Is $\frac{1}{\alpha} \in \mathbb{Q}[\alpha]$ for irrational $\alpha$? I have been trying to pick up abstract algebra and just attempted an exercise from Landin's An Introduction to Algebraic Structures which asks to prove whether $\frac{1}{\pi} \in \mathbb{Q}[\pi]$, and would like to ask a (slightly) more general question.

Let $\alpha$ be an irrational number. Is $\frac1\alpha \in \mathbb{Q}[\alpha]$?

$\mathbb{Q}[\alpha]$ being the ring of numbers of the form $a_0 +a_1\alpha+a_2\alpha^2\dots a_n\alpha^n$ with $a_i\in\mathbb{Q}$. I'm really not sure where to begin, and haven't found a similar question on MSE (perhaps because the solution is simpler than I realize.) I am stuck on what we can conclude about $\frac{1}{\alpha}$ other than that it is an irrational number greater than $1$. Clearly, $\alpha^k n \in \mathbb{Q}[\alpha]$, but I'm reluctant to make any claims about $\frac{1}{\alpha}$ and $\alpha^k n$, as I'm sure a solution would use other simpler methods.
*edited title and parts of the body because $\alpha$ need not be less than one.

Not, it's not true for all irrational numbers $0 \lt \alpha \lt 1$. To see this, assume it's true so you get
$$\frac{1}{\alpha} = \sum_{i=0}^{n}a_i \alpha^n \tag{1}\label{eq1A}$$
for some set of $a_i \in \mathbb{Q}$. Multiply by $\alpha$ on both sides and then subtract $1$ from both sides to get
$$\sum_{i=0}^{n}a_i \alpha^{n+1} - 1 = 0 \tag{2}\label{eq2A}$$
This means $\alpha$ is a root of the polynomial
$$p(x) = \sum_{i=0}^{n}a_i x^{n+1} - 1 \tag{3}\label{eq3A}$$
However, since all of the coefficients of the terms in $p(x)$ are rational, this can only happen with $\alpha$ being an algebraic number, so it's not true for all irrational, i.e., it doesn't hold for cases where $\alpha$ is a transcendental number.
