Finding the coefficients in a vector space of numbers Consider the vector space $V=\operatorname{span}_{\mathbb Q}(\sqrt2,\sqrt 3)$ of all rational linear combinations of $\sqrt 2$ and $\sqrt 3$ over $\mathbb Q$. If it makes things easier, take the module $\operatorname{span}_{\mathbb Z}(\sqrt 2,\sqrt 3)$.
My question is:

Given $\alpha = a\sqrt 2+b\sqrt 3$, how can we express the coefficients $a$ and $b$ in terms of $\alpha$?

Clearly by a simple linear algebra argument, the coefficients are uniquely determined for any $\alpha \in V$, but it's not obvious how to determine them given access to $\alpha$ alone (e.g. say we have access to arbitrarily many decimal digits $d_i\in\{0,\dots,9\}$ such that $\alpha = \sum_{i\in\mathbb Z}10^id_i$).
I suppose one can find them if we can define an inner product on $V$, but its not obvious how to define an inner product explicitly in terms of $\alpha$. For instance, I can easily define
$$\langle a\sqrt2+b\sqrt3,c\sqrt2+d\sqrt3\rangle = ac+bd,$$
and then we would get $a=\langle \alpha,\sqrt2\rangle$ and $b=\langle\alpha,\sqrt 3\rangle$, but obviously this is cheating.
I've tried messing around with different combinations of $\alpha$, $\alpha^2$, etc., but I can't seem to get anywhere. I appreciate any help with this.

Disclaimer
I suspect this is impossible, because it feels like if it were possible, then we could use something similar to express $\mathfrak{Re}(z)$ in terms of $z$ alone (without using $\bar z$ or anything like that), and I suspect that if that were possible, I would surely have come across it somewhere.
If it is impossible, a proof of impossibility would be nice, though I have no idea how one would go about such a thing.
 A: $\newcommand{\ZZ}{\mathbb Z}\newcommand{floor}[1]{\left\lfloor#1\right\rfloor}$As alluded to by Gerry Myerson, there are algorithms which take algebraic integers like $\alpha=a\sqrt2+b\sqrt3$ of arbitrary (say, decimal) precision and can output good (and small) $a,b\in\ZZ$ which approximate $\alpha.$
Even stronger, there are algorithms which can produce minimal polynomials: given a good enough approximation of a real number $\alpha$ which is the root of a polynomial $p(x)\in\ZZ[x]$ and given the degree $d:=\deg p,$ we can provide a good $p(x).$ Plugging in your $\alpha$ with $d=4$ into the algorithm would do what you want.
I will outline one way to do this. The idea is to use lattice basis reduction. Pick some very large integer $N$ representing precision of $\alpha$ and consider the set of $d$ vectors
$$\left\langle 1, 0, \cdots, 0, \floor{N\alpha^0}\right\rangle, \\
\left\langle 0, 1, \cdots, 0, \floor{N\alpha^1}\right\rangle, \\
\vdots \\
\left\langle 0, 0, \cdots, 1, \floor{N\alpha^{d-1}}\right\rangle.$$
These vectors define a lattice in $\mathbb R^{d+1}.$ After lattice basis reduction (say, with Lenstra–Lenstra–Lovász), we will have small vectors like
$$v=\left\langle a_0,a_1,\cdots,a_d,\sum_{k=0}^{d-1}a_k\floor{N\alpha^k}\right\rangle.$$
If the last component of $v$ is small enough, then we guess that it's supposed to be $0$ if we replaced the approximations $\floor{N\alpha^k}$ with actual $\alpha^k$s. In other words, we hope
$$\sum_{k=0}^{d-1}a_k\alpha^k=0.$$
Thus, we can read the minimal polynomial for $\alpha$ straight off the other coefficients $a_0,\ldots,a_d$ of $v.$
