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Lesson 1 — Number sets, intervals, notation

Rigorous mathematical language: number sets (ℕ, ℤ, ℚ, ℝ), intervals, set operations. Opening lesson of the program.

Used in: 1.º ano do EM (15 anos) · Equiv. Math I japonês · Equiv. Klasse 10 alemã

\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}

The number sets: each one contains all the elements of the previous one, and adds new objects. Naturals → integers (gains negatives) → rationals (gains fractions) → reals (gains irrationals like $\sqrt{2}$ and $\pi$ ).

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition

Fundamental number sets

"Every real number corresponds to a unique position on the number line. The converse is also true: each location on the number line corresponds to exactly one real number." — OpenStax College Algebra 2e, §1.1

Intervals

Definition· Types of interval

For $a, b \in \mathbb{R}$ with $a \leq b$ :

The four types of interval on the number line. Filled dot = endpoint included. Open dot = endpoint excluded.

[a, b] = \{x \in \mathbb{R} : a \leq x \leq b\}

what this means · Closed interval: both endpoints included.

(a, b) = \{x \in \mathbb{R} : a \lt x \lt b\}

what this means · Open interval: no endpoint included.

[a, b) = \{x : a \leq x \lt b\}, \quad (a, b] = \{x : a \lt x \leq b\}

what this means · Half-open intervals.

Infinite intervals: $[a, +\infty)$ , $(-\infty, b]$ , $(-\infty, +\infty) = \mathbb{R}$ .

Set operations

Worked examples

Example— 1· Listing elements of a set (application)

Problem: List, in brace notation, the set $A = \{x \in \mathbb{Z} : -2 < x \leq 3\}$ .

Strategy: The problem asks us to list the elements of a set. The set notation gives us two pieces of information:

Which number set the elements belong to: $\mathbb{Z}$ (integers).
Which conditions those elements must satisfy: $-2 < x \leq 3$ . We combine this information to identify all integers that fit.

Solution:

Identify the type of number: The elements $x$ must be integers ( $\mathbb{Z}$ ).
Analyze the first condition: $x > -2$ . This means $x$ must be greater than $-2$ . The integers satisfying this part are $-1, 0, 1, 2, 3, \ldots$ .
Analyze the second condition: $x \leq 3$ . This means $x$ must be less than or equal to $3$ . The integers satisfying this part are $\ldots, 1, 2, 3$ .
Combine the conditions: We need integers $x$ $x$ that are simultaneously greater than $-2$ $- 2$ AND less than or equal to $3$ $3$ .
- Integers greater than $-2$ : $\{-1, 0, 1, 2, 3, 4, \ldots\}$
- Integers less than or equal to $3$ : $\{\ldots, -1, 0, 1, 2, 3\}$
- The intersection of these two sets is $\{-1, 0, 1, 2, 3\}$ .

Verification: All listed numbers are integers.

$-1$ is greater than $-2$ and less than or equal to $3$ . (OK)
$0$ is greater than $-2$ and less than or equal to $3$ . (OK)
$1$ is greater than $-2$ and less than or equal to $3$ . (OK)
$2$ is greater than $-2$ and less than or equal to $3$ . (OK)
$3$ is greater than $-2$ and less than or equal to $3$ . (OK) No other integer between $-2$ and $3$ (excluding $-2$ , including $3$ ) was left out.

Source. OpenStax — College Algebra 2e §1.1 — CC-BY 4.0.

Example— 2· Intersection of intervals (intermediate)

Problem: Given the intervals $A = [-3, 7)$ and $B = (1, 10]$ , compute $A \cap B$ .

Strategy: The intersection of two sets includes only the elements that belong to both sets. Visualization on the number line is the most effective tool.

Solution:

Visualize interval $A$ : $A = [-3, 7)$ $A = [- 3, 7)$ goes from -3 (included) to 7 (excluded).
```
←—————[----------)————→
     -3          7
```
Visualize interval $B$ : $B = (1, 10]$ $B = (1, 10]$ goes from 1 (excluded) to 10 (included).
```
←——————————(----------]————→
           1          10
```
Combine for the intersection: We need to find the region where both intervals overlap.
- The leftmost point where both are present: $1$ . But $1$ is excluded in $B$ , so it is excluded from the intersection.
- The rightmost point where both are present: $7$ . But $7$ is excluded in $A$ , so it is excluded from the intersection.
- The overlap region is between $1$ (excluded) and $7$ (excluded).

$A \cap B = (1, 7)$

Verification:

A point in the intersection: $x=3$ . It is in $A$ (since $-3 \leq 3 < 7$ ) and in $B$ (since $1 < 3 \leq 10$ ). (OK)
A point only in $A$ : $x=-2$ . Not in $B$ . (OK)
A point only in $B$ : $x=8$ . Not in $A$ . (OK)
The endpoints: $1$ is not in $B$ , $7$ is not in $A$ . Both excluded from the intersection. The answer is consistent.

Source. Stitz–Zeager — Precalculus §1.2 — CC-BY-NC-SA.

Example— 3· Absolute value inequality, type |u| ≤ c (intermediate)

Problem: Solve the inequality $|x - 4| \leq 3$ and express the solution in interval notation.

Strategy: Absolute value inequalities of the type $|u| \leq c$ (where $c > 0$ ) can be rewritten as a compound inequality: $-c \leq u \leq c$ . We apply this property and then solve for $x$ .

Solution:

Rewrite the inequality: The absolute value property lets us transform $|x - 4| \leq 3$ into: $-3 \leq x - 4 \leq 3$
Isolate $x$ : To isolate $x$ , we add $4$ to all parts of the inequality: $-3 + 4 \leq x - 4 + 4 \leq 3 + 4$ $1 \leq x \leq 7$
Express in interval notation: The inequality $1 \leq x \leq 7$ means $x$ can be any real number between 1 and 7, including 1 and 7.

$\text{Solution: } [1, 7]$

Verification:

Test a point inside the interval: $x=5$ . $|5 - 4| = |1| = 1$ . $1 \leq 3$ . (OK)
Test an endpoint: $x=1$ . $|1 - 4| = |-3| = 3$ . $3 \leq 3$ . (OK)
Test the other endpoint: $x=7$ . $|7 - 4| = |3| = 3$ . $3 \leq 3$ . (OK)
Test a point outside the interval: $x=0$ . $|0 - 4| = |-4| = 4$ . $4 \not\leq 3$ . (OK)
Test a point outside the interval: $x=8$ . $|8 - 4| = |4| = 4$ . $4 \not\leq 3$ . (OK) The solution is correct.

Source. OpenStax — College Algebra 2e §1.7 — CC-BY 4.0.

Example— 4· Quadratic inequality (advanced)

Problem: Determine the solution set of the inequality $x^2 - x - 6 > 0$ in interval notation.

Strategy: To solve a quadratic inequality, we first find the roots of the associated quadratic equation ( $x^2 - x - 6 = 0$ ). These roots divide the real line into intervals. We then test a point in each interval to see where the inequality is satisfied.

Solution:

Find the roots of the associated equation: $x^2 - x - 6 = 0$ $x^{2} - x - 6 = 0$ .
- We can factor: $(x - 3)(x + 2) = 0$ .
- The roots are $x = 3$ and $x = -2$ .
Divide the real line into intervals: The roots $-2$ $- 2$ and $3$ $3$ divide the real line into three intervals:
- $(-\infty, -2)$
- $(-2, 3)$
- $(3, +\infty)$
Test a point in each interval:
- Interval $(-\infty, -2)$ : Choose $x = -3$ . $(-3)^2 - (-3) - 6 = 9 + 3 - 6 = 6$ . Since $6 > 0$ , this interval satisfies the inequality.
- Interval $(-2, 3)$ : Choose $x = 0$ . $(0)^2 - (0) - 6 = -6$ . Since $-6 \not> 0$ , this interval does not satisfy the inequality.
- Interval $(3, +\infty)$ : Choose $x = 4$ . $(4)^2 - (4) - 6 = 16 - 4 - 6 = 6$ . Since $6 > 0$ , this interval satisfies the inequality.
Combine the satisfying intervals: The solution is the union of the intervals where the inequality holds. Note that the inequality is strictly greater than 0, so the roots are not included.

$\text{Solution: } (-\infty, -2) \cup (3, +\infty)$

Verification:

The parabola $y = x^2 - x - 6$ opens upward (positive leading coefficient). This means the function values are positive "outside" the roots and negative "between" the roots. The solution found is consistent with the graph of a parabola.

Source. OpenStax — College Algebra 2e §3.6 — CC-BY 4.0.

Example— 5· Modeling with point exclusion (real-world modeling)

Problem: A temperature sensor operates correctly in an environment where the temperature $T$ (in Celsius) must be between $10\,°C$ (inclusive) and $40\,°C$ (exclusive). In addition, due to a specific requirement, the temperature cannot be exactly $25\,°C$ . Express the valid operating range for $T$ as a union of intervals.

Strategy: First, define the base operating interval. Then, remove the specific point that is not allowed. Removing a single point from an interval can result in two disjoint intervals.

Solution:

Define the base interval: The temperature must be between $10\,°C$ (inclusive) and $40\,°C$ (exclusive). This is represented by the interval $[10, 40)$ .
Identify the point to be excluded: The temperature cannot be $25\,°C$ .
Remove the point from the base interval: The number $25$ $25$ lies within the interval $[10, 40)$ $[10, 40)$ . Removing $25$ $25$ splits the interval into two parts:
- All numbers from $10$ (inclusive) to $25$ (exclusive): $[10, 25)$ .
- All numbers from $25$ (exclusive) to $40$ (exclusive): $(25, 40)$ .
Express as a union of intervals: The valid operating range is the union of these two disjoint intervals.

$\text{Operating range: } [10, 25) \cup (25, 40)$

Verification:

$T=10$ : Valid. In the first interval.
$T=20$ : Valid. In the first interval.
$T=25$ : Not valid. Not in either interval.
$T=30$ : Valid. In the second interval.
$T=40$ : Not valid. Excluded from the original interval. The solution is correct and satisfies all conditions.

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 1 — open license.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 22Understanding 11Modeling 5Challenge 2

Sources

Only books that directly fed the text and the exercises.

Precalculus / College Algebra / Trigonometry — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · ch. 1.
College Algebra 2e — OpenStax · 2022 · EN · CC-BY · §1.1, §1.7.
Book of Proof — Richard Hammack · 2018 · EN · free · chs. 1, 3, 6.
Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · free · ch. 1.
Matemática elementar — Wikilivros · live · PT-BR · CC-BY-SA.