Assuming the axiom of choice, how to find explicit group structure of a given set Let us assume the axiom of choice. This is equivalent to every nonempty set having group structure.
My question is, given some nonempty set, can we define the binary operator in a constructive way just by knowing that it's possible or do we have to be very clever for every case?
In particular, I'm interested in how could we can describe and define a group structure on $\mathbb{R} \setminus \mathbb{Q}$.
 A: There are two basic ideas:

*

*Given $X$, if it is finite then pick any bijection with $\Bbb Z/(n)$, and you have a finite group; otherwise consider $\Bbb Z[X]$, the ring of polynomials whose free variables are elements of the set $X$. We can prove, using the axiom of choice, that $\Bbb Z[X]$ has the same cardinality as $X$. Therefore there exists a bijection between the two sets, and that gives us a group structure on $X$.
Of course, neither of these are "constructive" or "canonical", but rather "definable from a fixed well-ordering" (and the problem is that usually the well-ordering is not canonical).


*In the case of $\Bbb{R\setminus Q}$ we know that this set has the same cardinality as $\Bbb R$, and we can even find explicit, even if quite convoluted, bijection between the two sets. Therefore we can transport the group structure of $\Bbb R$ (the additive group) to that set.
We don't need the axiom of choice for that. But again, this is "definable from a bijection". Only that in this case we can write the bijection.
A: In your particular case, you can get reasonably explicit by transferring the group structure over from a bijection between the irrationals and the reals. It's not hard to write down an explicit enumeration $q_n$ of $\mathbb Q$, and then we can map $\mathbb{R}\setminus \mathbb{Q}$ to $\mathbb{R}$ by $\sqrt{2}/2^{2k+1}\mapsto q_k$ and $\sqrt{2}/2^{2k}\mapsto \sqrt{2}/2^k$. Then you have a group structure on the irrationals that's usual addition away from $\{\sqrt 2/2^k\}_{k\in\mathbb N}$ and otherwise defined piecewise by something you could write down a closed formula for in a few minutes given a formula for your enumeration $q_n$.
Some of this hinges on what you mean by an "constructive" operation on an arbitrary set. There's no structure there, and any choice of such a structure seems tautologically to require the use of AC or well-ordering. For instance the usual group structure on an arbitrary infinite set $S$ is the disjoint union on its finite power set $P_f S$-but for that we need a bijection $S\to P_f S$, and there's nothing to work with in $S$ to describe such a bijection.
I don't think this is exactly the same issue as the linked MO question, whose answer is that ZF implies (AC iff group structure on every set). The question here is whether it's consistent with ZFC that ZF and group structure on every set not imply arbitrary choice. I doubt it, but anyway that's why I made the argument in the previous paragraph that there's certainly no hope of getting an affirmative answer of any useful sort. 
