Reflexive, symmetric, antisymmetric & transitive
Five properties classify almost every relation you will meet. R is reflexive if (a,a) ∈ R for every a; irreflexive if (a,a) ∈ R for no a; symmetric if (a,b) ∈ R forces (b,a) ∈ R; antisymmetric if (a,b) and (b,a) together force a = b; transitive if (a,b) and (b,c) force (a,c). These are not opposites — a relation can be neither symmetric nor antisymmetric, and the empty relation is both.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Reflexivity — check every elementEvery a in the underlying set must have (a,a) ∈ R. Missing even one loop kills it, and "irreflexive" needs all of them missing.
- Symmetry — check every pairFor each (a,b) ∈ R with a ≠ b, look for (b,a). One missing reverse pair is a counterexample.
- Antisymmetry — hunt for a two-cycleA relation fails antisymmetry only if some a ≠ b has both (a,b) and (b,a). Loops (a,a) never violate it.
- Transitivity — check every composable pairFor each chain a → b → c, verify a → c. This includes chains where b = a or c = a, which is where errors hide.
Worked example
On A = {1,2,3} let R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3)}. Which of the five properties hold?
- Reflexive: (1,1), (2,2), (3,3) all present ✓. Hence not irreflexive.
- Symmetric: (1,2) and (2,1) ✓, but (2,3) is present while (3,2) is not ✗.
- Antisymmetric: (1,2) and (2,1) both hold with 1 ≠ 2 ✗.
- Transitive: (1,2) and (2,3) hold but (1,3) does not ✗.
Answer. Reflexive only. It is neither symmetric nor antisymmetric — showing those two are not complements.
Where marks get dropped
These are the specific errors that cost credit on reflexive, symmetric, antisymmetric & transitive questions — QED's rubric penalises each of them separately.
- Treating antisymmetric as "not symmetric". They are independent: the identity relation is both, and R above is neither.
- Overlooking transitivity chains that revisit an element. From (a,b) and (b,a) transitivity demands (a,a).
- Declaring reflexivity from a few loops. Every element of the stated set needs one — and the stated set matters, not just the elements appearing in R.
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Reflexive, symmetric, antisymmetric & transitive — frequently asked questions
Can a relation be both symmetric and antisymmetric?
Yes. Any subset of the identity relation qualifies, including ∅ and the full identity, because there are no pairs (a,b) with a ≠ b to cause trouble.
Is the empty relation transitive?
Yes, vacuously — there are no chains to check. It is also symmetric and antisymmetric, but reflexive only on the empty set.
Why does antisymmetry matter?
It is what makes a partial order an order rather than a preorder: it prevents two distinct elements from each being "below" the other, so the Hasse diagram has no two-way arrows.
The rest of Relations
Properties of relations, composition, and representations. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Reflexive, symmetric, antisymmetric & transitive
- 2Checking properties for a given relation
- 3Composition R∘S & inverse R⁻¹
- 4Matrix & digraph representations
- 5Reflexive & transitive closures
- 6Relations as subsets of A × B
- 7Powers Rⁿ & reachability
- 8Warshall’s transitive closure algorithm
- 9Counting relations with a given property
- 10Union, intersection & complement of relations
- 11n-ary relations & the relational data model
- 12Restricting a relation to a subset
- 13Symmetric closure vs transitive closure
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