Lecture 1: The Number Line, Absolute Values, and Inequalities

1. The Hierarchy of Real Numbers

Real numbers ($\mathbb{R}$) are built in layers. Imagine a set of nesting dolls:

Real Numbers

Rational Numbers

78
13
0.97
0.21
Integers -3 -
82
Whole Numbers
5 82 √9
0
√
169

Irrational Numbers

√8
-√11
√1.6
π
√
25
0.3030030003...
Subset Chain
$$\mathbb{N} \subseteq \mathbb{W}\subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}.$$

Interactive Snapshot: Subsets of Real Numbers

Let's review how these subset layers serve as crucial upgrades to handle complex mathematical challenges. Click each cell under 'Interactive Insight' to discover their core motivations!

Subset & Symbol Core Elements / Form Interactive Insight & Motivation
Natural Numbers ($\mathbb{N}$) $\{1, 2, 3, 4, \dots\}$
Why isn't this enough? 🤔
Perfect for elementary counting! However, if you attempt to solve a simple algebraic problem like $x + 5 = 2$, no natural number can satisfy it. To fix this gap, mathematics needed to expand!
Whole Numbers ($\mathbb{W}$) $\{0, 1, 2, 3, \dots\}$
The Power of Nothingness 💡
Adds the element $0$, providing an additive identity ($a + 0 = a$). Positional numbering systems and modern computing architecture couldn't exist without this abstract concept of zero.
Integers ($\mathbb{Z}$) $\{\dots, -2, -1, 0, 1, 2, \dots\}$
The Symmetry of Balance ⚖️
The symbol comes from the German word Zahlen ("numbers"). It introduces negative numbers (additive inverses), which let us mathematically evaluate debts, opposing directions, and elegantly resolve equations like $x + 5 = 2 \implies x = -3$.
Rational Numbers ($\mathbb{Q}$) $\left\{\dfrac{p}{q} \;\middle|\; p, q \in \mathbb{Z}, q \neq 0\right\}$
The Infinite Dense Ribbon 📈
Represented by $\mathbb{Q}$ for Quotient. Their decimal expansions either terminate ($1/4 = 0.25$) or periodically repeat ($1/3 = 0.\overline{3}$). They are beautifully dense: between any two rational numbers, an infinite number of other rationals exist!
Irrational Numbers ($\mathbb{I}$ or $\mathbb{R} \setminus \mathbb{Q}$) $\sqrt{2}, \pi, e, \text{etc.}$
A Dangerous Ancient Scandal 🌊
These numbers feature infinite, completely non-repeating decimal paths. Legend says Hippasus was thrown off a boat by his fellow Pythagoreans because his rigorous proof that $\sqrt{2}$ is irrational shattered their dogmatic philosophy that the entire universe was composed strictly of clean ratios of whole numbers!

2. Visualizing the Real Line

Every real number corresponds to exactly one point on a line. The "Origin" is $0$.

Hover over the dots to see values like $-3,\; 0,\; 2/3,\; e,\; \pi$.

3. Real Numbers as a Field

The set $\mathbb{R}$ is a field under $+$ and $\cdot$
For all $a, b, c \in \mathbb{R}$, the following properties hold:
  • Associativity: $(a + b) + c = a + (b + c)$ and $(a \cdot b) \cdot c = a \cdot (b \cdot c)$
  • Commutativity: $a + b = b + a$ and $a \cdot b = b \cdot a$
  • Identities: $\exists\, 0, 1 \in \mathbb{R}$ s.t. $a + 0 = a$ and $a \cdot 1 = a.$
  • Inverses:
    • $\forall\, a \in \mathbb{R},\; \exists\, (-a) \in \mathbb{R}$ s.t. $a + (-a) = 0.$
    • $\forall\, a \in \mathbb{R} \setminus \{0\},\; \exists\, a^{-1} \in \mathbb{R}$ s.t. $a \cdot a^{-1} = 1.$
  • Distributivity: $a \cdot (b + c) = (a \cdot b) + (a \cdot c).$

4. Fundamental Order Properties

Law of Trichotomy
Given any two real numbers $a$ and $b$, exactly one of the following holds:
$a = b \quad$ or $\quad a > b \quad$ or $\quad a < b.$
How to prove $a > b$?
Show that $a - b > 0.$

Calculus relies on the "Order Properties" of real numbers. For any $a, b, c \in \mathbb{R}$:

  • Positivity: $a > b \iff a - b > 0.$
  • Addition: If $a > b$, then $a + c > b + c$ for any $c.$
    Proof: $(a+c)-(b+c) = a-b > 0.$
  • Multiplication:
    • If $a > b$ and $c > 0$, then $ac > bc.$
      Proof: $ac - bc = c(a-b).$ Since $c > 0$ and $a - b > 0$, we get $ac - bc > 0.$
    • If $a > b$ and $c < 0$, then $ac < bc$ (the inequality flips!).
  • Powers: If $a > b > 0$, then $a^n > b^n$ for any $n \in \mathbb{N}.$
    Proof: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1}) > 0.$
Exercise: Let $a, b \in \mathbb{R}.$ Show that $ab = 0 \iff a = 0$ or $b = 0.$
($\Leftarrow$) If $a = 0$ or $b = 0$, then clearly $ab = 0.$

($\Rightarrow$) Suppose $ab = 0.$ If $a = 0$, we are done. Suppose $a \neq 0.$ We need to show $b = 0.$
Hint: Since $a \neq 0$, the inverse $a^{-1}$ exists. Multiply both sides of $ab = 0$ by $a^{-1}.$
Exercise: Let $a, b \in \mathbb{R}.$ Show that $a^2 + b^2 = 0 \iff a = 0$ and $b = 0.$
($\Leftarrow$) If $a = 0$ and $b = 0$, then clearly $a^2 + b^2 = 0.$

($\Rightarrow$) Suppose $a^2 + b^2 = 0.$ Note that $$0 = a^2 + b^2 \geq a^2 \geq 0,$$ so $a^2 = 0,$ which gives $a = 0.$ Can you complete the argument for $b = 0$?
Self Assessment Quiz

Q1: If $a > b > 0$, prove that $0 < \dfrac{1}{a} < \dfrac{1}{b}.$

Show Hint
Hint: Show that $\dfrac{1}{b} - \dfrac{1}{a} > 0.$
Show Answer
$\dfrac{1}{b} - \dfrac{1}{a} = \dfrac{a - b}{ab}.$ Since $a > b > 0$, we have $a - b > 0$ and $ab > 0$, so $\dfrac{a-b}{ab} > 0.$

Q2: If $a^2 = a$, prove that either $a = 0$ or $a = 1.$

Show Hint
Hint: Recall that if $ab = 0$, then either $a = 0$ or $b = 0.$
Show Answer
$a^2 = a \Rightarrow a^2 - a = 0 \Rightarrow a(a - 1) = 0 \Rightarrow a = 0$ or $a = 1.$

Q3: If $a>b$ and $b>c$, then $a>c$

Show Hint
Hint: Consider $a-c.$ How to bring $b$?
Show Answer
$a-c=a-b+b-c=(a-b)+(b-c)>0$ as $a-b>0$ and $b-c>0.$

Q4: Let $a,b\in \mathbb{R}$ and $a < b.$ Then there exists $c\in \mathbb{R}$ such that $a < c < b.$
Suppose we proved this result, from this can we deduce:

  • There are infinite number of real numbers between two real numbers.
  • There are infinite number of rational numbers between two rational numbers.

Show Hint
Hint: Check is $\frac{a+b}{2}$ will work?
Show Answer
It is easy to see that $\frac{a+b}{2}-a=\frac{b-a}{2}=b-\frac{a+b}{2}.$ Since $b>a,$\;\; we have\;\; $\frac{b-a}{2}>0.$
That is $a < \frac{a+b}{2} < b.$ We proved more than required. The distance between $a$ and $b$ is $b-a.$
$ \frac{a+b}{2}$ is equidistant from $a$ and $b$ which is $\frac{b-a}{2}.$

Q5: Let $a$ be a rational number and $b$ is irrational. Then

  • $a+b$ is irrational.
  • if $a\ne 0$, then $ab$ is also irrational.

Show Hint
Proof by contradiction.
Show Answer
The proof is by contradiction. Let $a+b$ is rational. Then
$a+b=\frac{m}{n}$, where $m,n\in\mathbb{Z}, n\ne 0$. So if we write
$a=\frac{p}{q}$, where $p,q\in\mathbb{Z}, q\ne 0$, then
$b=\frac{m}{n}-\frac{p}{q}$ which is false as $b$ is irrational.
Similarly we can prove $ab$ is irrational.

5. Intervals and Bounds

We use intervals to describe "neighborhoods" around points.

NotationTypeConditionVisual
$[a,\, b]$Closed$a \leq x \leq b$Solid dots at both ends
$(a,\, b)$Open$a < x < b$Hollow circles at both ends
$[a,\, \infty)$Unbounded above$x \geq a$Solid dot at $a$, arrow right
$(-\infty,\, b]$Unbounded below$x \leq b$Arrow left, solid dot at $b$
Looking Ahead to Lecture 2
An interval like $(-\infty, b]$ is bounded above by $b$ — no number in the set exceeds $b$. In Lecture 2 we will ask: among all upper bounds of a set, is there a smallest one? That is the Least Upper Bound (supremum).
Press Enter or → to advance
0 / 0