Every property of a relation — reflexive, symmetric, transitive, antisymmetric — is a universal statement. It says "for all a in A..." or "for all a, b in A..." and carries a promise that the condition holds everywhere, no exceptions. That asymmetry between proving and disproving is one of the most leveraged shortcuts in JEE. To disprove a property, you need one counter-example. To prove it, you need an argument that works for every case at once. This single insight reshapes how you attack relation problems.
The asymmetry in one picture
The universal statements inside each property
Each property hides a universal quantifier — make it visible and the strategy becomes obvious.
- Reflexive: \forall a \in A,\ (a, a) \in R.
- Symmetric: \forall a, b \in A,\ (a, b) \in R \Rightarrow (b, a) \in R.
- Transitive: \forall a, b, c \in A,\ (a, b) \in R \land (b, c) \in R \Rightarrow (a, c) \in R.
- Antisymmetric: \forall a, b \in A,\ (a, b) \in R \land (b, a) \in R \Rightarrow a = b.
Every one is "\forall \ldots" — a universal promise. Its negation is "\exists \ldots" — one concrete instance where it fails. That single instance is your counter-example, and it is all you need.
Disproving: the speed advantage
When you suspect a property might fail, your only job is to find one bad triple. No algebra, no proof, no elegance — just a concrete witness.
Why one is enough: the property claims something holds for every element. Finding one element where it fails contradicts the "for every" directly. It does not matter if ten thousand other elements behave; the claim is broken the moment a single one misbehaves. This is logic's version of "one dead cockroach in the soup means the soup is bad."
Example. "Is R = \{(1,2), (2,1), (2,3)\} transitive on \{1, 2, 3\}?" You do not need to inspect every chain. Look for one failure: (1, 2) and (2, 3) are both in R, so transitivity demands (1, 3). Is (1, 3) \in R? No. Transitivity fails. Done — one bad chain. You spent ten seconds.
Proving: the burden is universal
To prove a property, "look, it holds on these three examples" is worthless. You must argue it holds on every element — possibly infinitely many, as with relations on \mathbb{Z} or \mathbb{R}. That is where algebra earns its keep: you introduce symbolic names ("let a \in A be arbitrary; then...") and write a single chain of reasoning that does not depend on which element you picked. The same proof then covers every element.
Example. On \mathbb{Z}, define a R b iff a - b is a multiple of 3. To prove reflexivity, you must show every integer a satisfies a R a. Argument: a - a = 0 = 3 \cdot 0, a multiple of 3. The value of a never appeared in the argument — it cancelled — so the same line covers every integer. That is how proofs of universal claims work.
The practical strategy — disproof-first
When faced with "is R reflexive / symmetric / transitive?", apply this pipeline:
- Hunt for a counter-example first. Try small, special elements: 0, 1, -1, a = b, a = b = c. Ten seconds of concrete arithmetic.
- If you find one, you are done — the property fails, and you have the witness.
- If you cannot find one after a few tries, switch to proof mode. Introduce symbolic a, b, c and argue algebraically.
This disproof-first habit saves you from wasting minutes proving something that is actually false. Many JEE relations are designed to fail in a subtle way; the question writers want you to find the counter-example, not to drift into a fake proof.
Counter-example patterns you should recognise
Certain failure modes show up repeatedly. Keep these in your holster.
| Property | Classic counter-example |
|---|---|
| Not reflexive | One specific element a_0 \in A with (a_0, a_0) \notin R |
| Not symmetric | One pair (a, b) \in R with (b, a) \notin R |
| Not transitive | A chain (a, b), (b, c) \in R with (a, c) \notin R |
| Not antisymmetric | Distinct a \neq b with (a, b) \in R and (b, a) \in R |
For each, you need one concrete instance. Never prose, never "in general it seems..." — write the actual pair.
Why quantifiers make proving slow
The universal quantifier forbids you from picking your favourite cases and calling it a day. "I showed it for a = 1, 2, 3" is no proof unless you can show the same argument works when a is replaced by any value. That is the mental discipline: when your variable is a, treat it as any element, not any specific one. If your argument uses a fact that is true only for positive a, your proof covers only positive a — and a negative-a counter-example can still destroy it.
The reverse trap — proving from examples
The other direction of the error is equally lethal. You check (1, 2) in R; see (2, 1) also in R; check (3, 5) in R; see (5, 3) also in R. "Symmetric ✓" — wrong. You checked two pairs. If R has fifty pairs and the forty-ninth is asymmetric, your check missed the counter-example. Examples suggest a property; they never prove it. Either exhaust the pair list (on finite R this is actually fine — list all pairs and check each one) or argue symbolically.
disproof beats elegance
Problem: On \mathbb{Z}, define a R b iff ab \geq 0. Is R transitive?
Temptation. You might start writing: "suppose ab \geq 0 and bc \geq 0. Multiply: ab \cdot bc = ab^2 c \geq 0. Since b^2 \geq 0..." and fumble your way toward a pseudo-proof that ac \geq 0. Wrong answer.
Disproof. Try a = 1, b = 0, c = -1. Then ab = 0 \geq 0 ✓ and bc = 0 \geq 0 ✓, but ac = -1 < 0, so (a, c) \notin R. Transitivity fails at this single triple. Ten seconds.
The counter-example kills the property. No elegant argument needed — and the "elegant argument" was about to be wrong.
The takeaway rule
Before you start proving, try to break the property. Pick small values, edge cases, zeros, identities. If the property falls to a counter-example, you save minutes and you save yourself from an incorrect proof. If the property survives your best attacks, only then shift to the symbolic proof — and write it so it does not depend on the specific elements you happened to pick.
Disprove in seconds. Prove in paragraphs. One is cheap, the other is not — use the cheap tool first.
Related: Relations · Test Abstract Relations With Small Numbers Before Proving · Check Reflexivity on an Infinite Set · Test the Condition, Not Individual Pairs · Check All Three Properties Methodically