MathematicsRelations and FunctionsJEE

Relations and Functions · Part 1 of 20

Relations: Definition, Domain and Range

A relation is any subset of A × B: no rule is needed. This part covers the Cartesian product, roster, set-builder and arrow-diagram forms, domain, range and codomain, counting relations, and the empty, universal, inverse and identity relations.

Video coming soon

Cartesian product

A × B

A × B is the set of all ordered pairs (a, b) with a ∈ A and b ∈ B. If n(A) = p and n(B) = q, then $n (A \times B) = pq$ . Order matters: (1, a) is not (a, 1).

A 2 by 3 grid of ordered pairs. — A = {1, 2}, B = {a, b, c}: six pairs.

Arrow diagram of x less than y. — Domain {1, 4, 5, 6}, range {2, 7}, codomain B.

Relation

Any subset of A × B

A = {1, 4, 5, 6}, B = {1, 2, 7}, x R y when x < y. Roster form: {(1, 2), (1, 7), (4, 7), (5, 7), (6, 7)}. Set-builder form: {(x, y) : x ∈ A, y ∈ B, x < y}. Each pair is one arrow. Each of the pq pairs is in or out, so there are $2^{pq}$ relations from A to B.

Domain and range

x² + y² = 25 on W

The whole-number points on the circle are (0, 5), (3, 4), (4, 3) and (5, 0), so the domain and the range are both {0, 3, 4, 5}. W includes 0: don't drop (0, 5) and (5, 0).

Circle of radius 5 with four lattice points marked. — Points with a negative coordinate are not in W.

5 by 5 grid with 12 dots. — Each dot is one pair (a, b).

Roster form

a R b: a divisible by b

On A = {1, 2, 3, 4, 6}: every number with 1 and itself (9 pairs), plus (4, 2), (6, 2), (6, 3). Twelve pairs; domain and range are all of A.

Summary

Key formulas

n (A \times B) = pq when n (A) = p, n (B) = q

Relation as a subset of the Cartesian product

relations from A to B = 2^{pq}

Relation as a subset of the Cartesian product

Worked examples

From the video

1. R = {(x, y) : x, y ∈ W, x² + y² = 25}. Find the domain and range.

Domain {0, 3, 4, 5}; range {0, 3, 4, 5}.

2. R₁ = {(x, y) : y = 2x + 7, −5 ≤ x ≤ 5}. Find the domain and range.

Domain [−5, 5]; range [−3, 17].

3. A = {2, 3, 4, 5, 6, 7}, x R y iff gcd(x, y) = 1. Find n(R).

22.

4. On {1, …, 50}, R₁ = {(p, pⁿ) : p prime, n ≥ 0}, R₂ = the same with n = 0 or 1. Find n(R₁ − R₂).

8: (2, 4), (2, 8), (2, 16), (2, 32), (3, 9), (3, 27), (5, 25), (7, 49).

JEE-style question

Your turn

n(A) = 3 and n(B) = 2. The number of relations from A to B is:

(a)6

(b)8

(c)32

(d)64

Show the answer and the traps

n(A × B) = 3 × 2 = 6, so the answer is 2⁶ = 64: option (d).

6 is the number of pairs, not the number of relations.

8 = 2³ uses only n(A), and 32 = 2⁵ adds 3 + 2 instead of multiplying.

Watch out

Common mistakes

Forgetting 0 when the set is WWhole numbers start at 0, so (0, 5) and (5, 0) count.

Counting (p, p⁰) twice(p, p⁰) = (p, 1) is in both R₁ and R₂, so it cancels in R₁ − R₂.

Thinking a reflexive relation must be the identityReflexive means every (a, a) is in R; R may contain more pairs.

Practice

Try these

1. n(A) = 4 and n(B) = 3. Find n(A × B) and the number of relations from A to B.

12 and 2¹² = 4096.

2. A = {1, 2, 3, 5}, B = {4, 6, 9}, R = {(x, y) : x − y is odd}. Write R in roster form.

{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

3. R = {(x, x³) : x is a prime less than 10}. Write R and its range.

R = {(2, 8), (3, 27), (5, 125), (7, 343)}; range {8, 27, 125, 343}.

4. R = {(1, 2), (2, 4), (3, 6)}. Find R⁻¹ and its domain and range.

R⁻¹ = {(2, 1), (4, 2), (6, 3)}; domain {2, 4, 6}, range {1, 2, 3}.