Module 0c — Real Numbers and Inequalities
Course: MAT1CJ101 — Differential Calculus (prerequisite refresher)
Status: Prerequisite background material — not part of the official 60 taught hours, not examined.
0. Why This Module Exists
Limits and continuity are stated using inequalities: “$\lvert f(x) - L \rvert < \varepsilon$ whenever $0 < \lvert x - c \rvert < \delta$.” If solving an inequality or manipulating absolute values isn’t automatic, that notation becomes a wall instead of a tool. This module drills the algebra of inequalities and absolute value until it’s routine, before you meet it embedded in a limit definition.
1. The Number Systems
| System | Symbol | Description | Example elements |
|---|---|---|---|
| Natural numbers | $\mathbb{N}$ | Counting numbers | $1, 2, 3, \ldots$ |
| Whole numbers | $W$ | $\mathbb{N}$ together with 0 | $0, 1, 2, 3, \ldots$ |
| Integers | $\mathbb{Z}$ | Whole numbers and their negatives | $\ldots, -2, -1, 0, 1, 2, \ldots$ |
| Rational numbers | $\mathbb{Q}$ | Ratios $p/q$ of integers, $q \neq 0$ | $1/2, -3, 0.75, 7$ |
| Irrational numbers | — | Real numbers not expressible as $p/q$ | $\sqrt{2}, \pi, e$ |
| Real numbers | $\mathbb{R}$ | Rationals together with irrationals | all of the above |
Containment chain: $\mathbb{N} \subset W \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$.
Worked Example 1: Classify each number into the smallest system it belongs to: $5, -3, 2/3, \sqrt{9}, \sqrt{2}, 0$.
$5$: natural ($\mathbb{N}$) — also in $W, \mathbb{Z}, \mathbb{Q}, \mathbb{R}$, but $\mathbb{N}$ is smallest
$-3$: integer ($\mathbb{Z}$) — negative, so not in $\mathbb{N}$ or $W$
$2/3$: rational ($\mathbb{Q}$) — not an integer
$\sqrt{9}$: $\sqrt{9} = 3$, a natural number ($\mathbb{N}$) — the radical simplifies away
$\sqrt{2}$: irrational — not exactly expressible as $p/q$ (proved by contradiction in most texts)
$0$: whole number ($W$) — not natural under the common convention that $\mathbb{N}$ starts at 1
1.1 Density of Q and R
Between any two distinct real numbers, no matter how close, there is always another real number (in fact infinitely many) — this is the density property of $\mathbb{R}$ (and of $\mathbb{Q}$ within $\mathbb{R}$). This is part of why “the next real number after 1” doesn’t exist — there’s always another one closer to 1 in between.
2. The Real Line, Intervals, and Order
2.1 Order Properties
For $a, b, c \in \mathbb{R}$:
| Property | Statement |
|---|---|
| Trichotomy | Exactly one of $a < b$, $a = b$, $a > b$ holds |
| Transitivity | $a < b$ and $b < c \implies a < c$ |
| Addition | $a < b \implies a + c < b + c$ (any $c$) |
| Multiplication (positive) | $a < b$ and $c > 0 \implies ac < bc$ |
| Multiplication (negative) | $a < b$ and $c < 0 \implies ac > bc$ — inequality flips |
The sign-flip rule for multiplying by a negative is the single most common source of errors when solving inequalities — flag it every time you multiply or divide by an unknown-sign expression.
2.2 Interval Notation
| Set description | Interval notation | Meaning |
|---|---|---|
| $\{x : a \leq x \leq b\}$ | $[a, b]$ | closed interval, both endpoints included |
| $\{x : a < x < b\}$ | $(a, b)$ | open interval, both endpoints excluded |
| $\{x : a \leq x < b\}$ | $[a, b)$ | half-open, left included |
| $\{x : a < x \leq b\}$ | $(a, b]$ | half-open, right included |
| $\{x : x \geq a\}$ | $[a, \infty)$ | unbounded above |
| $\{x : x < a\}$ | $(-\infty, a)$ | unbounded below |
$\infty$ and $-\infty$ are never included with a bracket — they aren’t actual numbers, just markers for “unbounded,” so they always get a parenthesis.
3. Solving Linear and Quadratic Inequalities
3.1 Linear Inequalities
Solve like an equation, remembering to flip the inequality when multiplying/dividing by a negative.
Worked Example 2: Solve $3x + 5 \leq 2x - 1$.
\(3x + 5 \leq 2x - 1\) \(3x - 2x \leq -1 - 5 \quad \text{(move $x$ terms left, constants right)}\) \(x \leq -6\)
Solution: $(-\infty, -6]$
Worked Example 3: Solve $-2x + 3 > 7$.
\[-2x + 3 > 7 \implies -2x > 4 \implies x < -2 \quad \text{(divided by } -2\text{, inequality FLIPPED)}\]Solution: $(-\infty, -2)$
Example 1 (Linear inequality solution): Does $x = -7$ solve $3x + 5 \leq 2x - 1$ (solution set $(-\infty, -6]$)?
$-7 \leq -6$? Yes. $\checkmark$ $x = -7$ is in $(-\infty,-6]$, so it must satisfy the original inequality. Check directly: $3(-7)+5 = -16$. $2(-7)-1 = -15$. Is $-16 \leq -15$? Yes. $\checkmark$ Confirmed.
Non-Example 1 (Linear inequality solution): Does $x = 0$ solve $3x + 5 \leq 2x - 1$?
Is $0 \leq -6$? No. $\times$ So $0$ is NOT in the solution set $(-\infty,-6]$. Check directly: $3(0)+5 = 5$. $2(0)-1 = -1$. Is $5 \leq -1$? No. $\times$ Confirmed — $x=0$ fails the original inequality, matching the fact that it’s outside the solution set.
3.2 Quadratic Inequalities — Sign Chart Method
Method: move everything to one side (so the inequality compares to 0), factor, find the roots, then use a sign chart to test each interval the roots create.
Worked Example 4: Solve $x^2 - x - 6 > 0$.
Factor: $x^2 - x - 6 = (x-3)(x+2)$
Roots (where expression $= 0$): $x = 3$, $x = -2$. These split $\mathbb{R}$ into three intervals: $(-\infty,-2)$, $(-2,3)$, $(3,\infty)$
Test a point in each interval: \(x = -3: \ (-3-3)(-3+2) = (-6)(-1) = 6 > 0 \ \checkmark \ \text{interval works}\) \(x = 0: \ (0-3)(0+2) = (-3)(2) = -6 < 0 \ \times\) \(x = 4: \ (4-3)(4+2) = (1)(6) = 6 > 0 \ \checkmark \ \text{interval works}\)
Since the inequality is strict ($> 0$), roots themselves are excluded.
Solution: $(-\infty, -2) \cup (3, \infty)$
Worked Example 5: Solve $x^2 + 2x - 3 \leq 0$.
Factor: $(x+3)(x-1) \leq 0$. Roots: $x = -3$, $x = 1$
Test: \(x = -4: \ (-1)(-5) = 5 > 0 \ \times \qquad x = 0: \ (3)(-1) = -3 \leq 0 \ \checkmark \qquad x = 2: \ (5)(1) = 5 > 0 \ \times\)
Inequality is non-strict ($\leq 0$), so roots are included.
Solution: $[-3, 1]$
Example 2 (Quadratic inequality solution): Does $x = 4$ solve $x^2 - x - 6 > 0$ (solution set $(-\infty,-2) \cup (3,\infty)$)?
$4 \in (3,\infty)$? Yes. $\checkmark$ Check directly: $4^2 - 4 - 6 = 16 - 10 = 6$. Is $6 > 0$? Yes. $\checkmark$ Confirmed.
Non-Example 2 (Quadratic inequality solution): Does $x = 3$ (a root of the equation) solve $x^2 - x - 6 > 0$?
$3$ is one of the boundary roots. The inequality is STRICT ($> 0$, not $\geq 0$), so roots are always excluded from the solution set. Check directly: $3^2 - 3 - 6 = 9 - 9 = 0$. Is $0 > 0$? No. $\times$ $x = 3$ does NOT solve the inequality — it only makes the expression exactly 0, not strictly positive.
3.3 Rational Inequalities
Same sign-chart idea, but critical points now include both the zeros of the numerator and the zeros of the denominator (which must always be excluded, since division by 0 is undefined).
Worked Example 6: Solve $(x - 2)/(x + 1) \geq 0$.
Critical points: numerator zero at $x = 2$; denominator zero at $x = -1$ (excluded from domain).
Intervals: $(-\infty,-1)$, $(-1,2)$, $(2,\infty)$
Test: \(x = -2: \ \frac{-2-2}{-2+1} = \frac{-4}{-1} = 4 > 0 \ \checkmark\) \(x = 0: \ \frac{0-2}{0+1} = \frac{-2}{1} = -2 < 0 \ \times\) \(x = 3: \ \frac{3-2}{3+1} = \frac{1}{4} > 0 \ \checkmark\)
$x = 2$ gives 0, and inequality is $\geq 0$, so $x = 2$ is included. $x = -1$ makes the expression undefined, so it is never included, regardless of the inequality symbol.
Solution: $(-\infty, -1) \cup [2, \infty)$
Example 3 (Rational inequality solution): Does $x = 5$ solve $(x-2)/(x+1) \geq 0$ (solution set $(-\infty,-1) \cup [2,\infty)$)?
$5 \in [2,\infty)$? Yes. $\checkmark$ Check directly: $(5-2)/(5+1) = 3/6 = 0.5$. Is $0.5 \geq 0$? Yes. $\checkmark$ Confirmed.
Non-Example 3 (Rational inequality solution): Does $x = -1$ solve $(x-2)/(x+1) \geq 0$?
$x = -1$ makes the denominator $(x+1) = 0$ — the expression is UNDEFINED there, not just “failing” the inequality. Undefined points are excluded from the solution set no matter what the inequality symbol is (even $\geq$, which usually includes boundary points). $x = -1$ does NOT solve the inequality — it isn’t even in the domain of the expression.
Common error: cross-multiplying a rational inequality by $(x + 1)$ as if it were an equation. This is invalid because the sign of $(x+1)$ is unknown — multiplying by it might flip the inequality (Section 2.1) or might not. Always move everything to one side and use the sign-chart method instead.
4. Absolute Value
4.1 Definition and Basic Properties
Definition: $\lvert x \rvert = x$ if $x \geq 0$, and $\lvert x \rvert = -x$ if $x < 0$. Geometrically, $\lvert x \rvert$ is the distance from $x$ to 0 on the number line; more generally, $\lvert a - b \rvert$ is the distance between $a$ and $b$.
| Property | Statement |
|---|---|
| Non-negativity | $\lvert x \rvert \geq 0$, and $\lvert x \rvert = 0$ iff $x = 0$ |
| Multiplicativity | $\lvert ab \rvert = \lvert a \rvert \cdot \lvert b \rvert$ |
| Symmetry | $\lvert -x \rvert = \lvert x \rvert$ |
| Squares | $\lvert x \rvert^2 = x^2$ |
| Triangle inequality | $\lvert a + b \rvert \leq \lvert a \rvert + \lvert b \rvert$ |
4.2 Absolute Value Equations
Key idea: $\lvert x \rvert = k$ (for $k \geq 0$) means $x = k$ or $x = -k$. If $k < 0$, there is no solution (distance can’t be negative).
Worked Example 7: Solve $\lvert 2x - 3 \rvert = 7$.
Case 1: $2x - 3 = 7 \to 2x = 10 \to x = 5$
Case 2: $2x - 3 = -7 \to 2x = -4 \to x = -2$
Check both: $\lvert 2(5)-3 \rvert = \lvert 7 \rvert = 7$ $\checkmark$ $\qquad$ $\lvert 2(-2)-3 \rvert = \lvert -7 \rvert = 7$ $\checkmark$
Solution: $x = 5$ or $x = -2$
| **Example 4 ( | x | = k has solutions)** is Worked Example 7 above: $\lvert 2x-3 \rvert = 7$ has $k = 7 > 0$, splitting into two solvable cases and giving $x = 5$ or $x = -2$. |
| **Non-Example 4 ( | x | = k has no solutions):** Solve $\lvert x + 5 \rvert = -3$. |
Here $k = -3 < 0$. But $\lvert \text{anything} \rvert \geq 0$ always (Section 4.1, non-negativity property) — an absolute value can never equal a negative number. No case-split is even needed: the equation is asking a distance to equal $-3$, which is impossible. $\times$ No solution (solution set is $\emptyset$), regardless of what’s inside the bars.
4.3 Absolute Value Inequalities
Key rules (for $k > 0$):
| Inequality | Equivalent to |
|---|---|
| $\lvert x \rvert < k$ | $-k < x < k$ |
| $\lvert x \rvert \leq k$ | $-k \leq x \leq k$ |
| $\lvert x \rvert > k$ | $x < -k$ or $x > k$ |
| $\lvert x \rvert \geq k$ | $x \leq -k$ or $x \geq k$ |
A useful way to remember which form you get: “less than” absolute value gives one connected interval (a “betweenness” statement); “greater than” absolute value gives two separate rays (an “outside” statement).
Worked Example 8: Solve $\lvert x - 4 \rvert < 3$.
\[\lvert x - 4 \rvert < 3 \iff -3 < x - 4 < 3\]Add 4 throughout: $1 < x < 7$
Solution: $(1, 7)$
Geometric check: this says “$x$ is within distance 3 of 4” — the interval centred at 4 with radius 3 is exactly $(4-3, 4+3) = (1,7)$. Matches.
| **Example 5 ( | x | < k membership):** Does $x = 3$ satisfy $\lvert x - 4 \rvert < 3$ (solution $(1,7)$)? |
$3 \in (1,7)$? Yes. $\checkmark$ Check directly: $\lvert 3-4 \rvert = \lvert -1 \rvert = 1$. Is $1 < 3$? Yes. $\checkmark$ Confirmed ($x=3$ is within distance 3 of 4, as the interval promises).
| **Non-Example 5 ( | x | < k membership):** Does $x = 8$ satisfy $\lvert x - 4 \rvert < 3$? |
$8 \in (1,7)$? No. $\times$ Check directly: $\lvert 8-4 \rvert = \lvert 4 \rvert = 4$. Is $4 < 3$? No. $\times$ Confirmed — 8 is distance 4 from 4, which is already outside the allowed “radius 3,” so it correctly fails.
Worked Example 9: Solve $\lvert 3x + 1 \rvert \geq 5$.
\[\lvert 3x+1 \rvert \geq 5 \iff 3x+1 \leq -5 \ \text{ or } \ 3x+1 \geq 5\]Case 1: $3x + 1 \leq -5 \to 3x \leq -6 \to x \leq -2$
Case 2: $3x + 1 \geq 5 \to 3x \geq 4 \to x \geq 4/3$
Solution: $(-\infty, -2] \cup [4/3, \infty)$
| **Example 6 ( | x | > k membership):** Does $x = -5$ satisfy $\lvert 3x + 1 \rvert \geq 5$ (solution $(-\infty,-2] \cup [4/3,\infty)$)? |
$-5 \in (-\infty,-2]$? Yes. $\checkmark$ Check directly: $\lvert 3(-5)+1 \rvert = \lvert -14 \rvert = 14$. Is $14 \geq 5$? Yes. $\checkmark$ Confirmed.
| **Non-Example 6 ( | x | > k membership):** Does $x = 0$ satisfy $\lvert 3x + 1 \rvert \geq 5$? |
Is $0 \in (-\infty,-2] \cup [4/3,\infty)$? $0$ is bigger than $-2$ and smaller than $4/3$, so it falls in the GAP $(-2, 4/3)$, which is excluded. $\times$ Check directly: $\lvert 3(0)+1 \rvert = \lvert 1 \rvert = 1$. Is $1 \geq 5$? No. $\times$ Confirmed — “greater than” absolute value excludes a middle band, and 0 sits inside it.
This exact skill — converting $\lvert x - c \rvert < \delta$ into an interval around $c$ — is precisely what you’ll do when unpacking the formal $\varepsilon$–$\delta$ definition of a limit in Module I.
4.4 The Triangle Inequality
Statement: for all real $a, b$: $\lvert a + b \rvert \leq \lvert a \rvert + \lvert b \rvert$.
Worked Example 10: Verify the triangle inequality for $a = 5, b = -8$, and for $a = 3, b = 4$.
\(a=5, b=-8: \quad \lvert a+b \rvert = \lvert 5-8 \rvert = \lvert -3 \rvert = 3 \qquad \lvert a \rvert+\lvert b \rvert = 5 + 8 = 13\) $3 \leq 13$ $\checkmark$ (strict — signs differ, big cancellation)
\(a=3, b=4: \quad \lvert a+b \rvert = \lvert 7 \rvert = 7 \qquad \lvert a \rvert+\lvert b \rvert = 3+4 = 7\) $7 \leq 7$ $\checkmark$ (equality — happens exactly when $a,b$ have the same sign, or one is 0)
Example 7 (Triangle inequality — equality case): $a = 2, b = 6$ (same sign). Does $\lvert a+b \rvert = \lvert a \rvert+\lvert b \rvert$?
$\lvert a+b \rvert = \lvert 2+6 \rvert = \lvert 8 \rvert = 8$. $\lvert a \rvert+\lvert b \rvert = 2+6 = 8$. $8 = 8$. $\checkmark$ Equality holds — as predicted, $a$ and $b$ have the same sign, so no cancellation happens when adding them.
Non-Example 7 (Triangle inequality — NOT equality, only strict inequality): $a = 6, b = -2$ (opposite signs). Does $\lvert a+b \rvert = \lvert a \rvert+\lvert b \rvert$?
$\lvert a+b \rvert = \lvert 6-2 \rvert = \lvert 4 \rvert = 4$. $\lvert a \rvert+\lvert b \rvert = 6+2 = 8$. Is $4 = 8$? No. $\times$ Equality fails here — the triangle inequality still HOLDS ($4 \leq 8$ is true), but only as a strict inequality, not equality, because opposite signs cause cancellation inside $\lvert a+b \rvert$ that $\lvert a \rvert+\lvert b \rvert$ doesn’t experience.
A useful corollary, sometimes called the reverse triangle inequality: $\big\lvert \lvert a \rvert - \lvert b \rvert \big\rvert \leq \lvert a - b \rvert$. This is used constantly in limit proofs to bound how much $\lvert f(x) \rvert$ can differ from $\lvert L \rvert$ once you know how close $f(x)$ is to $L$.
5. Summary
| Concept | Key Idea | Worked Example |
|---|---|---|
| Number systems | $\mathbb{N} \subset W \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$ | $\sqrt{2}$ is irrational, not in $\mathbb{Q}$ |
| Order + sign flip | Multiplying/dividing by negative flips inequality | $-2x+3>7 \implies x<-2$ |
| Linear inequality | Solve like an equation, watch sign flips | $3x+5\leq2x-1 \implies x\leq-6$ |
| Quadratic inequality | Factor, find roots, sign chart | $x^2-x-6>0 \implies (-\infty,-2)\cup(3,\infty)$ |
| Rational inequality | Sign chart with numerator AND denominator zeros | $(x-2)/(x+1)\geq0 \implies (-\infty,-1)\cup[2,\infty)$ |
| $\lvert x \rvert = k$ | $x = \pm k$ ($k\geq0$) | $\lvert 2x-3 \rvert=7 \implies x=5$ or $-2$ |
| $\lvert x \rvert < k$ | $-k<x<k$, one interval | $\lvert x-4 \rvert<3 \implies (1,7)$ |
| $\lvert x \rvert > k$ | $x<-k$ or $x>k$, two rays | $\lvert 3x+1 \rvert\geq5 \implies (-\infty,-2]\cup[4/3,\infty)$ |
| Triangle inequality | $\lvert a+b \rvert \leq \lvert a \rvert+\lvert b \rvert$ | Equality iff same sign |
6. Practice Problems
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Classify each of the following into the smallest number system ($\mathbb{N}, W, \mathbb{Z}, \mathbb{Q}$, or irrational/$\mathbb{R}$) it belongs to: $-7, 4/9, \sqrt{16}, \sqrt{5}, 0, 2.5$.
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Solve $5 - 3x \geq 2x + 15$, and express the solution in interval notation.
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Solve $-4 < 2x - 6 \leq 10$ (a compound inequality) and express the solution in interval notation.
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Solve $x^2 - 5x + 6 < 0$ using the sign-chart method.
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Solve $x^2 + x - 12 \geq 0$.
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Solve $(2x + 1)(x - 4) > 0$.
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Solve $(x + 3)/(x - 1) < 0$, being careful about the excluded point.
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Solve $(x - 5)/(x + 2) \leq 1$. (Hint: move everything to one side and combine into a single fraction before making a sign chart — don’t cross-multiply.)
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Solve $\lvert x + 6 \rvert = 4$.
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Solve $\lvert 5 - 2x \rvert = 9$.
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Solve $\lvert x - 2 \rvert \leq 5$, and describe the solution set as “$x$ is within __ of __.”
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Solve $\lvert 4x + 3 \rvert > 9$.
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Solve $\lvert x \rvert^2 - 3\lvert x \rvert - 4 = 0$. (Hint: let $u = \lvert x \rvert$ and solve the quadratic in $u$ first, then recover $x$.)
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Verify the triangle inequality $\lvert a + b \rvert \leq \lvert a \rvert + \lvert b \rvert$ for $a = -6, b = -2$, and identify whether equality holds. Explain in one sentence why it should.
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(Harder) Solve $\lvert x - 1 \rvert + \lvert x + 2 \rvert = 5$. (Hint: consider the cases $x < -2$, $-2 \leq x < 1$, and $x \geq 1$ separately, since the expressions inside each absolute value change sign at different points.)
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(Harder) Prove that for $k > 0$, $\lvert x \rvert < k$ implies $-k < x < k$, directly from the definition of absolute value (split into the cases $x \geq 0$ and $x < 0$).