Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My university "textbook" for discrete math is Schaum's Outline. In this outline he goes over Equivalence Relations and Partitions, and I got confused at a particular theorem.

From the book:

Theorem 2.6: Let R be an equivalence relation on a set S. Then S/R is a partition of S. Specifically:

(i) For each a in S, we have a ∈ [a].

(ii) [a] = [b] if and only if (a,b) ∈ R. (iii) If $[a] \ne [b]$, then [a] and [b] are disjoint. Conversely, given a partition {Ai} of the set S, there is an equivalence relation R on S such that the sets Aiare the equivalence classes. This important theorem will be proved in Problem 2.17.

EXAMPLE 2.13

(a) Consider the relation R = {(1,1),(1,2),(2,1),(2,2),(3,3)} on S = {1,2,3}. One can show that R is reflexive, symmetric, and transitive, that is, that R is an equivalence relation.

Also: [1] = {1,2},[2] = {1,2},[3] = {3} Observe that [1] = [2] and that S/R = {[1],[3]} is a partition of S. One can choose either {1,3} or {2,3} as a set of representatives of the equivalence classes.

My confusion arises from the S/R = {[1],[3]}. I don't understand how one can subtract a relation from a set of integers. What fundamental understanding am I missing?

share|improve this question
    
This is related –  Pedro Tamaroff Oct 17 '12 at 22:50

2 Answers 2

There's no subtraction there. Set subtraction is denoted by a backslash. The forward slash denotes forming the quotient set.

share|improve this answer

Numbers origin as cardinality of sets (how many elements it has).

  • Addition on numbers corresponds to disjoint union
  • Subtraction corresponds to subtraction of a subset
  • Multiplication corresponds to direct product ($A\times B$ contains all $\langle a,b\rangle$ pairs)

And, somehow Division corresponds to a quotient by a regular equivalence relation i.e. such that each partition has the same cardinality.

share|improve this answer

Your Answer

 
discard

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.