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 $A_i$ of the set $S$, there is an equivalence relation $R$ on $S$ such that the sets $A_i$ are the equivalence classes. This important theorem will be proved in Problem 2.17.


(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?

  • $\begingroup$ This is related $\endgroup$
    – Pedro
    Oct 17, 2012 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.

  • $\begingroup$ ( 10 years later... ) - OP likely used the word "subtract" incorrectly. I assume he meant something like deduce or create. $\endgroup$ Apr 15 at 18:21

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.


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