A JEE question defines R on \mathbb{Z} by a cryptic condition — say "a \mathrel{R} b \iff a^2 + b^2 = some even number" or "a \mathrel{R} b \iff a - b is a multiple of ab." You stare at the condition and start writing a general proof. Ten minutes later, you have two contradictory lines and no idea which step is wrong. The fix is a single habit: before you prove anything abstract, plug in 0, 1, 2 (or the smallest concrete numbers allowed). If a property fails on small numbers, you kill the property in seconds. If it survives, your intuition is now sharp enough to write a clean general proof.

The rule in one line

Pick a, b, c \in \{0, 1, 2, -1, -2\} — whichever subset the relation is defined on — and run the property tests concretely. Small numbers are cheap; algebra on unknowns is expensive. Use the cheap tool first.

Why tiny integers: they expose the edge cases that abstract manipulation glides over. Zero is special (additive identity, annihilator for multiplication). One is special (multiplicative identity). Two is the smallest non-trivial case. If a property breaks on one of these, no amount of general algebra will resurrect it — so the 30-second concrete check saves you from writing a wrong "proof."

Why small numbers are so effective

The three equivalence properties are universally quantified — they say "for all a" or "for all a, b, c." A single concrete counterexample kills the universal claim. So when you suspect a property might fail, the best weapon is concrete arithmetic, not abstract manipulation.

Even when the property does hold, computing examples teaches you the pattern. You see how a - b behaves, what a^2 + b^2 reduces to, when the condition is tight and when it is loose. By the time you write the general proof, you know which algebraic move to make.

The three-column workbench

Three column concrete testing workbenchA table-like figure with three columns labelled Reflexive, Symmetric, and Transitive. Under each column heading is a row showing the concrete test with small numbers — (0,0) and (1,1) for reflexive, (1,2) and (2,1) for symmetric, (1,2) and (2,3) and (1,3) for transitive. A footer note reads "one yes or no per column decides that property." Reflexive Symmetric Transitive (0, 0)? (1, 1)? (2, 2)? all in R? (1, 2)? then (2, 1)? or another small pair (1, 2), (2, 3)? then (1, 3)? three small numbers in a row one counterexample per column kills the property
Three columns, three concrete tests. Run each test on the smallest allowed values; if any fails, the property is dead. If all pass, you still need a general proof — but now you know what to prove.

Reflexive column. Plug in a = 0. Does the relation hold? Plug in a = 1. Does it hold? If either fails, reflexivity fails, full stop.

Symmetric column. Pick a small (a, b) in R. Check if (b, a) is also in R. Try (1, 2) first; then (2, 1); then (0, 1) if you need a failure.

Transitive column. Pick two overlapping pairs: (a, b) and (b, c) with tiny a, b, c. Check if (a, c) is also in R. A popular choice is a = 1, b = 2, c = 3 — three consecutive integers.

Worked example: concrete test catches a trap

Problem: On \mathbb{R}, define a \mathrel{R} b \iff a \leq b^2. Is R an equivalence relation?

Abstract attempt (slow). Reflexive: is a \leq a^2? Hmm, depends on a. You start writing cases and get tangled.

Concrete test (fast). Reflexive: try a = 0.5. Is 0.5 \leq (0.5)^2 = 0.25? No, 0.5 > 0.25. Reflexivity fails at a = 0.5. R is not an equivalence relation. Done in ten seconds.

Try a = 1: 1 \leq 1^2 = 1? Yes. Try a = 2: 2 \leq 4? Yes. If you had only tried 1 and 2, you would have wrongly concluded reflexivity holds. The critical value was a fraction. This is why probing 0, 1, -1, 0.5 — numbers that break different kinds of algebraic patterns — matters.

Worked example: concrete test confirms, then guides proof

Problem: On \mathbb{Z}, define a \mathrel{R} b \iff a \equiv b \pmod{4} (that is, a - b is a multiple of 4). Is R an equivalence relation?

Concrete tests. Reflexive: 0 - 0 = 0, multiple of 4 ✓. 1 - 1 = 0 ✓. 2 - 2 = 0 ✓. Passes.

Symmetric: pick a = 1, b = 5: 1 - 5 = -4, multiple of 4 ✓; reverse 5 - 1 = 4, multiple of 4 ✓. Passes.

Transitive: pick a = 1, b = 5, c = 9: 1 - 5 = -4 ✓, 5 - 9 = -4 ✓, 1 - 9 = -8 = 4 \cdot (-2) ✓. Passes.

All three small-number tests pass. Now write the general proof — but your concrete work has already shown you how. Reflexive: a - a = 0 = 4 \cdot 0. Symmetric: a - b = 4k \Rightarrow b - a = -4k = 4(-k). Transitive: a - b = 4k and b - c = 4m \Rightarrow a - c = 4(k + m). The general proof is essentially the concrete pattern "0 - 0, -(-4) = 4, -4 + -4 = -8" re-written with letters.

The special values to try

Different numbers probe different failure modes. Have this short list ready.

Number What it probes
0 Additive identity behaviour; multiplicative annihilator
1 Multiplicative identity behaviour
-1 Sign flip; useful for multiplicative conditions
2 Smallest non-trivial positive
1/2 (if reals allowed) Fraction vs integer; catches a \leq a^2-type traps
a = b (reflexive probe) Checks what happens when the two variables collide

For integer relations, start with \{-1, 0, 1, 2\}. For real relations, add 0.5. For relations on \mathbb{N} (positive integers), start with \{1, 2, 3\}.

Common JEE traps that small numbers catch

Trap 1: conditions that accidentally require a = b. Example: a \mathrel{R} b \iff a^2 = b^2. Concrete test: (1, -1): 1^2 = 1 = (-1)^2 ✓, so 1 \mathrel{R} (-1) and by symmetry (-1) \mathrel{R} 1. Fine for equivalence. But (1, 2): 1^2 = 1 \neq 4 = 2^2, so not related. This is an equivalence relation partitioning \mathbb{Z} by absolute value.

Trap 2: conditions with hidden division. Example: a \mathrel{R} b \iff a/b \in \mathbb{Q} on \mathbb{R}. Concrete test: (0, 0): 0/0 is undefined, so is (0, 0) \in R or not? This is the trap — the definition is ill-posed at 0, and the concrete test exposes it before you write a whole proof assuming reflexivity holds.

Trap 3: conditions that look symmetric but are not. Example: a \mathrel{R} b \iff a \mid b (divides). Concrete test: (1, 2): 1 \mid 2 ✓; reverse (2, 1): does 2 \mid 1? No. Symmetry fails. Takes five seconds.

JEE-style abstract relation

Problem: On \mathbb{Z}, define a \mathrel{R} b \iff a - b is an odd integer. Is R an equivalence relation?

Concrete test first.

  • Reflexive: a = 0. Is 0 - 0 = 0 odd? No, 0 is even. Reflexivity fails at a = 0 (and at every a).
  • Verdict: Not an equivalence relation. Killed in one line.

Without the concrete test you might start writing a symmetric proof: "if a - b is odd, then b - a = -(a - b) is also odd" — correct, but irrelevant, because the relation already failed at step 1. You would have finished an "elegant" symmetric and transitive analysis (actually both fail; check: 1 \mathrel{R} 0 and 0 \mathrel{R} (-1), so (1, -1) should be related — but 1 - (-1) = 2 is even, so transitivity fails too) and arrived at the same answer after thirty times the effort.

The full habit

Whenever you see a relation on \mathbb{Z}, \mathbb{N}, \mathbb{Q}, or \mathbb{R}:

  1. Before anything else, pick three to five small values from \{0, 1, 2, -1, 0.5\} (as allowed).
  2. Run each property test concretely — pencil and arithmetic, no letters.
  3. If any test fails, write "fails at a = \ldots" and stop. The property is dead.
  4. If all tests pass, write the general proof — your concrete work has already shown the algebraic structure.

The concrete check is free. Skipping it costs you minutes when a property fails silently, and it costs you intuition when a property holds.

Related: Relations · Equivalence Relations · Check All Three Properties Methodically · Test the Condition, Not Individual Pairs