If you don't know a particular integer $a$, but you know several residues ($r_i$) of $a$ modulo several mutually prime integers ($m_i$), you can use the Chinese Remainder Theorem to find $r$ such that $a = r + k\prod{m_i}$, varying over $k$, with $0 \le r < \prod{m_i}$. If you know $0 \le a < \prod{m_i}$, then you have $a = r$.
Every rational number $a/b$ can be projected onto a prime finite field $\mathbb{Z}/\mathbb{Z}_p$ (given that $b$ is invertible, i.e. $\gcd(p,b) = 1)$, for many $p$.
Given a collection of these projections, $i$ pairs $(r_i, p_i)$ where $a/b \equiv r_i \mod p_i$, is there some process similar to the Chinese Remainder Theorem by which you can take that collection of pairs and work backwards to find $a/b$?
For example, given the list of pairs $[(2,3), (3,5), (6,11), (500000004, 1000000007)]$, can you work backwards to find $1/2$?
I realize that there will always be an infinity of integer and rational solutions for every collection of residues, but there will always be a smallest solution corresponding to the $0 \le r < \prod{m_i}$ above, perhaps a $a/b$ minimizing $|ab|$ or $a^2b^2$, and I would like to find that.
If you need to, add in the ability to generate new residues for arbitrary given primes instead of working from a fixed input collection of pairs. These rational numbers I'm trying to solve for are solutions to some large linear equations, and these equations are convenient to solve over "small" finite fields ($p \le 2^{32}$), but inconvenient to solve over infinite-precision rationals.