Similar questions have been asked without great success to answer. I read what I could about the problem, but no idea. I wrote to university professor, no response.
There has been question about examples of associative binary operation which are not closed. No example was given, I found none.
If definition of associativity is standard as we all know it then all not closed operations must be non-associative - and this is in stark contrast with articles on wikipedia like: https://en.wikipedia.org/wiki/Semigroup
Lets have Cayley table defined on $\{a, b, c\}$, operation $*$ which is commutative and: $$aa = a, bb = b, cc = c, ab = b, ac = c, bc = Y$$
If I try to prove associativity, then it fails with:
$$a*(b*c) = a*Y = ???$$
$$(a*b)*c = Y$$
Sides are simply not equal. First part cannot defined on my structure, there is no answer.
And this in general happens whenever there is element outside the set I my humble opinion. If I am mistaken, I would be very happy to understand what am I missing. So far I have found no concrete counterexample.
One solution would be that binary operation must be closed, then there is conflict with table of structures on wikipedia page.
Other solution would be, that these instances where there is undefined operations, are simply left out. Then we would work only with associative triples where both sides are defined.
Thank you all kindly.
You are talking about partial semigroups. These can be defined as sets with partial binary operation such that one of the three conditions holds:
1 $a(bc)=(ab)c$ provided both sides are defined
2 If $a(bc)$ is defined then $ab$ and $(ab)c$: are defined and $a(bc)=(ab)c$.
3 Similar to 2 with left and right sides switched.
So there are three (nonequivalent) definitions of partial semigroups. Here is an old paper about partial semigroups: Hrmová, Renáta Partial groupoids with some associativity conditions. Mat. Časopis Sloven. Akad. Vied 21 (1971), 285–311.
One easy way to obtain a partial semigroup in the sense 1 is to take a (finite) semigroup, and delete a few entries of its multiplication table. There exist results about when we can reconstruct the original multiplication table then.
This is an even older paper about this Kozlov, K. P. Existence of rigid cells in the Cayley table of a semigroup. Leningrad. Gos. Ped. Inst. Učen. Zap. 387 1968 131–133.