Check All Three Properties Methodically — Don't Skip Transitivity

A question asks: "Show that R is an equivalence relation on A." You stare at R, notice it looks symmetric, see a few self-loops, and write "therefore R is an equivalence relation." Marks lost. The habit that saves you is boring but bulletproof: check all three properties, in order, each one as a separate labelled step — never as a bundle and never by feel.

The three-box checklist

Every "is this an equivalence relation?" question should, on your rough paper, look like this skeleton:

1. Reflexive?  For all a in A, is (a, a) in R?     YES / NO
2. Symmetric?  If (a, b) in R, is (b, a) in R?     YES / NO
3. Transitive? If (a, b) and (b, c) in R,
               is (a, c) in R?                     YES / NO

Three boxes, three yes/no answers. The relation is an equivalence relation if and only if all three are YES. One NO anywhere — even if the other two are confident YES — and the answer is "not an equivalence relation."

Why the order matters: reflexive is the fastest to falsify (one counterexample kills it), symmetric comes next, and transitivity is the most laborious. Doing the cheap checks first means that when a relation fails reflexivity, you save yourself from writing a paragraph about transitivity that nobody needed.

Why "two out of three" is a trap

The three properties are logically independent — no pair forces the third. Knowing this, examiners design questions that pass two and fail one, counting on students to miss the subtle failure.

All three boxes must be ticked. Two of three is the commonest examiner trap — a relation can be reflexive and symmetric but fail transitivity, and vice versa.

Two-of-three counterexamples (memorise these):

Reflexive + symmetric, not transitive: on \{1, 2, 3\}, take R = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}. Reflexive ✓, symmetric ✓, but (1, 2) \in R and (2, 3) \in R while (1, 3) \notin R. Transitivity fails.
Reflexive + transitive, not symmetric: "less than or equal to" on any set of numbers. Every a \leq a, and a \leq b \land b \leq c \Rightarrow a \leq c, but 1 \leq 2 does not give 2 \leq 1.
Symmetric + transitive, not reflexive: on \{1, 2, 3\}, take R = \{(1,1), (1,2), (2,1), (2,2)\}. Symmetric ✓ (check), transitive ✓ (check), but (3, 3) \notin R, so not reflexive.

If you skip even one box, an examiner armed with one of these will catch you.

The per-property technique

Each box has its own technique. Use the right tool for the right property.

Reflexive. List the elements of A. For each one a, check (a, a) \in R. Finite set → just enumerate. Infinite set → prove the general claim "(a, a) \in R for all a \in A" using the defining condition of R.

Symmetric. Pick a generic (a, b) \in R — do not pick a specific one. Ask: "using only the fact that (a, b) \in R, can I derive (b, a) \in R?" If yes, symmetry holds. If no, try to find one concrete (a, b) \in R with (b, a) \notin R — one counterexample is enough to falsify.

Transitive. The three-variable property is the slowest, and the one most often skimmed. Assume (a, b) \in R and (b, c) \in R for generic a, b, c, and try to derive (a, c) \in R. If you cannot, hunt for a concrete triple where (a,b) and (b,c) are in R but (a, c) is not.

A full worked check

Problem: Let R on \mathbb{Z} be defined by a \mathrel{R} b \iff a - b is divisible by 5. Is R an equivalence relation?

Box 1 — Reflexive. For any a \in \mathbb{Z}, is a \mathrel{R} a? That asks: is a - a = 0 divisible by 5? Yes, 0 = 5 \cdot 0. So (a, a) \in R for all a. ✓

Box 2 — Symmetric. Assume a \mathrel{R} b, so a - b = 5k for some integer k. Then b - a = -(a - b) = -5k = 5(-k), which is divisible by 5. So b \mathrel{R} a. ✓

Box 3 — Transitive. Assume a \mathrel{R} b and b \mathrel{R} c. So a - b = 5k and b - c = 5m for integers k, m. Add: (a - b) + (b - c) = a - c = 5k + 5m = 5(k + m). Divisible by 5. So a \mathrel{R} c. ✓

All three boxes ticked. R is an equivalence relation. This is "divisibility by 5" — also called congruence modulo 5 — and it partitions \mathbb{Z} into 5 equivalence classes.

Catching a sneaky failure

Problem: On the set of people in Delhi, define R by "a \mathrel{R} b iff a and b share at least one common friend." Is this an equivalence relation?

Box 1 — Reflexive. Does a share a friend with a? You share all your friends with yourself, so yes — as long as you have at least one friend. Assuming everyone in the set has at least one friend: ✓. (If someone has zero friends, reflexivity fails for them; this is worth flagging.)

Box 2 — Symmetric. If a shares a friend with b, then b shares a friend with a — same friend, viewed from the other side. ✓

Box 3 — Transitive. a shares a friend with b, b shares a friend with c. Must a share a friend with c? No. Concrete counterexample: a and b share friend X; b and c share friend Y; a and c have no friend in common. Transitivity fails. ✗

Verdict: Not an equivalence relation. Had you stopped at box 2 — "sounds symmetric, sounds like it should be" — you would have written the wrong conclusion. Box 3 saved you.

The rubric your examiner uses

For a "prove R is an equivalence relation" question, the standard rubric awards marks as: 1 for reflexive, 1 for symmetric, 2 for transitive, 1 for the concluding statement. Skipping a property does not cost you half a mark — it costs you the full mark for that property, because the examiner cannot award marks for working that is not there. Writing "all three hold, therefore equivalence" without per-property work picks up the concluding mark only.

The boring three-box skeleton costs you thirty seconds of writing. It earns the full four marks. Train the reflex once: see "equivalence relation?" → draw three boxes → fill them in → conclude.