Concerning decidability of a problem on real numbers This question is an outgrowth of a certain maths problem I've been thinking about.

Suppose you use an oracle to represent a real number. The oracle is of the following form: you give it an integer and it returns you an integer back (for example: given input $n$ the oracle representing $r$ returns integer $k$ such that $k/n \leq r &lt; (k+1)/n$; but it can be any other oracle complying to the aforementioned rule).

Now, given two oracles like that you can't decide if they represent equivalent numbers (if the numbers are equal the algorithm won't halt in 'most' cases). So you are given another oracle that takes two real-number oracles as input and tells if they represent equal numbers.

Given two real numbers $r$ and $s$ represented by such oracles plus the oracle that tells us if two real numbers are equal is it possible to decide if $s = qr + p$, where $q$ and $p$ are rational?

Please provide a formal proof (or at least an idea that can be easily formalized into a proof).

Update: There is a problem with the oracle comparing two real numbers - see comments. Intuitively what I want is the answer to the following question: is deciding that two real numbers belong to the equivalence class defined by $s=qr+p$ harder than deciding if they are equal. I'll close this question and restate it.
I think it's undecidable, but I don't have a proof. Intuitively, if $s\neq qr+p$ for all $p,q\in\mathbb{Q}$ (i.e. if the answer is "NO"), then you will be always looking for a "next" pair $(q,p)$ such that $|s-(qr+p)|$ is smaller than before. You don't really have an argument to decide when to stop and return "NO" because, for all you know, there is a better approximation to $s$ of the form $qr+p$ than the one you already have.