This is a question I've had for a while and really don't know how to go about finding an answer:


Does there exist a pair of binary operations, $\boxplus$ and $\boxtimes$, other than the usual $+$ and $\times$, such that $(\mathbb{Q}, \boxplus, \boxtimes)$ forms a ring?


I realize that there's probably some "axiom of choice" proof that constructs unintelligible binary operations or even a construction using some other countably infinite ring and a bijection to the rationals. So more importantly I ask:


Does there exist such a $\boxplus$ and $\boxtimes$ such that $a \boxplus b$ and $a \boxtimes b$ can be computed from (closed?) formulas that only involve $+$ and $\times$ (or other related properties of $a$ and $b$ such as prime factors, divisors, gcd, partitions, etc.)?


My apologies if there's some sort of easy example out there that I'm missing. (Though in that case I'll push further and ask if it can be generalized to a larger class of examples.)


Thanks.



Other Ring Structures on $\mathbb{Q}$
Given any bijection $f : \mathbb{Q} \to R$ where $(R,\oplus,\otimes)$ is some (necessarily countable) ring, you'll be able to get a new ring structure $(\mathbb{Q},\boxplus,\boxtimes)$ isomorphic to $(R,\oplus,\otimes)$, by setting:


$a \boxplus b = f^{-1}(f(a)\oplus f(b))$  

$a \boxtimes b = f^{-1}(f(a)\otimes f(b))$ 


The nicer $f$ is, the nicer the expressions for $\boxplus$ and $\boxtimes$ will be. Perhaps the simplest examples are if $p \in \mathbb{Q}^\times,\ q\in \mathbb{Q}$, $R = \mathbb{Q}$, then $f(x) = px+q$ will work. This generalizes Neil's and Pace's answers.


The "converse" is trivially true, in that if $f : (\mathbb{Q},\boxplus,\boxtimes) \to (R,\oplus,\otimes)$ is an isomorphism from some ring structure on $\mathbb{Q}$ to a ring $R$, then $f$ is a bijection $\mathbb{Q} \to R$ and 


$a \boxplus b = f^{-1}(f(a)\oplus f(b))$  

$a \boxtimes b = f^{-1}(f(a)\otimes f(b))$


So in some sense, the above method for getting a ring structure on $\mathbb{Q}$ is the only way to do it. The question (more or less) boils down to, "for which rings $(R,\oplus,\otimes)$ is there a 'nice' bijection $\mathbb{Q} \to R$?" It depends, of course, on what you think "nice" means.


