Relations Calculator

Test a relation on a set: reflexive, symmetric, transitive and antisymmetric, with a counterexample for each property that fails, the equivalence classes, the transitive closure, and whether a relation is a one-one or onto function.

Set A	1, 2, 3
Relation R on A (ordered pairs)	(1,1), (2,2), (3,3), (1,2), (2,1)
Set B, to test R as a function A → B (optional)

Set A

Relation R on A (ordered pairs)

Pairs (a, b) meaning a R b.

Set B, to test R as a function A → B (optional)

R is

An equivalence relation

R is reflexive, symmetric and transitive, so it's an equivalence relation, and it splits A into the equivalence classes listed.

Reflexive: Yes
Symmetric: Yes
Transitive: Yes
Antisymmetric: No: both (1, 2) and (2, 1)
Equivalence classes: {1, 2}, {3}

Relation matrix (row a, column b: 1 if a R b)

	1	2	3
1	1	1	0
2	1	1	0
3	0	0	1

About the Relations Calculator

A relation R on A is reflexive if a R a for every a, symmetric if a R b always gives b R a, and transitive if a R b and b R c give a R c. A relation with all three is an equivalence relation; it splits A into disjoint equivalence classes.

A function from A to B relates each element of A to exactly one element of B; it is one-one if different inputs have different outputs and onto if every element of B is an output.

The formula

Equivalence = reflexive + symmetric + transitive

Worked example

A = {1, 2, 3}, R = {(1,1), (2,2), (3,3), (1,2), (2,1)}

Reflexive, symmetric and transitive
Classes {1, 2} and {3}

Questions people ask

Can a relation be both symmetric and antisymmetric?

Yes, if its only pairs are (a, a): the identity relation is both.

Is the empty relation transitive?

Yes: there are no pairs to break the rule (it's transitive 'vacuously').