MathIsimple
Chapter 1 • Course 3
3-4 Hours

Properties of Real Numbers

Foundation Level
Bounds & Inequalities
Essential for Analysis
Learning Objectives
  • Understand and apply absolute value properties in proofs
  • Define bounded, unbounded, and finite sets precisely
  • Compute supremum and infimum of various sets
  • Apply the completeness principle to prove existence results
  • Master essential inequalities: AM-GM, Bernoulli, Cauchy-Schwarz
  • Use the Archimedean property to prove density results
  • Apply the approximation property of supremum in limit arguments
  • Recognize when equality holds in classical inequalities

Absolute Value

The distance function on the real line

Definition

x={xif x0xif x<0| x | = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

The absolute value |x| represents the distance of x from zero on the number line. It is always non-negative.

Key Properties

x0|x| \geq 0

Non-negativity: absolute value is always non-negative

x=0x=0|x| = 0 \Leftrightarrow x = 0

Zero property: only zero has absolute value zero

x=x|-x| = |x|

Symmetry: opposite numbers have equal absolute values

xy=xy|xy| = |x||y|

Multiplicativity: absolute value of product is product of absolute values

x/y=x/y|x/y| = |x|/|y|

Divisibility: absolute value of quotient (y ≠ 0)

x+yx+y|x + y| \leq |x| + |y|

Triangle inequality: fundamental for analysis

xyxy||x| - |y|| \leq |x - y|

Reverse triangle inequality

xy=yx|x - y| = |y - x|

Distance symmetry: distance is symmetric

x2=x2|x|^2 = x^2

Square relation: useful for algebraic manipulations

xxx-|x| \leq x \leq |x|

Bounds: x is bounded by its absolute value

The Triangle Inequality is Fundamental

The triangle inequality |x + y| ≤ |x| + |y| is one of the most important inequalities in analysis. It's used in proofs involving limits, continuity, and metric spaces. The name comes from the fact that in any triangle, the sum of two sides is at least as long as the third side.

Bounded Sets

Bounded Above

A set S ⊆ ℝ is bounded above if there exists M ∈ ℝ such that x ≤ M for all x ∈ S.

Example:

S = (-∞, 5) is bounded above by 5 (or any M ≥ 5)

Bounded Below

A set S ⊆ ℝ is bounded below if there exists m ∈ ℝ such that x ≥ m for all x ∈ S.

Example:

S = [3, ∞) is bounded below by 3 (or any m ≤ 3)

Bounded Sets

A set is bounded if it is both bounded above and bounded below. Equivalently:

S is boundedM>0:xS,xMS \text{ is bounded} \Leftrightarrow \exists M > 0: \forall x \in S, |x| \leq M

Common Examples

SetBounded?SupInfNotes
[0, 1]Yes1 (attained)0 (attained)Closed bounded interval
(0, 1)Yes1 (not attained)0 (not attained)Open bounded interval
{1/n : n ∈ ℕ}Yes1 (attained)0 (not attained)Sequence tending to 0
[0, ∞)Below only0 (attained)Unbounded above
No-∞Unbounded in both directions
{(-1)ⁿ/n : n ∈ ℕ}Yes1/2 (attained)-1 (attained)Alternating sequence
{r ∈ ℚ : r² < 2}Yes√2 (not in set)-√2 (not in set)ℚ lacks sup (√2 ∉ ℚ)

Supremum & Infimum

The least upper bound and greatest lower bound

Supremum (sup)

α=supS\alpha = \sup S means:

  • 1. α is an upper bound: ∀x ∈ S, x ≤ α
  • 2. α is the smallest such: if β < α, then ∃x ∈ S with x > β

Equivalent:

ε>0,xS:x>αε\forall \varepsilon > 0, \exists x \in S: x > \alpha - \varepsilon

Infimum (inf)

β=infS\beta = \inf S means:

  • 1. β is a lower bound: ∀x ∈ S, x ≥ β
  • 2. β is the largest such: if γ > β, then ∃x ∈ S with x < γ

Equivalent:

ε>0,xS:x<β+ε\forall \varepsilon > 0, \exists x \in S: x < \beta + \varepsilon

Completeness Principle (Supremum Property)

Every non-empty subset of ℝ that is bounded above has a supremum in ℝ. Every non-empty subset of ℝ that is bounded below has an infimum in ℝ.

This is the Key Property!

The completeness principle distinguishes ℝ from ℚ. It guarantees that limits exist, that continuous functions on closed intervals attain their bounds, and is the foundation of calculus. Without it, we couldn't define integrals or prove the Intermediate Value Theorem.

Equivalent Forms of Completeness

Five equivalent ways to express the completeness of ℝ

Least Upper Bound Property

Every non-empty set bounded above has a supremum

Standard definition of completeness

Monotone Convergence

Every bounded monotonic sequence converges

Essential for series convergence proofs

Nested Interval Property

Nested closed intervals with lengths → 0 have exactly one common point

Bisection method convergence

Bolzano-Weierstrass

Every bounded sequence has a convergent subsequence

Compactness arguments

Cauchy Completeness

Every Cauchy sequence converges

Metric space completeness

Heine-Borel Theorem

Closed bounded sets are compact

Covering arguments in analysis

Essential Inequalities

AM-GM Inequality

For non-negative reals a₁, a₂, ..., aₙ:

a1+a2++anna1a2ann\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}

Equality Condition:

Equality holds if and only if a₁ = a₂ = ⋯ = aₙ

Example:

For a = 3, b = 12: (3+12)/2 = 7.5 ≥ √36 = 6 ✓

Bernoulli Inequality

For x ≥ -1 and n ∈ ℕ:

(1+x)n1+nx(1 + x)^n \geq 1 + nx

Equality Condition:

Equality holds if and only if x = 0 or n = 1

Example:

For x = 0.5, n = 3: 1.5³ = 3.375 ≥ 1 + 1.5 = 2.5 ✓

Cauchy-Schwarz Inequality

For reals a₁, ..., aₙ and b₁, ..., bₙ:

(i=1naibi)2(i=1nai2)(i=1nbi2)\left(\sum_{i=1}^n a_i b_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)

Equality Condition:

Equality holds if aᵢ = k·bᵢ for some constant k

Example:

For (1,2) and (3,4): (1·3+2·4)² = 121 ≤ (1+4)(9+16) = 125 ✓

Power Mean Inequality

For positive reals, as r increases, the r-power mean increases:

minHMGMAMQMmax\min \leq \text{HM} \leq \text{GM} \leq \text{AM} \leq \text{QM} \leq \max

Equality Condition:

All means are equal if and only if all values are equal

Example:

HM ≤ GM ≤ AM generalizes to any power means

Worked Examples

Example 1: Finding Supremum and Infimum

Problem:

FindsupSandinfSwhereS={nn+1:nN}Find sup S and inf S where S = \{\frac{n}{n+1} : n \in \mathbb{N}\}

Solution:

First, list some elements: S = {1/2, 2/3, 3/4, 4/5, ...}

As n → ∞, n/(n+1) → 1, so 1 is an upper bound

For any ε > 0, choose n > 1/ε - 1, then n/(n+1) > 1 - ε

So 1 is the least upper bound: sup S = 1

Note: 1 ∉ S, so max S does not exist

The smallest element is 1/2 when n = 1

So inf S = min S = 1/2

Example 2: Proving AM-GM for Two Numbers

Problem:

Prove that for a, b ≥ 0: (a+b)/2 ≥ √(ab)

Solution:

Start with (√a - √b)² ≥ 0 (always true for reals)

Expand: a - 2√(ab) + b ≥ 0

Rearrange: a + b ≥ 2√(ab)

Divide by 2: (a+b)/2 ≥ √(ab) ∎

Equality holds when √a = √b, i.e., a = b

Example 3: Proving Bernoulli Inequality by Induction

Problem:

Prove (1+x)ⁿ ≥ 1 + nx for x ≥ -1, n ∈ ℕ

Solution:

Base case (n = 1): (1+x)¹ = 1 + x ≥ 1 + 1·x ✓

Inductive step: Assume (1+x)ᵏ ≥ 1 + kx for some k

(1+x)ᵏ⁺¹ = (1+x)ᵏ · (1+x)

≥ (1 + kx)(1 + x) by inductive hypothesis (note 1+x ≥ 0)

= 1 + x + kx + kx²

= 1 + (k+1)x + kx²

≥ 1 + (k+1)x since kx² ≥ 0 ∎

Example 4: Applying Triangle Inequality

Problem:

Prove that ||a| - |b|| ≤ |a - b|

Solution:

From triangle inequality: |a| = |(a-b) + b| ≤ |a-b| + |b|

So |a| - |b| ≤ |a-b|

Similarly: |b| = |(b-a) + a| ≤ |b-a| + |a| = |a-b| + |a|

So |b| - |a| ≤ |a-b|, i.e., -(|a| - |b|) ≤ |a-b|

Combining: -|a-b| ≤ |a| - |b| ≤ |a-b|

Therefore: ||a| - |b|| ≤ |a-b| ∎

Example 5: Using AM-GM to Minimize a Sum

Problem:

Find the minimum value of x + 4/x for x > 0

Solution:

By AM-GM inequality: (a + b)/2 ≥ √(ab) for positive a, b

Let a = x and b = 4/x

Then (x + 4/x)/2 ≥ √(x · 4/x) = √4 = 2

So x + 4/x ≥ 4

Equality holds when x = 4/x, i.e., x² = 4, x = 2 (since x > 0)

Minimum value is 4, achieved at x = 2 ∎

Example 6: Proving the Cauchy-Schwarz Inequality

Problem:

Prove that (∑aᵢbᵢ)² ≤ (∑aᵢ²)(∑bᵢ²) for real numbers

Solution:

Consider f(t) = ∑(aᵢ + tbᵢ)² ≥ 0 for all t ∈ ℝ

Expand: f(t) = ∑aᵢ² + 2t∑aᵢbᵢ + t²∑bᵢ²

Let A = ∑aᵢ², B = ∑aᵢbᵢ, C = ∑bᵢ²

So f(t) = Ct² + 2Bt + A ≥ 0

A non-negative quadratic has discriminant ≤ 0

(2B)² - 4AC ≤ 0

4B² ≤ 4AC, so B² ≤ AC

Therefore (∑aᵢbᵢ)² ≤ (∑aᵢ²)(∑bᵢ²) ∎

Example 7: Finding Infimum of an Open Set

Problem:

Find inf S and determine if it's a minimum, where S = {1/n : n ∈ ℕ, n ≥ 1}

Solution:

Elements of S: 1, 1/2, 1/3, 1/4, ...

All elements are positive, so 0 is a lower bound

For any ε > 0, by Archimedean property ∃n with 1/n < ε

So 0 is the greatest lower bound: inf S = 0

But 0 ∉ S (no n gives 1/n = 0)

Therefore inf S = 0, but min S does not exist ∎

Example 8: Applying Weighted AM-GM

Problem:

Find the maximum of x²y³ subject to 2x + 3y = 10, x,y > 0

Solution:

By AM-GM: (a₁ + a₂ + ... + aₙ)/n ≥ (a₁a₂...aₙ)^(1/n)

Write 2x + 3y = x + x + y + y + y = 10

Apply AM-GM to these 5 terms: (x + x + y + y + y)/5 ≥ (x·x·y·y·y)^(1/5)

So 2 ≥ (x²y³)^(1/5), meaning x²y³ ≤ 32

Equality when x = y, from 2x + 3y = 10 we get 5x = 10, x = 2

Maximum is 32, achieved at x = y = 2 ∎

Important Theorems

Approximation Property of Sup

If α = sup S and ε > 0, then ∃x ∈ S: α - ε < x ≤ α

Application: Used to show sequences converging to supremum exist

Monotone Convergence Property

Every bounded monotonic sequence converges

Application: Fundamental for series convergence and integration

Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence

Application: Essential for compactness arguments and analysis

Generalized AM-GM

For positive reals and weights summing to 1: weighted AM ≥ weighted GM

Application: Extends to weighted averages and probability

Power Mean Inequality

For r < s: M_r ≤ M_s where M_r = (∑xᵢʳ/n)^(1/r)

Application: Generalizes AM-GM to continuous scale of means

Young's Inequality

For a,b ≥ 0 and p,q > 1 with 1/p + 1/q = 1: ab ≤ aᵖ/p + bᵠ/q

Application: Used in proving Hölder's inequality

Triangle Inequality Generalization

|∑xᵢ| ≤ ∑|xᵢ| for any finite collection

Application: Error bounds in numerical computations

Proof Techniques

ε-δ Arguments for Supremum

To prove α = sup S, show: (1) α is an upper bound, (2) for any ε > 0, ∃x ∈ S with x > α - ε

Example: Proving sup(0,1) = 1: 1 bounds (0,1), and 1-ε/2 ∈ (0,1) for any ε > 0

Squaring Technique

Many inequalities become easier after squaring both sides (when both sides are non-negative)

Example: To prove AM ≥ GM, start with (√a - √b)² ≥ 0

Induction for Inequalities

Prove inequality for n=1, then show if true for n, it holds for n+1

Example: Bernoulli's inequality (1+x)ⁿ ≥ 1+nx is proven by induction on n

Contradiction via Archimedean Property

To disprove 'x is infinitely small', find n with 1/n < x

Example: There is no smallest positive real: for any x > 0, x/2 is smaller and positive

Telescoping Products

Rewrite products as telescoping to simplify bounds

Example: ∏(1+1/k²) telescopes nicely when written as ∏((k²+1)/k²)

Jensen's Inequality Technique

For convex f: f(∑λᵢxᵢ) ≤ ∑λᵢf(xᵢ) when ∑λᵢ = 1

Example: Proving AM-GM using convexity of -ln(x)

Lagrange Multipliers for Equality

Equality in AM-GM achieved when all variables equal

Example: min(x+y+z) s.t. xyz=1: use x=y=z=1, minimum is 3

Historical Background

Augustin-Louis Cauchy (1789-1857)

Developed rigorous definitions for limits and continuity, introduced the ε-δ formalism

His work on limits laid the foundation for modern analysis and the formal treatment of inequalities

Karl Weierstrass (1815-1897)

Perfected Cauchy's definitions and proved the Bolzano-Weierstrass theorem

Established the importance of supremum and infimum in analysis, formalized the concept of bounded sets

Bernhard Bolzano (1781-1848)

Early work on the least upper bound property and intermediate value theorem

Recognized the need for a rigorous foundation of real numbers before Dedekind and Weierstrass

Real-World Applications

Optimization

AM-GM inequality is used to find optimal dimensions in engineering and economics

Example: Minimizing surface area of a box with fixed volume: make it a cube

Statistics

Cauchy-Schwarz inequality bounds correlations and covariances

Example: Proving that correlation coefficient |ρ| ≤ 1

Physics

Triangle inequality describes the metric structure of space

Example: Proving the uncertainty principle in quantum mechanics

Computer Science

Bounds on approximations and error analysis

Example: Analyzing floating-point arithmetic errors using absolute value inequalities

Frequently Asked Questions

What's the difference between supremum and maximum?

The supremum (least upper bound) always exists for non-empty bounded-above sets of reals, but may not be an element of the set. The maximum is the largest element of the set and must belong to the set. If the maximum exists, it equals the supremum. Example: S = (0,1) has sup S = 1 but no maximum.

Why is the completeness axiom needed for supremum existence?

In ℚ, the set {x ∈ ℚ : x² < 2} is bounded above but has no supremum in ℚ (would need √2). The completeness axiom guarantees that every non-empty bounded-above set of reals HAS a supremum, which is essential for limits, continuity, and integration.

How do I determine if a set is bounded?

A set S is bounded above if ∃M: ∀x ∈ S, x ≤ M. Bounded below if ∃m: ∀x ∈ S, x ≥ m. Bounded means both. To show unboundedness, prove that for any M, you can find x ∈ S with x > M (or x < M for below).

When does equality hold in AM-GM?

Equality in AM-GM holds if and only if all numbers are equal: a₁ = a₂ = ⋯ = aₙ. This is crucial for optimization: to minimize a sum given a fixed product (or vice versa), make all terms equal.

How is Cauchy-Schwarz used in practice?

Cauchy-Schwarz is extremely versatile. It's used to prove other inequalities, bound dot products, in optimization problems, and even in probability (correlation bounds). The key insight: it measures how 'aligned' two vectors are.

What is the approximation property of supremum?

If α = sup S, then for any ε > 0, there exists x ∈ S such that α - ε < x ≤ α. This says we can always find elements of S arbitrarily close to the supremum. This property is often used in proofs involving limits.

How does the Archimedean property relate to supremum?

The Archimedean property (for any x, ∃n ∈ ℕ with n > x) follows from the completeness axiom. It's equivalent to saying ℕ is unbounded above, which prevents 'infinitely large' or 'infinitely small' real numbers from existing.

What are some applications of these inequalities?

AM-GM is used in optimization (minimizing sums with fixed products). Cauchy-Schwarz appears in statistics (covariance bounds), physics (uncertainty principles), and machine learning (cosine similarity). Bernoulli's inequality is key for proving limits involving powers.

Practice Quiz

Test your understanding of real number properties

Properties of Real Numbers Practice
10
Questions
0
Correct
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Accuracy
1
What is the supremum of the set S={xR:x<3}S = \{x \in \mathbb{R} : x < 3\}?
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2
Which property does the absolute value satisfy?
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3
For positive reals a and b, the AM-GM inequality states:
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4
A set S is bounded if and only if:
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5
The Bernoulli inequality states that for x1x \geq -1 and nNn \in \mathbb{N}:
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6
If supS=α\sup S = \alpha, which statement must be true?
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7
The reverse triangle inequality states:
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8
When does equality hold in the Cauchy-Schwarz inequality?
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9
What is the infimum of {1/n:nN,n1}\{1/n : n \in \mathbb{N}, n \geq 1\}?
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10
The completeness principle guarantees that:
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