Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

By "compatible" I mean, if $R$ is my relation and $+$ my operation, then if $aRb$ and $cRd$, then $(a+c)R(b+d)$. For an congruence relation this is claimed, but also that $R$ must be an equivalence relation, but there are relations, like $<$ that fulfill the first statement, but are not equivalence relation's. Are there any terms for such relations and are they studied in mathematics?

share|cite|improve this question
You have chosen the right word, compatible is a standard technical term for the notion. – André Nicolas Feb 1 '13 at 19:33
up vote 3 down vote accepted

...are they studied in mathematics?

Considering a functional relation $R$, the compatability says, in functional notation: $R(a+c)=R(a)+R(c)$. In other words, the function preserves that operation. Of course we study examples of such relations all the time whenever we are looking at homomorphisms!

If you're asking if relations in general are studied this way, then maybe so. While searhing I came across a paper by Saunders MacLane which looks like a relevant example:

He talks about regarding a homomorphism of things in a category with an operation $S\rightarrow B/L$ instead as a "many valued function" from $S$ into $B$ (that is, every pre-image is related to the multiple things in the coset of the image). It remains "compatible", but is no longer a true function.

share|cite|improve this answer
Another way to look at R preserves the operation is that it commutes with it. But is this the same setup as the OP's? $aRb \wedge cRd \to (a+c)R(b+d)$ versus $R(a+b)=R(a)+R(b)$? There are obvious distinctions such as the number of arguments, implication versus equality, and also in the first case there is a conjunction and $+$, while in the latter only $+$ operator. – alancalvitti Feb 2 '13 at 2:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.