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This repository was archived by the owner on Feb 8, 2022. It is now read-only.
Consider the non-negative reals (or computable reals if you're a constructivist).
You can have inverses in multiplication, even if you don't have them for addition. In fact, the positive (so here we are excluding 0) reals form an Abelian group under multiplication/division.
This might be a semifield [1] but I'm not sure (look at the ring theory definition).
I've heard of division rings/skew fields (fields without commutativity) and near fields (division rings with only one distributive law) but am not confident about what this would be.
In general, we should consider units, i.e. elements in a ring that have inverses. This is for example the case for matrix rings, where the subset of invertible matrices form a multiplicative group.
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johnynek commentedon Nov 17, 2015
Of course the above is also true of the rationals.
non commentedon Nov 17, 2015
This might be a semifield [1] but I'm not sure (look at the ring theory definition).
I've heard of division rings/skew fields (fields without commutativity) and near fields (division rings with only one distributive law) but am not confident about what this would be.
[1] https://en.wikipedia.org/wiki/Semifield
denisrosset commentedon Dec 9, 2015
In general, we should consider units, i.e. elements in a ring that have inverses. This is for example the case for matrix rings, where the subset of invertible matrices form a multiplicative group.