Mathematical-Reasoning · Unit 4 · Video 3 · Interactive Practice

Sets Defined by a Rule: Set Builder Notation, the Empty Set, and Truth Sets That Cover Everything

IKey Formulas

Notation	Name	Meaning
$\{x \in U \mid P(x)\}$	Set builder (rule form)	all $x$ in $U$ for which $P(x)$ is true; the bar reads “such that”
$\{3n + 1 \mid n \in \mathbb{N}\} = \{4, 7, 10, 13, \ldots\}$	Expression form	substitute every allowed $n$ into the expression
$\{x \in \mathbb{Q} \mid x^2 = -2\} = \emptyset$	Empty set	no element of $U$ makes $P(x)$ true
$\{x \in \mathbb{R} \mid x + 0 = x\} = \mathbb{R}$	Whole universal set	every element of $U$ makes $P(x)$ true

Key Insight: a truth set depends on the predicate and the universal set: $\{x \in \mathbb{R} \mid x^2 \le 9\} = \{x \in \mathbb{R} \mid -3 \le x \le 3\}$ , but $\{x \in \mathbb{Z} \mid x^2 \le 9\} = \{-3, -2, -1, 0, 1, 2, 3\}$ . In this course $\mathbb{N} = \{1, 2, 3, \ldots\}$ .

IIOne Predicate, Three Universal Sets

Changing the universal set changes the truth set, and one well-chosen element proves two truth sets differ.

Predicate P(x) Universal set U Test element t 1/2

💡 Challenge: for $x^2 - 3x - 10 = 0$ exactly one of the three universal sets gives a different truth set; find it, and the single element that proves the difference.

IIIExpression and Range, Chosen Together

In the expression-first form $\{f(n) \mid n \in S\}$ , the expression and the range of $n$ together decide which roster you get.

Target roster Expression Range of n

💡 Challenge: two different expressions in the list, both with $n \in \mathbb{Z}$ , produce $\{\ldots, -11, -7, -3, 1, 5, 9, 13, \ldots\}$ ; find both.

IVQuiz Questions

Problem 1 · From Rule to Roster

Given: the set $\{x \in \mathbb{Z} \mid x^2 < 10\}$ . Which roster describes it?

✅ Correct! The integers whose squares are less than

10

run from

-3

3

, because

3^2 = 9 < 10

but

4^2 = 16 \ge 10

❌ Not quite. The universal set is

\mathbb{Z}

, so negative integers count too:

(-2)^2 = 4 < 10

, so

-2

belongs to the set.

❌ Not quite. Those are the squares

x^2

, not the elements

x

. The set collects the integers

x

themselves, and

4

is not one of them since

4^2 = 16

❌ Close, but… that is the truth set over

\mathbb{N}

. Over

\mathbb{Z}

the integers

0

-1

-2

-3

also have squares less than

10

Show solution

The set contains every integer $x$ for which $x^2 < 10$ is true. Check integers outward from $0$ :

0^2 = 0,\quad (\pm 1)^2 = 1,\quad (\pm 2)^2 = 4,\quad (\pm 3)^2 = 9 \quad\text{are all } < 10,

(\pm 4)^2 = 16 \ge 10,

and larger integers (in absolute value) have even larger squares. So

\{x \in \mathbb{Z} \mid x^2 < 10\} = \{-3, -2, -1, 0, 1, 2, 3\}.

Problem 2 · Where Does the List Start?

Given: the roster $\{1, 4, 7, 10, \ldots\}$ , with $\mathbb{N} = \{1, 2, 3, \ldots\}$ . Which set builder description equals it?

✅ Correct! The step is

3

, and

n = 1

must give

1

, so the expression is

3n - 2

: it produces

1, 4, 7, 10, \ldots

❌ Close, but… with

n \in \mathbb{N}

the first value is

3 \cdot 1 + 1 = 4

, so this set is

\{4, 7, 10, \ldots\}

and misses

1

. The expression

3n + 1

would need

n

to start at

0

❌ Not quite. Letting

n

run over

\mathbb{Z}

also produces negative values such as

3(-1) + 1 = -2

, which the roster does not contain.

❌ Not quite. The expression

n + 3

increases by

1

each time, giving

\{4, 5, 6, \ldots\}

; the roster increases by

3

Show solution

Consecutive elements differ by $3$ , which suggests an expression $3n + c$ . With $n$ running through $\mathbb{N}$ , the first element must come from $n = 1$ :

3 \cdot 1 + c = 1 \quad\Longrightarrow\quad c = -2.

Check: $3 \cdot 1 - 2 = 1$ , $3 \cdot 2 - 2 = 4$ , $3 \cdot 3 - 2 = 7$ , $3 \cdot 4 - 2 = 10$ . So $\{1, 4, 7, 10, \ldots\} = \{3n - 2 \mid n \in \mathbb{N}\}$ .

For comparison, $\{3n + 1 \mid n \in \mathbb{N}\} = \{4, 7, 10, \ldots\}$ is missing $1$ ; the expression $3n + 1$ produces the roster only if $n$ ranges over the nonnegative integers $0, 1, 2, \ldots$

Problem 3 · A Roster Running Both Ways

Given: the roster $\{\ldots, -9, -5, -1, 3, 7, 11, \ldots\}$ , continuing in both directions. Which set builder description equals it?

✅ Correct!

n = 0

gives

3

n = -1

gives

-1

n = 1

gives

7

, and each step in

n

moves by

4

in both directions.

❌ Close, but… the expression fits, but

n \in \mathbb{N}

only produces

\{3, 7, 11, \ldots\}

. The roster also runs to the left, so

n

must range over all of

\mathbb{Z}

❌ Not quite.

\{4n + 1 \mid n \in \mathbb{Z}\} = \{\ldots, -3, 1, 5, 9, \ldots\}

. The step of

4

is right, but

3

is not of the form

4n + 1

for an integer

n

❌ Not quite. The coefficient of

n

must be the step between neighbours, which is

4

;

3n + 4

steps by

3

and gives

\{\ldots, -2, 1, 4, 7, \ldots\}

Show solution

Adding $4$ to any element gives the next one, so try an expression $4n + c$ with $n \in \mathbb{Z}$ (the roster runs in both directions).

Look at the element $3$ : $3 = 4 \cdot 0 + 3$ , so $c = 3$ . Test at both ends:

4(-3) + 3 = -9, \qquad 4 \cdot 2 + 3 = 11.

Both match, so the roster equals $\{4n + 3 \mid n \in \mathbb{Z}\}$ . Starting from $-1 = 4 \cdot 0 - 1$ instead gives $\{4n - 1 \mid n \in \mathbb{Z}\}$ , which is the same set: the value $4n + 3$ equals $4(n + 1) - 1$ . With $n$ restricted to $\mathbb{N}$ , as in (a), only the part $\{3, 7, 11, \ldots\}$ appears.

Problem 4 · The Two Extremes

Part 1: for which universal set $U$ is $\{x \in U \mid 2x = 3\} = \emptyset$ ? Part 2: with universal set $\mathbb{R}$ , which predicate has truth set equal to $\mathbb{R}$ ?

Part 1 · Which universal set gives the empty set?

Part 2 · Which predicate is true for every real number?

✅ Correct! The only solution of

2x = 3

\tfrac{3}{2}

, which is rational but not an integer; and

(x + 1)^2 = x^2 + 2x + 1

holds for every real

x

❌ Not quite. Solve

2x = 3

first:

x = \tfrac{3}{2}

. The set is empty exactly when

\tfrac{3}{2}

is not in

U

❌ Not quite. Solving

2x = 3

gives

x = \tfrac{3}{2}

, which is a real number, so

\{x \in \mathbb{R} \mid 2x = 3\} = \{\tfrac{3}{2}\}

, not

\emptyset

❌ Not quite.

\tfrac{3}{2}

is a quotient of integers, so it is rational and

\{x \in \mathbb{Q} \mid 2x = 3\} = \{\tfrac{3}{2}\}

, not

\emptyset

❌ Not quite. The truth set depends on the universal set too:

\tfrac{3}{2}

is a solution in

\mathbb{R}

and

\mathbb{Q}

, but it is not an integer.

❌ Not quite. Substitute a few real numbers, including

0

; the truth set is

\mathbb{R}

only if every substitution gives a true statement.

❌ Close, but…

x^2 > 0

is true for every nonzero real number, but at

x = 0

it reads

0 > 0

, which is false. Its truth set is all real numbers except

0

❌ Not quite. The left side

x \cdot 0

is always

0

, so the equation holds only when

x = 0

. Its truth set is

\{0\}

❌ Not quite. That is the opposite extreme:

x + 1 = x

would force

1 = 0

, so no real number satisfies it and its truth set is

\emptyset

Show solution

Part 1. Over $\mathbb{R}$ , $2x = 3$ has the single solution $x = \tfrac{3}{2}$ . This number is rational, so $\{x \in \mathbb{R} \mid 2x = 3\} = \{x \in \mathbb{Q} \mid 2x = 3\} = \{\tfrac{3}{2}\}$ . It is not an integer, so

\{x \in \mathbb{Z} \mid 2x = 3\} = \emptyset.

Part 2. Test each predicate:

$x^2 > 0$ : false at $x = 0$ , so its truth set is not $\mathbb{R}$ .
$(x + 1)^2 = x^2 + 2x + 1$ : expanding the left side gives the right side, so it is true for every real $x$ . Truth set $\mathbb{R}$ .
$x \cdot 0 = x$ : the left side is always $0$ , so it is true only at $x = 0$ . Truth set $\{0\}$ .
$x + 1 = x$ : subtracting $x$ gives $1 = 0$ , false for every $x$ . Truth set $\emptyset$ .

Even though $(x + 1)^2 = x^2 + 2x + 1$ is true for every real number, on its own (with no quantifier stated) it is still a predicate.

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