You see this in JEE every year: "R = \{(a, b) \in \mathbb{Z} \times \mathbb{Z} \mid a - b \text{ is divisible by } 3\}. Is R an equivalence relation?" A student's first instinct is to list some pairs — (0, 0), (0, 3), (3, 0), (1, 4), \ldots — and check whether reflexivity, symmetry, transitivity hold for those specific pairs. This approach fails badly. For infinite sets there is no way to list, and even for finite ones the pair-by-pair method is the slow path. The condition is the whole relation; work with the condition.
The mindset shift
When R is given by a condition C(a, b) (read: "a \mathrel{R} b iff C(a, b)"), the pair (a, b) belongs to R precisely when C(a, b) is true. The set of pairs is just the graph of C. So the three property checks become:
| Property | What to prove using only the condition |
|---|---|
| Reflexive | For all a, the statement C(a, a) is true |
| Symmetric | For all a, b, if C(a, b) is true, then C(b, a) is true |
| Transitive | For all a, b, c, if C(a, b) and C(b, c) are both true, then C(a, c) is true |
No pair-listing anywhere. You manipulate the condition algebraically.
Why this works: the condition encodes the relation completely. If you prove the condition-level claim "for all a, C(a, a) is true," you have simultaneously verified (a, a) \in R for every element of the base set — one proof covers infinitely many pairs at once.
The three-step pattern
Every condition-based question follows the same choreography.
Step 1: Write down the condition C(a, b) clearly. If it is "a - b is divisible by 3," rewrite it algebraically: "a - b = 3k for some integer k."
Step 2: Substitute the pattern required by each property.
- Reflexive: set b = a. What does the condition become?
- Symmetric: assume the condition for (a, b); try to derive it for (b, a).
- Transitive: assume it for (a, b) and (b, c); try to derive it for (a, c).
Step 3: Simplify. If the target condition drops out, the property holds. If you can produce a concrete counterexample against the target, the property fails.
Worked example: "same remainder on division by 3"
Problem: On \mathbb{Z}, define a \mathrel{R} b \iff a and b leave the same remainder on division by 3. Is R an equivalence relation?
Extract the condition. C(a, b): "a \bmod 3 = b \bmod 3." Equivalently, a - b is divisible by 3.
Reflexive. Substitute b = a: does a and a leave the same remainder? Trivially yes — every number has the same remainder as itself. Or algebraically, a - a = 0 = 3 \cdot 0. ✓
Symmetric. Assume a \bmod 3 = b \bmod 3. Then b \bmod 3 = a \bmod 3 by the symmetry of equality itself. ✓
Transitive. Assume a \bmod 3 = b \bmod 3 and b \bmod 3 = c \bmod 3. By transitivity of equality, a \bmod 3 = c \bmod 3. ✓
Three ticks. R is an equivalence relation — it is the famous congruence modulo 3. The entire proof used the condition, not a single pair.
Worked example: a condition that fails
Problem: On \mathbb{R}, define a \mathrel{R} b \iff a + b = 0. Is R an equivalence relation?
Condition. C(a, b): "a + b = 0" — that is, b = -a.
Reflexive. Substitute b = a: is a + a = 0? Only when a = 0. So (a, a) \in R fails for every non-zero a. ✗
As soon as one box fails, you stop. R is not an equivalence relation, and you do not need to check symmetry or transitivity to conclude. (If the question asked "which properties does R have?", you would continue — symmetric holds because a + b = 0 \iff b + a = 0; transitive fails because 1 \mathrel{R} (-1) and (-1) \mathrel{R} 1 but 1 \mathrel{R} 1 is false.)
The condition is cleaner than the pair list
Look at what the listing approach would have cost in these two examples:
- "Divisible by 3" on \mathbb{Z}: the pair list is infinite. Listing cannot even begin.
- "a + b = 0" on \mathbb{R}: the pair list is uncountable. Listing is impossible.
Even when R is finite and the list is manageable, the condition-level proof is usually 3-4 lines while the pair-checking is a table of 9 or 16 rows. Algebra scales; enumeration does not.
When conditions hide inside prose
JEE questions sometimes state the condition in English. You must rewrite it algebraically before you can manipulate it.
| Prose | Algebraic condition |
|---|---|
| "a divides b" (on \mathbb{N}) | b = ka for some k \in \mathbb{N} |
| "a and b have the same parity" | a - b is even, i.e. a - b = 2k |
| "a and b live in the same city" | \text{city}(a) = \text{city}(b) |
| "$ | a - b |
| "a and b are both even" | a = 2k and b = 2m — note: this is not about a-vs-b directly |
The last row is a trap. "a and b are both even" is a condition on a and on b separately — it is not really a pairing rule. It gives a relation, but the relation factors as "a is even AND b is even," which makes symmetry and reflexivity weirdly trivial or weirdly false (reflexivity fails for odd a). Condition-hunting exposes this structure; pair-listing obscures it.
JEE-style condition question
Problem: On \mathbb{R}, define a \mathrel{R} b \iff 1 + ab > 0. Show R is reflexive and symmetric but not transitive.
Condition. C(a, b): "1 + ab > 0."
Reflexive. Substitute b = a: is 1 + a^2 > 0? Since a^2 \geq 0 always, 1 + a^2 \geq 1 > 0. ✓
Symmetric. Assume 1 + ab > 0. Then 1 + ba = 1 + ab > 0 since multiplication commutes. ✓
Transitive. Need a counterexample. Try a = 1, b = 0, c = -1. Check: 1 + (1)(0) = 1 > 0 ✓, 1 + (0)(-1) = 1 > 0 ✓, 1 + (1)(-1) = 0, which is not > 0. So a \mathrel{R} b and b \mathrel{R} c but not a \mathrel{R} c. Transitivity fails. ✗
Three properties — three algebraic tests on the condition. No pair-listing.
The habit to train
Every time you see a relation defined by a rule, force yourself to:
- Write the rule as C(a, b) = \text{(algebraic expression)}.
- For each of the three properties, substitute the required pattern.
- Either derive the target algebraically (property holds), or hunt for a concrete counterexample (property fails).
Pair-listing is for when R is given by a list, not by a rule. When the question specifies "all pairs such that some condition," the condition is the relation — work with it.
Related: Relations · Equivalence Relations · Check All Three Properties Methodically · Test Abstract Relations With Small Numbers First