Once you know what reflexive, symmetric, and transitive mean, the next skill is checking all three at once on a relation you have never seen before. The three properties combine in eight ways (2^3 = 8), and exam questions routinely ask "which of these does R satisfy?" This article is a guided quiz: six relations on the same four-element set, and the full classification for each.
The setup
Every relation below lives on A = \{1, 2, 3, 4\}. For each relation R, you check four questions:
- Reflexive? Is (a, a) \in R for every a \in A?
- Symmetric? Whenever (a, b) \in R, is (b, a) \in R?
- Transitive? Whenever (a, b) \in R and (b, c) \in R, is (a, c) \in R?
- Equivalence? Is it all three at once?
Case 1 — R = \{(1,1), (2,2), (3,3), (4,4)\} (the identity)
Every element related to itself, nothing else.
- Reflexive? Yes — all four diagonal pairs present.
- Symmetric? Yes — the reverse of (a, a) is (a, a), always in.
- Transitive? Yes — if (a, b) and (b, c) are both in, then a = b = c, and (a, c) = (a, a) is in.
- Equivalence? Yes. The classes are \{1\}, \{2\}, \{3\}, \{4\} — four singleton classes.
Why: the identity relation is the smallest possible equivalence relation. It distinguishes every element from every other element.
Case 2 — R = \{(1,2), (2,1)\} (one mutual pair, nothing else)
Two arrows between 1 and 2, no self-loops.
- Reflexive? No — (1,1) \notin R, for starters.
- Symmetric? Yes — (1,2) \in R and (2,1) \in R. Both pairs match.
- Transitive? No — (1, 2) \in R and (2, 1) \in R would require (1, 1) \in R by transitivity, but it is not. Fails.
- Equivalence? No (fails reflexivity and transitivity).
Case 3 — R = \{(1,2), (2,3), (1,3)\} (a directed chain)
Three arrows in one direction, forming a path 1 \to 2 \to 3 plus the shortcut 1 \to 3.
- Reflexive? No — no diagonal pairs.
- Symmetric? No — (1, 2) \in R but (2, 1) \notin R.
- Transitive? Yes — the only chain is (1, 2) \in R and (2, 3) \in R, which requires (1, 3) \in R, and indeed (1, 3) \in R.
- Equivalence? No (fails both reflexive and symmetric).
Case 4 — R = \{(1,1), (2,2), (3,3), (4,4), (1,2), (2,1)\} (identity plus a mutual pair)
- Reflexive? Yes — all diagonal pairs present.
- Symmetric? Yes — (1, 2) \in R and (2, 1) \in R.
- Transitive? Yes — the only non-trivial chain is (1, 2) \in R and (2, 1) \in R requiring (1, 1) \in R, which holds. And (2, 1) \in R and (1, 2) \in R requiring (2, 2) \in R, which also holds.
- Equivalence? Yes. The classes are \{1, 2\} and \{3\} and \{4\} — three classes, two of size one and one of size two.
Case 5 — R = \{(1,2), (2,1), (1,3), (3,1)\} (two mutual pairs sharing a hub)
- Reflexive? No — no diagonals.
- Symmetric? Yes — every arrow has its reverse.
- Transitive? No — (2, 1) \in R and (1, 3) \in R require (2, 3) \in R, but (2, 3) \notin R.
- Equivalence? No.
This is the trap everyone falls into at least once: the relation is symmetric, but not transitive. Two symmetric pairs sharing a middle element would have to be chained, and the chain demands a new pair that is not in R.
Case 6 — R = \{(1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (3,4), (4,3)\} (identity plus two mutual pairs)
- Reflexive? Yes.
- Symmetric? Yes — every non-diagonal pair is paired with its reverse.
- Transitive? Yes. The chains to check: (1, 2) and (2, 1) need (1, 1) — present. (2, 1) and (1, 2) need (2, 2) — present. (3, 4) and (4, 3) need (3, 3) — present. And there is no chain that crosses between the \{1, 2\} block and the \{3, 4\} block (since there is no arrow between them), so transitivity is trivially satisfied there.
- Equivalence? Yes. The classes are \{1, 2\} and \{3, 4\} — two classes of size two.
The eight possible combinations
Across three independent properties, eight combinations exist. In practice, the ones you encounter most often are these four:
| Reflexive | Symmetric | Transitive | Example |
|---|---|---|---|
| Yes | Yes | Yes | Identity, Case 4, Case 6 (any equivalence relation) |
| Yes | No | Yes | Divisibility on \{1, 2, 3, 4\} (a partial order) |
| No | Yes | No | Case 2, Case 5 (common trap) |
| No | No | No | A random chain like Case 3 |
The other four combinations exist but are less common. For example, "reflexive and symmetric but not transitive" requires a relation like \{(1,1), (2,2), (3,3), (1, 2), (2, 1), (2, 3), (3, 2)\} — the transitivity fails at the chain 1 \to 2 \to 3 without the pair (1, 3).
The reflex you build
After walking through the six cases, a pattern crystallises:
- For reflexivity: scan the diagonal cells (a, a) for every a. If any are missing, the relation is not reflexive.
- For symmetry: scan each non-diagonal pair (a, b) and check (b, a). If any is missing, not symmetric.
- For transitivity: this is the trickiest. For every chain (a, b) and (b, c) both in R, check that (a, c) is in. Most failures of transitivity come from two symmetric pairs sharing a hub — Case 5 is the prototype.
The fastest order to check is reflexive first (cheapest), then symmetric (cheap), then transitive (usually the one that fails in a problem trying to trick you). If any of the first three fail, the relation is not an equivalence relation — you can stop checking and declare "no."
Why: an equivalence relation needs all three properties simultaneously. Missing any one is enough to disqualify it. The property cheapest to check and most commonly missing is reflexivity (the diagonal), so start there.
The goal of working through six cases is not to memorise them — it is to train the pattern recognition so that when you see a new relation, your eye immediately scans the diagonal, the pairing, and the chains in that order. Once the reflex is wired, classifying a relation is a twenty-second task.
Related: Relations · Equivalence Relations · Symmetry Check: Reverse Arrows · Equivalence-Class Partition · Ordered-Pair Plotter