Find all rings $R$ satisfying $\mathbb Z \subset R \subset \mathbb Z[\frac {1}{2}]$.
Solution: Let $f \in R$. Then $f= \frac {m}{2^n}$ for some integers $m,n$. Since $2^{n-1} \in R$, hence $\frac {m}{2} \in R$. Hence one gets that $ \frac {1}{2} \in R$, so $R=\mathbb Z[\frac {1}{2}]$, so there is no ring $R$ satisfying given conditions.
What are the other possible solutions of this problem?
P.S: You can give a solution using any theorem from commutative algebra also.
Subrings $\,R\subseteq \Bbb Q$ are characterized by the subset of $S$ of primes that become invertible in $\,R,\,$ i.e. $\,R = \Bbb Z[1/p\ :\ p\in S]$ is obtained by adjoining the inverses of all primes in $S$. One easily checks that this is a ring since fractions having denominators being products of such primes are closed under addition and multiplication.
Conversely, let $\,R\,$ be such a proper subring $\,\Bbb Z\subsetneq R\subset \Bbb Q,\,$ and let $\,r = a/b\in R\,$ be a noninteger. Wlog $\,(a,b) = 1\,$ so by Bezout $\,aj+bk = 1\,$ for some $\,j,k\in\Bbb Z.\,$ Thus $\,j(a/b) + k = 1/b\in R.\,$ Hence $\,a/b\in R\iff 1/b\in R,\,$ so $R$ is generated by adjoining inverses of integers to $\,\Bbb Z.\,$ But $n$ is invertible iff all its prime factors are invertible, so we can restrict to inverses of primes.
Remark $\ $ The above generalizes to any Bezout domain, i.e. rings like $\,\Bbb Z\,$ where gcds enjoy a Bezout identity. Said in the language of commutative algebra, every overring (of fractions) of a Bezout domain is a localization, i.e. is generated by adjoining inverses of elements. See here for further discussion, including literature references.
Note: we eliminate Bezout by instead using Euclid: $\,(a,b) = 1,\ b\mid a\,\Rightarrow\, b\mid 1,\,$ i.e. $\,1/b\in R.$
