6 min readConcept sheetPart I (1 mark) and Part II (2 marks); verifying a property such as A × (B ∪ C) = (A × B) ∪ (A × C) is a Part III (5 marks) sum

Ordered pairs and the Cartesian product

An ordered pair (a, b) keeps its order. The Cartesian product A × B collects every pair with first entry from A and second from B.

DefinitionA × B = {(a, b) | a ∈ A, b ∈ B}

Number of pairsn(A × B) = n(A) × n(B)

Order matters(a, b) = (c, d) only when a = c and b = d; A × B ≠ B × A unless A = B or one set is empty

Three setsA × B × C is a set of ordered triplets (a, b, c); n(A × B × C) = n(A) × n(B) × n(C)

What an ordered pair is

Two objects written in a fixed order, like (2, 5), form an ordered pair. The first number is the first coordinate and the second is the second coordinate. (2, 5) and (5, 2) are different pairs. A cinema seat works the same way: (Row F, Seat 12) names one seat, and the order of the two parts is fixed.

Rules of the Cartesian product

A × B = {(a, b) | a ∈ A, b ∈ B}. Read '|' as 'such that' and '∈' as 'belongs to'.
If A has p elements and B has q elements, A × B has p × q ordered pairs.
A × B ≠ B × A in general, but n(A × B) = n(B × A).
If A = ∅ (the empty set) or B = ∅, then A × B = ∅.
A × B × C = {(a, b, c) | a ∈ A, b ∈ B, c ∈ C}, a set of ordered triplets.
Distributive properties: A × (B ∪ C) = (A × B) ∪ (A × C) and A × (B ∩ C) = (A × B) ∩ (A × C).

Example 1: A × B and B × A

Problem

If A = {5, 7} and B = {2, 4, 6}, find A × B and B × A. Show that A × B ≠ B × A.

Working

A × B pairs each element of A (first) with each element of B (second).
A × B = {(5, 2), (5, 4), (5, 6), (7, 2), (7, 4), (7, 6)}
B × A = {(2, 5), (2, 7), (4, 5), (4, 7), (6, 5), (6, 7)}
n(A × B) = 2 × 3 = 6 and n(B × A) = 3 × 2 = 6.
(5, 2) ∈ A × B, but (5, 2) ∉ B × A, because 5 ∉ B.

Answer

A × B ≠ B × A, although both have 6 ordered pairs.

Example 2: verifying a distributive property

Problem

Let A = {1, 2}, B = {2, 3} and C = {3, 4}. Verify that A × (B ∪ C) = (A × B) ∪ (A × C).

Working

B ∪ C = {2, 3, 4}
LHS: A × (B ∪ C) = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4)}
A × B = {(1, 2), (1, 3), (2, 2), (2, 3)}
A × C = {(1, 3), (1, 4), (2, 3), (2, 4)}
RHS: (A × B) ∪ (A × C) = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4)}
Both sides have the same 6 pairs.

Answer

LHS = RHS, so A × (B ∪ C) = (A × B) ∪ (A × C) is verified.

In real life

A school canteen offers {idli, dosa, pongal} with {sambar, chutney}. Every possible plate is an ordered pair in this Cartesian product, so there are 3 × 2 = 6 combinations.

Exam angle

2 marks: “If A = {2, 3} and B = {1, 3, 5}, find A × B and n(A × B).” A × B = {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)}; n(A × B) = 2 × 3 = 6.

2 marks: “If B × A = {(−1, 2), (−1, 6), (0, 2), (0, 6)}, find A and B.” First coordinates belong to B, so B = {−1, 0}. Second coordinates belong to A, so A = {2, 6}.

5 marks: “Let A = {x ∈ N | x < 3}, B = {3, 4} and C = {4, 5}. Verify that A × (B ∩ C) = (A × B) ∩ (A × C).” A = {1, 2}, B ∩ C = {4}, so LHS = {(1, 4), (2, 4)}. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)} and A × C = {(1, 4), (1, 5), (2, 4), (2, 5)}, so RHS = {(1, 4), (2, 4)}. LHS = RHS, verified.

Remember it

A × B: A first, B second. Count the pairs by multiplying: n(A) × n(B).