Sets and their algebra

• ›a set is a well-defined collection; it is written in roster form {1, 2, 3} or set-builder form {x : x is a natural number less than 4}
• ›the empty set ∅ has no element; a set is finite or infinite
• ›if a set has n elements, it has 2ⁿ subsets and 2ⁿ − 1 proper subsets; the set of all subsets is the power set P(A)

Operations

De Morgan's laws — (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.

Counting — n(A ∪ B) = n(A) + n(B) − n(A ∩ B); for three sets, add the three, subtract the three pairwise intersections, and add back n(A ∩ B ∩ C).

Example — in a class of 60, 35 take Hindi, 30 take Urdu and 12 take both; those taking at least one = 35 + 30 − 12 = 53, and 7 take neither.

The trap: a 4-element set has 16 subsets — 15 is the count of proper subsets.

Relations

• ›the Cartesian product A × B is the set of ordered pairs (a, b); if n(A) = p and n(B) = q, then n(A × B) = pq
• ›a relation from A to B is any subset of A × B, so there are 2^(pq) relations from A to B
• ›its domain is the set of first elements, its range the set of second elements, and B is the codomain

Types of relation on a set A

• ›an equivalence relation splits A into disjoint equivalence classes; 'a − b is divisible by 3' on the integers gives three classes — remainders 0, 1 and 2
• ›'is less than' on the real numbers is transitive but neither reflexive nor symmetric; 'is perpendicular to' among lines in a plane is symmetric only

The trap: the number of relations from A to B is 2^(pq), not pq — pq counts the ordered pairs.

Functions

• ›a function f : A → B assigns to every element of A exactly one element of B

Counting functions — with n(A) = m and n(B) = n:

• ›all functions from A to B: nᵐ
• ›one-one functions (m ≤ n): n!/(n − m)!
• ›bijections of an n-element set onto itself: n!

Standard real functions

Composition — (g ∘ f)(x) = g(f(x)); in general g ∘ f ≠ f ∘ g. If f(x) = x + 1 and g(x) = x², then (g ∘ f)(x) = (x + 1)² but (f ∘ g)(x) = x² + 1.

The trap: nᵐ counts functions from an m-element set into an n-element set — the exponent is the size of the domain.

Inverse trigonometric functions

The trigonometric functions are not one-one on ℝ, so each is restricted to a principal value branch before it is inverted.

Identities

• ›sin⁻¹ x + cos⁻¹ x = π/2, for x in [−1, 1]
• ›tan⁻¹ x + cot⁻¹ x = π/2
• ›sin⁻¹(−x) = −sin⁻¹ x, but cos⁻¹(−x) = π − cos⁻¹ x

Working through a principal value — sin⁻¹(sin 2π/3) is not 2π/3, because 2π/3 lies outside [−π/2, π/2]. Since sin 2π/3 = √3/2, the answer is the angle in the principal range with that sine: π/3.

The trap: sin⁻¹(sin θ) = θ only when θ is already in [−π/2, π/2] — otherwise, find the principal-range angle with the same sine.

Points to Remember

• ›n elements → 2ⁿ subsets, 2ⁿ − 1 proper · power set
• ›De Morgan — (A ∪ B)′ = A′ ∩ B′ · n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
• ›n(A × B) = pq · relations from A to B = 2^(pq)
• ›reflexive, symmetric, transitive → equivalence → disjoint classes
• ›one-one, onto, bijective — only bijections have inverses
• ›functions A → B: nᵐ · one-one: n!/(n − m)! · bijections on n elements: n!
• ›∣x∣ → [0, ∞) · signum → {−1, 0, 1} · [x] → integers
• ›g ∘ f ≠ f ∘ g in general
• ›principal ranges — sin⁻¹: [−π/2, π/2] · cos⁻¹: [0, π] · tan⁻¹: (−π/2, π/2)
• ›sin⁻¹ x + cos⁻¹ x = π/2 · cos⁻¹(−x) = π − cos⁻¹ x
• ›Exam tip: in a counting question on relations or functions, first write down which set is the domain — almost every wrong answer swaps m and n