Sets and logic · Sets & logic

This lesson builds on sets and logic.

Complements and De Morgan’s laws

The complement $A'$ is everything in the universal set $U$ that is not in $A$ . Two rules link complements with unions and intersections:

(A ∪ B)′ = A′ ∩ B′outside the union: in neither set

(A ∩ B)′ = A′ ∪ B′not in both: outside at least one

De Morgan's lawsPick a law

{7, 8}(A ∪ B)′{7, 8}A′ ∩ B′

U = {1, …, 8}, A = {1, 2, 3, 4}, B = {3, 4, 5, 6}. (A ∪ B)′ and A′ ∩ B′ both shade the same region, {7, 8}. Outside the union means outside A and outside B.

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q1

Let $U = \{e, f, g, h, i\}$ be a universal set, and $X = \{e, g\}$ and $Y = \{g, h\}$ subsets of $U$ .

Draw a Venn diagram to represent the information.

Use the Venn diagram to find (i) $X'$ ; (ii) $(X \cup Y)'$ ; (iii) $X' \cap Y'$ .

The Venn diagram
- $g$ is in both; $e$ is in $X$ only; $h$ is in $Y$ only; $f$ and $i$ are outside both.
Think first. Which element is in both X and Y?
The three sets
- ${X' = \{f, h, i\}}$ .
- ${X \cup Y = \{e, g, h\}}$ , so ${(X \cup Y)' = \{f, i\}}$ .
- ${Y' = \{e, f, i\}}$ , so ${X' \cap Y' = \{f, i\}}$ : the same set, as De Morgan’s law says.

More: unions, intersections and complements

WAEC 2022 · Paper 1 · Q15Given that $P = \{x : x \text{ is a multiple of } 5\}$ , $Q = \{x : x \text{ is a multiple of } 3\}$ and $R = \{x : x \text{ is an odd number}\}$ …

Counting with two sets

Members of both sets are counted twice in $n(A) + n(B)$ , so

n(A \cup B) = n(A) + n(B) - n(A \cap B)

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q10 (b)

In a junior secondary school, 60 students play table tennis or basketball. The number who play table tennis is 7 more than three times the number who play basketball. If 3 students play both games and every student plays at least one game, how many students play table tennis?

Two equations
- ${T + B - 3 = 60}$ , so ${T + B = 63}$ .
- Also ${T = 3B + 7}$ .
Think first. The 3 who play both are counted in T and in B.
Solve
- ${3B + 7 + B = 63}$ , so ${4B = 56}$ and ${B = 14}$ .
- ${T = 3(14) + 7 = 49}$ students play table tennis.

More: counting with two sets

WAEC 2008 · Paper 2 · Q5In a hotel the breakfast menu is a choice between yam ( $Y$ ) or plantain ( $P$ ) or both. The Venn diagram shows the choices …

Sets defined by inequalities

A set such as $\{x : 2x^2 - x - 10 \le 0\}$ is a range of numbers. Solve each inequality (see quadratic inequalities), draw both ranges on one number line, then read off the union or intersection.

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q3

Given that $P = \{x : x \in \mathbb{R},\ 2x^2 - x - 10 \le 0\}$ and $Q = \{x : x \in \mathbb{R},\ 4x^2 - 1 \ge 0\}$ , find $P \cap Q$ .

The set P
- ${2x^2 - x - 10 = (x + 2)(2x - 5)}$ , with roots ${-2}$ and ${\frac52}$ .
- ${\le 0}$ is between the roots: ${-2 \le x \le \frac52}$ .
Think first. Factorise: roots, then between or outside?
The set Q
- ${4x^2 - 1 = (2x - 1)(2x + 1)}$ , with roots ${\pm\frac12}$ .
- ${\ge 0}$ is outside the roots: ${x \le -\frac12}$ or ${x \ge \frac12}$ .
The intersection
- ${P \cap Q = \{x : -2 \le x \le -\frac12\} \cup \{x : \frac12 \le x \le \frac52\}}$ .
Think first. Which parts of P are also in Q?

More: sets defined by inequalities

Statements in symbols

$\sim p$ means “not $p$ ”, $p \wedge q$ means ” $p$ and $q$ ”, and $p \vee q$ means ” $p$ or $q$ ”. “Neither $p$ nor $q$ ” means “not $p$ and not $q$ ”.

p	q	p ∧ q
T	T	T
T	F	F
F	T	F
F	F	F

p ∧ qTrue only when both are true

p	q	p ∨ q
T	T	T
T	F	T
F	T	T
F	F	F

p ∨ qFalse only when both are false

$p \Rightarrow q$ means “if $p$ , then $q$ ”, and $p \Leftrightarrow q$ means ” $p$ if and only if $q$ ”. Both, and how to pick a valid conclusion, are taught in logic.

p	q	p ⇒ q
T	T	T
T	F	F
F	T	T
F	F	T

p ⇒ qFalse only when p is true and q is false

p	q	p ⇔ q
T	T	T
T	F	F
F	T	F
F	F	T

p ⇔ qTrue when p and q match

More: statements, implications and conclusions

Your turn

NECO 2023 · Paper 1 · Q20

Sets & logic

Let $p$ denote “The cost of living is high” and $q$ denote “The standard of living is low”. Which of the following describes “Neither the cost of living is high nor the standard of living is low”?

Worked solution (try it first)

"Neither $p$ nor $q$ " means "not $p$ and not $q$ ".
"Not" is $\sim$ and "and" is $\wedge$ , so the statement is $\sim p \wedge \sim q$ , option C.

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