Checking properties for a given relation
When a relation is defined by a rule — "a R b iff a − b is divisible by 3", or "a R b iff a ≤ b²" — you cannot list the pairs, so each property becomes a small proof or a search for one counterexample. The discipline is fixed: to confirm a property, argue for arbitrary elements; to refute it, produce specific numbers and show the definition fails.
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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.
- Restate the rule as a conditionWrite "a R b means …" explicitly. Half of all errors come from misreading the rule, especially with inequalities.
- Prove positively with arbitrary elementsFor symmetry, assume a R b, unfold the definition, and derive b R a using algebra that works for all values.
- Refute with the smallest witnessTry 0, 1, 2 and small negatives first. Give the pair and show the definition fails explicitly.
- Answer each of the four separatelyMarks are allocated per property. State a verdict and a justification for every one, even the ones that hold trivially.
Worked example
On ℤ, define a R b iff a ≤ b². Determine reflexivity, symmetry, antisymmetry and transitivity.
- Reflexive? Need a ≤ a² for every integer a. If a ≤ 0 then a ≤ 0 ≤ a²; if a ≥ 1 then a² = a·a ≥ a·1 = a. So reflexive ✓.
- Symmetric? Take a = 0, b = 3: 0 ≤ 9 so 0 R 3. But 3 ≤ 0 is false, so 3 R 0 fails ✗.
- Antisymmetric? Take a = −1, b = −2: −1 ≤ 4 ✓ and −2 ≤ 1 ✓, so both hold with a ≠ b ✗.
- Transitive? Take a = 4, b = −2, c = 1: 4 ≤ 4 ✓ and −2 ≤ 1 ✓, but 4 ≤ 1 is false ✗.
Answer. Reflexive only; symmetry, antisymmetry and transitivity all fail, each with an explicit counterexample.
Where marks get dropped
These are the specific errors that cost credit on checking properties for a given relation questions — QED's rubric penalises each of them separately.
- Testing only positive integers. Negative values and zero break most inequality-based relations, and that is exactly where examiners look.
- Justifying a property with one confirming example. Confirmation needs a general argument; only refutation may use a single case.
- Forgetting that a counterexample must satisfy the hypotheses. For transitivity you need both a R b and b R c to actually hold.
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Checking properties for a given relation — frequently asked questions
How many counterexamples should I give?
One per failed property, fully verified. Showing that both hypotheses hold and the conclusion fails is what earns the mark.
Do I need to prove properties that clearly hold?
Yes — a one-line general argument. "Reflexive since a − a = 0 is divisible by 3 for every a" is enough, but the reason must appear.
What if the relation is on a restricted domain?
The domain changes the answer. a R b iff a ≤ b² is reflexive on ℤ but fails on the reals, since ½ > ¼. Always work in the stated set.
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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