Related to "division with remainder property" This is a standard theorem in probably many books at introductory level to number theory:

For $a,b \in \mathbb Z$ with $b >0$, there exist unique $q,r \in \mathbb Z$ such that $a=bq+r$ and $0 \leq r < b$.

If we agree that from now on the multiplication and addition and ordering are the "usual" ones over $\mathbb R$ then I would like to write of what I am thinking about.
I am thinking of does this theorem holds over some other sets than $\mathbb Z$ , that is:

If for every $a,b \in S \subseteq \mathbb R$ and $b>0$ there exist unique  $q,r \in S$ such that $a=bq+r$ and $0 \leq r < b$ is then necessarily $S \subseteq \mathbb Z$? Does $S$ needs to be equal to $\mathbb Z$? Are there some sets that are not subsets of $\mathbb Z$ over which this holds? Are there some exotic ones?

 A: So, not an anwser but just a special case:
If $S$ is a subring of $\mathbb{R}$, then it must be discrete (and therefore be $\mathbb{Z}$), since if there is $0 < s < 1$ in $S$, then $s = 1\times 0 + s = 1 \times s + 0$ with $0 \leq 0,s < 1$.
A: This is not a definite answer, but a collection of observations, examples and ideas, some of them more worked out than others. Any feedback, refinements, counterexamples etc. are greatly appreciated.

Observation: Every $S\subseteq \Bbb{R}_{\le 0}$ is an example because the condition is vacuously true.
So from now on, we assume $S^+ :=S \cap \Bbb{R}_{>0} \neq \emptyset$. Let $b :=\inf S^+$.
Lemma: Either $b=\min S^+ = 1$ (case 1), or $b = 0$ (case 2). In case 1, necessarily $0\in S$.
Proof: Let $x\in S^+$. First assume $b \ge 1$. By hypothesis there are $q \in S, r\in S^+\cup \{0\}, r<x$ with $x=qx+r$. Since $r<x$ we get $q\in S^+$, hence $q \ge b$, hence the contradiction
$$x = qx+r\quad \ge \quad bx +r \; \stackrel{b>1 \; \text{or} \;b=1, r>0}> \;x,$$
unless $q=b=1$ and $r=0$, which is case 1.
Now assume $0\le b<1$. For every $\epsilon >0$, there is $x_\epsilon\in S^+$ with  $x_\epsilon<1$ and $x_\epsilon-b<\epsilon$. We have $x_\epsilon = q_\epsilon x_\epsilon +r_\epsilon$ with $r_\epsilon<x_\epsilon$, so necessarily $q_\epsilon\in S^+$ and hence $q_\epsilon x_\epsilon \ge b^2$. If $r_\epsilon=0$, we get $q_\epsilon=1$, so $S$ contains the three elements $r_\epsilon = 0<x_\epsilon<1 = q_\epsilon$, which contradicts the argument in nombre's answer. So $r_\epsilon>0$ and thus $r_\epsilon>b$. Now if $b >0$, any $0< \epsilon < b^2$ gives the contradiction $q_\epsilon x_\epsilon = x_\epsilon-r_\epsilon \le \epsilon$. QED.

See below for many different examples for case 1, but as of now, there are no examples for case 2.
Conjecture: Case 2 is impossible.  
The argument in nombre's answer shows that an $S$ in case 2 could not contain both $0$ and $1$. Such an example would (obviously) contain arbitrarily small positive numbers, but also (as one sees via "dividing" by such a small number) arbitrarily big numbers.

Here are some examples for case 1 containing elements outside of $\mathbb{Z}$:


*

*For any real $x>1$, $S = \{0,1,x\}$.


What about $\{0,1,x,y\}$ with $1<x<y$? One checks that for existence of mutual divisions with remainder, one needs $y=x^2$ or $y=x+1$, or $y=x^2+1$. For uniqueness in the first two cases, one needs however that $x^2 \neq x+1$ or in other words, $x \neq \frac{1}{2}(1+\sqrt5)$. (Because otherwise, $x\cdot x + 0 = 1\cdot x + 1$ are two different ways to "divide $x^2$ (resp. $x+1$) by $x$ with remainder $<x$".) So:


*

*For any real $1<x$, $S = \{0,1,x, x^2+1\}$.

*For any real $1<x\neq \frac{1}{2}(1+\sqrt5)$, $S = \{0,1,x, x+1\}$.

*For any real $1<x\neq \frac{1}{2}(1+\sqrt5)$, $S = \{0,1,x, x^2\}$.


Generalising the third example, for any $x>1$ and $k \in \Bbb{N}$,


*

*$S = \{0,1,x, x^2, ..., x^k\}$


as well as the infinite


*

*$S = \{0,1,x, x^2, ...\}$


work, if one makes sure there is no relation of the form $x^a = x^b +x^c$ with $c<b<a$ among the occurring exponents. In particular, each $x\ge 2$, and also each transcendental as well as each rational $x>1$, works.
A general idea here would be to start with a set $\hat S \,(\supseteq \{0,1\})$ of polynomials in $\mathbb{Q}[X]$ (or $\mathbb{Z}[X]$; actually, we want to restrict to non-negative coefficients, $\mathbb{N}_0[X]$) for which we have unique division with remainder (with respect to the degree valuation), and then evaluate those polynomials at $x\in \mathbb{R}$. As long as $\hat S$ is finite, it seems as if by choosing $x$ big enough one can make sure that the order on $\mathbb{R}$ refines the order given by the degree, and the "evaluated set" $S = \hat S(X=x)$ would be an example. But there are some subtleties here I am not sure about.
Finally, here is a non-discrete example which further generalises the last one:
Let $R$ be a non-discrete additive subgroup of $\Bbb{Q}$, and $R_{\ge 0}$ its non-negative elements. Then for each choice of $x>1$, except possibly countably many algebraic ones,


*

*$S= \{0\} \cup \{x^r: r\in R_{\ge 0}\}$


is an example. Namely, again one just has to make sure that there is no relation of the form $x^a = x^b +x^c$ with $c<b<a$ among the occurring exponents. But every $x$ that satisfies such a relation is algebraic (as well as $\notin \Bbb{Q}$).

Finally, even though the OP asks about sets $S$ that are larger than $\mathbb{Z}$, I find it already interesting to think about which subsets $S\subset \Bbb{Z}$ have this property, and, to keep it in the spirit of the considerations above, I want to restrict to those which contain $0$ and $1$ (which makes sense if we want to write every element $a \in S$ as $1\cdot a +0$), and whose elements are non-negative.
The good thing in this situation that one does not need to worry about uniqueness of the division with remainder, since that is obviously inherited from $\mathbb{Z}$. So one just has to make sure that all quotients and remainders of pairs are in the set. Here are some examples, the first four being just special cases of the more general ones above:


*

*For every $n, k \in \Bbb{N}$, $S = \{0,1, n, n^2, ..., n^k\}$.

*For every $n \in \Bbb{N}$, $S = \{0,1, n, n^2, ...\}$.

*For every $n \in \Bbb{N}$, $S = \{0,1, n, n+1\}$.

*For every $n \in \Bbb{N}$, $S = \{0,1, n, n^2+1\}$.

*For every $n \in \Bbb{N}$, $S = \{0,1, n, n^2+1, n(n^2+1)\}$.

*For every $n \in \Bbb{N}$, $S = \{0,1,2, ..., n\} \cup B$ where $B$ is an arbitrary subset of $\{n+1, ..., 2n+1\}$. (Proposed by Glen Whitney, see his comment).


Note how the last example feels different from the others, as it seems to be not "induced" by certain polynomials, as all the ones before.
One way to think about this problem is to define the following "closure" operator on any subset $\{0,1\} \subseteq T \subseteq \mathbb{Z}$:
$$c(T) := T \cup \{\lfloor a/b\rfloor, a-\lfloor a/b \rfloor b :a,b \in T, 0\neq b\}$$
which adds all quotients and remainders of elements in $S$ to it. A set $S$ that satisfies your property then is one with
$$c(S) =S$$
and one could ask whether one always reaches such a "$c$-closed" set after finitely many iterations of applying $c$. This is certainly true if $T$ is finite, since all elements that get possibly added are bounded by the largest element of $T$. For example, if I start with $T = \{0,1,7,9\}$, I get $c(T) = \{0,1,2,7,9\}, c^2(T) = \{0,1,2,3,4,7,9\}$ and this last set is $c$-closed,


*

*$S = \{0,1,2,3,4,7,9\}$


which is a special case of the last example above. On the other hand, the $c$-closure of $\{0,1,2,10\}$ is just


*

*$S = \{0,1,2,5,10\}$


which is of the type of the penultimate example.
It would be very nice if one had an alternative explicit description of such $c$-closed sets. Also, whose sets' $c$-closure ends up being of what kind?
