infty k_n(x-c)^n$? Or does this not count as "integer coefficients" anymore? $\endgroup$ – Paŭlo Ebermann Sep 9 '18 at 1:56
  • $\begingroup$ So are all transcendental numbers the root of a power series with integer coefficients but not necessarily centered at zero? But can't this be remedied by expanding the binomial? Unless this becomes a problem in an infinite series... $\endgroup$ – Ozaner Hansha Sep 9 '18 at 5:08
  • $\begingroup$ @alphacapture I think this problem goes away if you allow rational coefficients, which is probably the more interesting question here. $\endgroup$ – Jack M Sep 9 '18 at 6:30
  • $\begingroup$ @PaŭloEbermann I suppose whether that counts as a series depends on whether the OP wants to allow series centered around other points. OzanerHansha You can't expand it because you would need an infinite sum to find each of the coefficients, and the infinite sum in question does not necessarily converge. $\endgroup$ – alphacapture Sep 10 '18 at 16:49

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