Relations

What Is a Relation?

Take P = {a, b, c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}. The product $P \times Q$ has $3 \times 5 = 15$ ordered pairs — every letter paired with every name. Most of those pairs are meaningless. But select just the pairs satisfying a rule:

$$R = \{(x, y) : x \text{ is the first letter of the name } y\}$$

and suddenly the pairs carry meaning: R = {(a, Ali), (b, Bhanu), (b, Binoy), (c, Chandra)}.

Key Point (Definition): A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product $A \times B$. The subset is obtained by describing a relationship between the first element and the second element of the pairs. The second element is called the image of the first.

That is the entire definition — a relation is nothing more mysterious than a chosen set of pairs. Any subset qualifies, whether or not it follows a neat rule.

A relation can be presented three ways:

1. Roster form — list the pairs: R = {(1, 2), (2, 3), (3, 4)}.
2. Set-builder form — state the rule: $R = \{(x, y) : y = x + 1,\ x, y \in A\}$.
3. Arrow diagram — draw both sets and an arrow for each pair.

[Board Important] A relation from A to A is called a relation on A. The arrow diagram above shows $R = \{(x, y) : y = x + 1\}$ on A = {1, …, 6}.

Domain, Codomain and Range

Three vocabulary words describe every relation R from A to B:

• Domain — the set of all first elements of the pairs in R (the elements that actually send an arrow).
• Range — the set of all second elements of the pairs in R (the elements that actually receive an arrow).
• Codomain — the whole set B, whether or not its elements are used.

Key Point: Range $\subset$ codomain, always. The codomain is the target announced in advance; the range is the part of the target actually hit.

Worked illustration: for $R = \{(x, y) : y = x + 1\}$ on A = {1, 2, 3, 4, 5, 6}, the pairs are (1,2), (2,3), (3,4), (4,5), (5,6). So domain = {1, 2, 3, 4, 5} (6 sends no arrow), range = {2, 3, 4, 5, 6} (1 receives no arrow), codomain = {1, 2, 3, 4, 5, 6}.

The picture shows another standard relation — "x is the square of y" from P = {4, 9, 25} to Q = {$-5$, $-3$, $-2$, 1, 2, 3, 5} — as highlighted cells of the full $P \times Q$ lattice: R = {(4, 2), (4, $-2$), (9, 3), (9, $-3$), (25, 5), (25, $-5$)}, with domain {4, 9, 25} and range {$-5$, $-3$, $-2$, 2, 3, 5}. Note 1 receives no arrow: it is in the codomain but not the range.

[JEE Tip] When a relation is given by a formula on a restricted set, compute the pairs first and only then read off domain and range — elements whose images fall outside the target set must be dropped. This "image escapes the set" trap is the most common source of wrong domains.

Counting Relations

Since a relation from A to B is any subset of $A \times B$, counting relations is counting subsets:

Key Point: If $n(A) = p$ and $n(B) = q$, then $n(A \times B) = pq$, and the total number of relations from A to B is $$2^{pq}$$

Each of the pq pairs makes an independent in-or-out choice — the same $2^n$ logic as the power set (because the set of relations from A to B IS the power set of $A \times B$).

Example: A = {1, 2}, B = {3, 4}. Then $A \times B$ has 4 pairs and there are $2^4 = 16$ relations — from the empty relation $\phi$ (no pairs at all) up to $A \times B$ itself (every pair).

Two extreme relations are always available:

• The empty relation $\phi$ — no element related to any element.
• The universal relation $A \times B$ — every element related to every element.

[JEE Tip] Count relations satisfying a condition by counting the free pairs. Example: relations from A to B that contain a specific pair $(a_1, b_1)$: fix it in, the other $pq - 1$ pairs are free — $2^{pq-1}$. The "fix and count the rest" trick from the power set chapter transfers wholesale.

Solved Examples

Example 1. Let A = {1, 2, 3, 4, 5, 6} and define R from A to A by $R = \{(x, y) : y = x + 1\}$. (i) Write R in roster form. (ii) Give its domain, codomain and range.

Solution.

Step 1 — generate pairs, checking each image stays in A. $x = 1 \to y = 2$ ✓; $x = 2 \to 3$ ✓; $x = 3 \to 4$ ✓; $x = 4 \to 5$ ✓; $x = 5 \to 6$ ✓; $x = 6 \to 7$ — but $7 \notin A$, so this pair is DROPPED.

Step 2 — roster form. R = {(1,2), (2,3), (3,4), (4,5), (5,6)}.

Step 3 — read off the three sets. Domain = first elements = {1, 2, 3, 4, 5}. Range = second elements = {2, 3, 4, 5, 6}. Codomain = the whole target set A = {1, 2, 3, 4, 5, 6}.

Takeaway: the formula wanted to send 6 to 7 but the codomain refused — ALWAYS check that images stay inside the target set before listing a pair.

Example 2. A relation between P = {9, 4, 25} and Q = {5, 3, 2, 1, $-2$, $-3$, $-5$} is given by arrows joining each square to its square roots. Write it in set-builder and roster form; give the domain and range.

Solution.

Step 1 — name the rule. Each first element is the square of its partner: $R = \{(x, y) : x \text{ is the square of } y,\ x \in P,\ y \in Q\}$.

Step 2 — roster form, element by element. $9 = 3^2 = (-3)^2$: pairs (9, 3), (9, $-3$). $4 = 2^2 = (-2)^2$: (4, 2), (4, $-2$). $25 = 5^2 = (-5)^2$: (25, 5), (25, $-5$). Six pairs.

Step 3 — domain and range. Domain = {4, 9, 25} (every element of P sends arrows). Range = {$-5$, $-3$, $-2$, 2, 3, 5}. Note $1 \in Q$ receives no arrow: it sits in the codomain but NOT in the range.

Takeaway: one first element may send several arrows (9 relates to both 3 and $-3$) — perfectly legal for a relation, and exactly what will FAIL the function test next section.

Example 3. Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.

Solution.

Step 1 — size of the product. $n(A \times B) = 2 \times 2 = 4$: the pairs (1,3), (1,4), (2,3), (2,4).

Step 2 — relations are subsets. Every subset of $A \times B$ is one relation, and a 4-element set has $2^4 = 16$ subsets.

Step 3 — conclude. There are 16 relations — from the empty relation $\phi$ up to the universal relation $A \times B$ itself.

Takeaway: the formula $2^{pq}$ in action with $p = q = 2$ — relations are counted by the power set of the product.

Example 4. Let A = {1, 2, 3, 4, 6} and R = {(a, b) : a, b $\in$ A, b is exactly divisible by a}. (i) Write R in roster form. (ii) Find the domain and (iii) the range.

Solution.

Step 1 — list row by row (fix a, scan b through A). a = 1 divides everything: (1,1), (1,2), (1,3), (1,4), (1,6) — 5 pairs. a = 2 divides 2, 4, 6: (2,2), (2,4), (2,6) — 3 pairs. a = 3 divides 3, 6: (3,3), (3,6) — 2 pairs. a = 4 divides only 4: (4,4). a = 6 divides only 6: (6,6).

Step 2 — count. $5 + 3 + 2 + 1 + 1 = 12$ pairs in R.

Step 3 — domain. Every element divides itself, so every a sends atleast one arrow: domain = {1, 2, 3, 4, 6} = A.

Step 4 — range. Every element is divisible by 1, so every b receives an arrow: range = A as well.

Takeaway: the row-by-row discipline (fix a, sweep b) prevents undercounting — 12 pairs here are easy to miss without it.

Example 5. A = {1, 2, 3, …, 14}. R is defined from A to A by $R = \{(x, y) : 3x - y = 0\}$. Write down its domain, codomain and range.

Solution.

Step 1 — solve the rule for y. $3x - y = 0 \iff y = 3x$.

Step 2 — generate pairs with BOTH entries in A. $x = 1 \to 3$ ✓; $x = 2 \to 6$ ✓; $x = 3 \to 9$ ✓; $x = 4 \to 12$ ✓; $x = 5 \to 15 \notin A$ — stop (larger x only overshoot further).

Step 3 — read the three sets. R = {(1,3), (2,6), (3,9), (4,12)}. Domain = {1, 2, 3, 4}; codomain = A = {1, …, 14}; range = {3, 6, 9, 12}.

Takeaway: the formula runs out of room inside A quickly — the domain ends where the image escapes, so test every candidate x against the bound.

Example 6. Define R on $\mathbb{N}$ by R = {(x, y) : y = x + 5, x a natural number less than 4}. Write R in roster form; give its domain and range.

Solution.

Step 1 — enumerate the domain condition. Naturals less than 4: x = 1, 2, 3.

Step 2 — apply the formula. $y = x + 5$ gives the pairs (1, 6), (2, 7), (3, 8).

Step 3 — read off. Domain = {1, 2, 3}; range = {6, 7, 8}; the codomain is all of $\mathbb{N}$.

Takeaway: with a small finite condition, plain enumeration is the whole method — condition first, formula second.

Example 7. A = {1, 2, 3, 5}, B = {4, 6, 9}. R = {(x, y) : the difference between x and y is odd, $x \in A$, $y \in B$}. Write R in roster form.

Solution.

Step 1 — convert the condition to parity. A difference is odd exactly when one number is even and the other odd (odd − odd and even − even are both even).

Step 2 — sort the sets by parity. A: odd = {1, 3, 5}, even = {2}. B: even = {4, 6}, odd = {9}.

Step 3 — pair across parities. Odd x with even y: (1,4), (1,6), (3,4), (3,6), (5,4), (5,6) — six pairs. Even x with odd y: (2,9) — one pair.

Step 4 — sanity check one exclusion. (1, 9): both odd, $|9 - 1| = 8$ even — correctly excluded ✓. Final: R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}, seven pairs ($3 \times 2 + 1 \times 1 = 7$ ✓).

Takeaway: translate the condition into parity FIRST; the listing then takes seconds and the product-of-counts check confirms it.

Example 8. Determine the domain and range of the relation R = {(x, x + 5) : x $\in$ {0, 1, 2, 3, 4, 5}}.

Solution.

Step 1 — list the pairs. (0,5), (1,6), (2,7), (3,8), (4,9), (5,10).

Step 2 — domain. All listed x-values qualify (no image escapes): domain = {0, 1, 2, 3, 4, 5}.

Step 3 — range. The images: {5, 6, 7, 8, 9, 10} — the domain shifted up by 5.

Takeaway: when the rule applies cleanly to every listed x, domain = the given set and range = its image under the formula.

Example 9. Write the relation R = {(x, $x^3$) : x is a prime number less than 10} in roster form.

Solution.

Step 1 — enumerate the condition. Primes below 10: 2, 3, 5, 7 (1 is not prime; 9 = 3 × 3 is composite).

Step 2 — apply the formula. $2^3 = 8$, $3^3 = 27$, $5^3 = 125$, $7^3 = 343$.

Step 3 — roster form. R = {(2, 8), (3, 27), (5, 125), (7, 343)}.

Takeaway: two-step recipe — enumerate the domain condition, then apply the formula; compute $7^3 = 343$, never guess.

Example 10. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Solution.

Step 1 — size of the product. $n(A \times B) = 3 \times 2 = 6$.

Step 2 — count the subsets. Each of the 6 pairs is independently in or out: $2^6 = 64$ relations.

Step 3 — contrast with functions. The number of FUNCTIONS from A to B is $2^3 = 8$ (each of 3 inputs picks one of 2 images) — functions are a small minority of the 64 relations, as the next section explains.

Takeaway: relations count as $2^{pq}$, functions as $q^p$ — keep the two formulas firmly apart.

Example 11. R is defined on $\mathbb{Z}$ by R = {(a, b) : a, b $\in \mathbb{Z}$, $a - b$ is an integer}. Find the domain and range of R.

Solution.

Step 1 — test the condition honestly. Take ANY integers a and b: their difference $a - b$ is automatically an integer — the condition never fails.

Step 2 — identify the relation. Every pair qualifies: R = $\mathbb{Z} \times \mathbb{Z}$, the universal relation on $\mathbb{Z}$.

Step 3 — read off. Domain = $\mathbb{Z}$ (every integer sends arrows), range = $\mathbb{Z}$ (every integer receives them).

Takeaway: some conditions are no restriction at all — recognising a universal relation instantly saves time and shows understanding.

Example 12. Let R be a relation from $\mathbb{Q}$ to $\mathbb{Q}$ defined by R = {(a, b) : a, b $\in \mathbb{Q}$ and $a - b \in \mathbb{Z}$}. Show that (i) (a, a) $\in$ R for all $a \in \mathbb{Q}$ (ii) (a, b) $\in$ R implies (b, a) $\in$ R (iii) (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R.

Solution.

(i) Step 1. Compute the difference for a pair (a, a): $a - a = 0$, and $0 \in \mathbb{Z}$. So (a, a) ∈ R for every rational a. ✓

(ii) Step 2. Given $a - b \in \mathbb{Z}$. Then $b - a = -(a - b)$ is the negative of an integer, hence an integer. So (b, a) ∈ R. ✓

(iii) Step 3. Given $a - b \in \mathbb{Z}$ and $b - c \in \mathbb{Z}$. Add them: $(a - b) + (b - c) = a - c$ — a sum of two integers, hence an integer. So (a, c) ∈ R. ✓

Takeaway: the three properties just verified are reflexivity, symmetry and transitivity — the trio defining an equivalence relation in Class 12; the telescoping sum in (iii) is the standard transitivity move.

Example 13. R is defined from $\mathbb{N}$ to $\mathbb{N}$ by R = {(a, b) : $a = b^2$}. Are these true? (i) (a, a) $\in$ R for all a (ii) (a, b) $\in$ R implies (b, a) $\in$ R (iii) (a, b), (b, c) $\in$ R implies (a, c) $\in$ R.

Solution.

Step 1 — strategy. To defeat a universal claim, one numeric counterexample suffices — hunt among small squares.

(i) Step 2 — false. Try a = 2: (2, 2) ∈ R would need $2 = 2^2 = 4$. Counterexample found.

(ii) Step 3 — false. $(4, 2) \in R$ since $4 = 2^2$; but (2, 4) ∈ R would need $2 = 4^2 = 16$.

(iii) Step 4 — false. $(16, 4) \in R$ ($16 = 4^2$) and $(4, 2) \in R$ ($4 = 2^2$); but (16, 2) ∈ R would need $16 = 2^2 = 4$.

Takeaway: the chain 2 → 4 → 16 of repeated squares supplies every counterexample — small numbers do all the work.

Example 14. A = {1, 2, 3}, B = {a, b}. How many relations from A to B contain the pair (1, a)?

Solution.

Step 1 — count all pairs. $n(A \times B) = 3 \times 2 = 6$.

Step 2 — apply the constraint. The pair (1, a) is FORCED in; the remaining $6 - 1 = 5$ pairs each choose freely, in or out.

Step 3 — count. $2^5 = 32$ relations.

Takeaway: conditioned counting = fix the forced pairs, free the rest — the same $2^{n-k}$ logic as constrained subset counts.