# Does it make sense to examine properties of $(G,\circ)$ if we, in the meantime prove that $(G,\circ)$ is not a groupoid?

If, for example, we are given a set $G$ and an operation $\circ$ and we have to examine properties of that operation on that set (closeness, associativity, commutativity, exsistance of neutral, exsistance of inverse), should we continue with examination if we find out that the given set is not closed for that operation?
That would mean that the ordered pair $(G,\circ)$ is not groupoid. Does it make sense to examine other properties if $(G,\circ)$ is not groupoid?

• Why should it stop making sense? -- If you want to answer the question" Is this a groupoid?" then you can give a shortcut answer such as "No, $a\circ b\notin G$ for the following specific choice of $a,b\in G$ ...". If your task is explicitly to investigate associativity, commutativity, etc., then in order to complete the task, you have to complete the task (though that may be difficult in the case of associativity when e.g. $(a\circ b)\circ c$ is sometimes not even defined). In particular, there may still be a neutral element! Sep 23, 2017 at 18:10
• Note that scalar product is a non-closed operation on vectors, and division isn't closed on the integers. They are still very interesting operations. Sep 23, 2017 at 18:15
• @HagenvonEitzen that is what I thought. My task had me examine the properties and not if it was gruopid and I did (as you said it turned out to have a neutral element). So it was weird for me when I saw the solution stop when it proved that $(G,\circ)$ is not groupoid, like it doesn't make sense to do other properties after it. That's why I asked the question Sep 23, 2017 at 18:25
• I'm not sure it makes sense to answer this question without some other kind of context. Mathematicians study plenty of things that aren't groupoids, so clearly it can be worth studying something that fails to be a groupoid. Whether it makes sense for certain applications is a different matter. Probably the reason the solution stopped at failure of groupoid-ness is because it's in a context where being a groupoid is important? Sep 23, 2017 at 18:28
• @MaliceVidrine It didn't say anything about groupoid being important, that's why it's weird, it just said examine the properties of given operation on given set. But now I'm assuming it's just a bad written task/solution. Sep 23, 2017 at 18:35