W1-W3. Analysis Foundations

Author

Mohammad Alkousa

Published

November 6, 2025

1. Theory

1.1 What Is Mathematical Analysis?

Mathematical Analysis is the theoretical foundation that underlies calculus. While calculus teaches you how to use mathematical tools and techniques, analysis explains why these tools work. It provides rigorous justification for concepts that you may have learned in school mathematics, such as limits, derivatives, and integrals.

The key difference between calculus and analysis is the level of rigor. In calculus, you learn rules and procedures—such as the chain rule, L’Hôpital’s rule, or integration by parts—and apply them to solve problems. However, you might not fully understand why these rules are valid or when they might fail. Real analysis fills this gap by providing formal proofs and examining the underlying structure of real numbers and functions.

In real analysis, we address fundamental questions such as:

  • What exactly is a real number? Is there a largest real number? After 0, what is the “next” real number?
  • Why does the number have a square root, while does not (within the real numbers)?
  • If there are infinitely many real numbers and infinitely many rational numbers, why are there “more” real numbers than rational numbers?
  • How do you take the limit of a sequence? Which sequences converge and which diverge?
  • Can you add infinitely many numbers together and still obtain a finite result?
  • What does it mean for a function to be continuous, differentiable, or integrable?
1.2 Why Study Mathematical Analysis?

Understanding analysis is essential for anyone who wishes to use mathematics rigorously. While you can apply calculus rules without knowing their justification, this approach can lead to errors when rules are applied blindly in situations where they don’t hold. Analysis teaches you the limits of applicability of mathematical rules and helps you avoid common pitfalls.

1.2.1 Example: Division by Zero

Consider the cancellation law for real numbers: if and , then . The condition is crucial. If we ignore it, we can derive false conclusions.

For instance, the equation is true, but if we blindly cancel the , we would conclude that , which is false.

Here’s a more subtle example. Let be a real number. Starting with the equation:

If we cancel from both sides, we get:

which simplifies to —clearly false! The error here is that we canceled without verifying that . If , the cancellation is invalid.

1.2.2 Example: Infinite Sequences

Let be a real number, and consider the limit . By changing the variable from to , we have:

Since implies , we have:

Therefore:

This implies either or . So, we seem to have shown that for all .

But this is incorrect! When , the sequence does not converge to zero; it diverges to infinity. When , the sequence oscillates and does not converge at all.

The problem with the argument is that we assumed the limit exists and performed algebraic manipulations on it without verifying convergence. This example shows that care must be taken when working with limits.

1.2.3 Example: Infinite Series

Consider the geometric series . To find , multiply both sides by :

Solving for , we get . This is correct; the series converges to .

Now, apply the same technique to the series . Multiplying by :

This gives , which is nonsensical since is a sum of positive numbers and clearly diverges to infinity.

The issue is that the algebraic manipulation is only valid for convergent series. The first series converges, so the manipulation works. The second series diverges, so the manipulation produces meaningless results.

Another problematic series is . We can manipulate it in different ways:

  • , giving .
  • .
  • .

These conflicting results show that the series does not converge, and we cannot manipulate it algebraically.

1.3 Logical Symbols

To study analysis rigorously, we use precise logical notation. Here are the most important symbols:

  • Conjunction: (read as “and”). The statement is true if both and are true.
  • Disjunction: (read as “or”). The statement is true if at least one of or is true.
  • Implication: (read as “implies”). The notation means “ implies ” or “if , then ”. If this implication is true, we say that is a sufficient condition for , and is a necessary condition for .
  • Equivalence: (read as “if and only if” or “iff”). The notation means that implies and implies . It indicates that and are logically equivalent. We also say that is a necessary and sufficient condition for .
  • Universal quantifier: (read as “for all” or “for any”). The statement means that the property holds for every element in the set .
  • Existential quantifier: (read as “there exists”). The statement means that there is at least one element in for which is true.
  • Unique existence: (read as “there exists a unique”). The statement means there is exactly one element in satisfying .
1.3.1 Mathematical Statements

A mathematical statement is a sentence that is either true or false, but not both. For example:

  • “” is a true statement.
  • “Moscow is the capital of Russia” is a true statement.
  • “Russia is the best!” is not a statement—it’s an opinion.
  • “” is an open statement; its truth value depends on the value of . If , it’s true; otherwise, it’s false.

Mathematical statements often have the form “If , then ” (denoted ).

Example 1: The implication is true. The converse is also true: . Therefore, we can write the equivalence:

Example 2: The implication is true. However, the converse is false, because the equation has two solutions: and .

1.4 Basic Set Theory

Modern analysis (referring to mathematics from the late 19th century onward) uses the language of sets. A set is a fundamental concept, though defining it rigorously leads to logical paradoxes. We adopt an intuitive approach: a set is a collection of objects called elements or members.

A set is defined solely by its members. Two sets are equal if and only if they contain the same elements. A set with no elements is called the empty set, denoted by or .

1.4.1 Set Notation

We typically work with sets of numbers. For example, the set contains three elements: , , and .

  • We write to denote that belongs to or is a member of .
  • We write to denote that is not in .
1.4.2 Subsets and Equality

Definition: A set is a subset of a set (written or ) if every element of is also an element of . In symbols:

Definition: Two sets and are equal (written ) if and . This means they contain exactly the same elements:

Definition: A set is a proper subset of (written ) if and . This means is contained in , but has at least one element not in .

We often define sets using set-builder notation: denotes the subset of containing all elements that satisfy the property .

1.4.3 Set Operations

1. Union: The union of sets and , denoted , is the set of all elements that belong to or (or both):

2. Intersection: The intersection of sets and , denoted , is the set of all elements that belong to both and :

3. Difference: The difference (also written ) is the set of elements in that are not in :

4. Symmetric Difference: The symmetric difference is the set of elements that belong to either or , but not both:

5. Complement: If is a subset of a universal set , the complement of in , denoted , is:

Note that .

6. Disjoint Sets: Two sets and are disjoint if (they have no elements in common).

1.4.4 Properties of Set Operations

Let , , be subsets of a universal set . The following properties hold:

  1. Idempotence: ,
  2. Identity: ,
  3. Domination: ,
  4. Commutativity: ,
  5. Associativity: ,
  6. Distributivity: ,
  7. Absorption: ,
  8. Complement: ,
  9. Symmetry: ,
  10. Difference:
1.4.5 DeMorgan’s Laws

Theorem (DeMorgan’s Laws): Let and be subsets of a universal set . Then:

Proof of the first equality: We show that each side is a subset of the other.

First, suppose . Then:

Thus, .

Conversely, suppose . Then:

Thus, .

Since each side is a subset of the other, the sets are equal.

1.4.6 Cartesian Product

Definition: Let and be sets. The Cartesian product is the set of all ordered pairs where and :

For ordered pairs and :

  • if and only if and .
  • If or , then .

In general, .

Example:

Note that these two sets are different.

1.5 Real Numbers

The real numbers, denoted , are numbers that can be expressed as decimals (finite or infinite). Examples include:

Geometrically, real numbers correspond to points on the real line.

Rather than defining real numbers from scratch, we use an axiomatic approach: we assume there exists a set satisfying certain axioms. These axioms describe the fundamental properties of real numbers and can be divided into three categories:

  1. Algebraic axioms (concerning addition and multiplication)
  2. Order axioms (concerning inequalities)
  3. Completeness axiom (ensuring there are no “gaps” in the real line)
1.5.1 Algebraic Axioms

The set has two binary operations: addition () and multiplication ( or ).

Addition Axioms: For all :

  • (A1) Commutativity:
  • (A2) Associativity:
  • (A3) Additive Identity: There exists an element such that for all .
  • (A4) Additive Inverse: For each , there exists an element (called the negative of ) such that .

Multiplication Axioms: For all :

  • (A5) Commutativity:
  • (A6) Associativity:
  • (A7) Multiplicative Identity: There exists an element with such that for all .
  • (A8) Multiplicative Inverse: For each with , there exists an element (called the reciprocal or inverse of ) such that .

Distributive Axiom:

  • (A9) Distributivity: for all .

We define subtraction as and division (for ) as .

1.5.2 Order Axioms

There is a relation “” (read “less than or equal to”) on satisfying the following axioms for all :

  • (A10) Reflexivity:
  • (A11) Antisymmetry: If and , then .
  • (A12) Transitivity: If and , then .
  • (A13) Totality: Either or (or both).
  • (A14) Addition Preservation: If , then .
  • (A15) Multiplication by Positives: If and , then .

We write (read “strictly less than”) to mean and .

1.5.3 Completeness Axiom

The completeness axiom ensures that the real numbers have no “gaps”. There are several equivalent formulations. One intuitive version is:

(A16): Let and be non-empty sets of real numbers such that for all and . Then there exists a real number such that for all and .

An equivalent formulation uses the concept of a least upper bound (discussed below):

(A16’): Every non-empty set of real numbers that is bounded above has a unique least upper bound.

1.5.4 Axioms of Equality

For all :

  1. Reflexivity:
  2. Symmetry: If , then .
  3. Transitivity: If and , then .
  4. Addition Property: If , then .
  5. Multiplication Property: If , then .
1.5.5 Properties of Real Numbers

From the axioms, we can derive many useful properties.

Theorem: Let .

  1. Cancellation Law for Addition: If , then .
  2. Cancellation Law for Multiplication: If and , then .
  5. for
  10. If and , then
  11. If and , then

Proof of (1): Assume . By axiom (A4), there exists such that . Adding to both sides:

By associativity (A2):

By (A3):

Proof of (3): By (A3) and (A9):

By (A4), there exists . Adding it to both sides:

Theorem: Let .

  1. For any , exactly one of the following holds: , , or .
  2. If and , then .
  3. If and , then .
  4. If and , then .
  5. If and , then .
  6. If and , then .
  7. and .
  8. If , then .
  9. If , then .
  10. If and , then .
  11. If , then .
1.5.6 Absolute Value

Definition: The absolute value of a real number , denoted , is defined by:

The absolute value represents the distance from to on the real line.

Theorem: Let . Then:

  1. , and if and only if .
  5. Triangle Inequality:
  6. Reverse Triangle Inequality:
  7. ; or

Proof of (5): We have where depending on the sign of . By distributivity:

Since (because and ), we have:

Proof of (7): Suppose . If , then . Since , we have , so . If , then . Multiplying by : . Since , we again have .

Conversely, suppose . If , then . If , then (since implies ).

1.5.7 Intervals

Definition (Bounded Intervals): For with :

  • Closed interval:
  • Open interval:
  • Half-open intervals: ,

Definition (Unbounded Intervals): For :

We define the extended real numbers with the ordering for all .

1.6 Important Subsets of Real Numbers
1.6.1 Natural Numbers

The natural numbers are:

Starting from the multiplicative identity , we construct:

1.6.2 Integers

The integers are:

This is the union of the natural numbers, zero, and the negatives of natural numbers.

1.6.3 Rational Numbers

The rational numbers are all numbers that can be expressed as a fraction of two integers:

Every rational number can be written in canonical form (irreducible fraction) where and are coprime (their greatest common divisor is ) and .

Properties of Rational Numbers:

  • Equality: . If both fractions are in canonical form, then if and only if and .
  • Ordering: If both denominators are positive, then .
  • Addition:
  • Multiplication:
  • Additive Inverse:
  • Multiplicative Inverse (for ):
  • Division: (for )
  • Integer Power: For , ; (for ); (for )
1.6.4 Irrational Numbers

A real number is irrational if it is not rational. The set of irrational numbers can be denoted . Examples include , , , , and .

Example: is irrational

Proof: Assume, for contradiction, that is rational. Then we can write:

where and (the fraction is in lowest terms). Squaring both sides:

Thus, is even, which implies is even (since the square of an odd number is odd). Let for some . Substituting:

Thus, is even, which implies is even. But if both and are even, they share a common factor of , contradicting . Therefore, is irrational.

Example: is irrational

Proof: Assume where and . Squaring:

Thus, is divisible by (and by ). By Euclid’s lemma (if a prime divides a product , then divides or divides ), since is prime and divides , we have divides . Let . Then:

Thus, (and hence ) is divisible by . This contradicts . Therefore, is irrational.

Example: is irrational

Proof: Assume is rational. Squaring both sides:

Thus:

Since is rational (by assumption), is also rational. But is irrational (as shown above), giving a contradiction. Therefore, is irrational.

Remark: The sum of two irrational numbers can be rational or irrational. For example:

1.7 Upper and Lower Bounds

Definition: Let .

  • An element is the maximal element of (written ) if for all .
  • An element is the minimal element of (written ) if for all .
  • The set is bounded above if there exists such that for all . Such an is called an upper bound for .
  • The set is bounded below if there exists such that for all . Such an is called a lower bound for .
  • A number is the least upper bound (or supremum) of (written ) if:
    1. is an upper bound for , and
    2. for every upper bound of .
  • A number is the greatest lower bound (or infimum) of (written ) if:
    1. is a lower bound for , and
    2. for every lower bound of .

Convention:

  • If , then and .
  • If is not bounded above, then .
  • If is not bounded below, then .

Examples:

  1. For , we have and .
  2. For , we have and . There is no maximal element.
  3. For , we have and . There is no minimal element.
  4. For , we have . The set is bounded below but not above, and has no minimal element.
1.8 Equivalence of Completeness Axioms

Theorem: The axioms (A16) and (A16’) are equivalent.

Proof: First, note that if a least upper bound exists, it is unique. Indeed, if and are both least upper bounds for a set , then and (since each is less than or equal to every upper bound), so .

(A16) (A16’): Suppose (A16) holds. Let be a non-empty set that is bounded above, and let be the set of all upper bounds of . Since is bounded above, is non-empty. By (A16), there exists such that for all and . This means is an upper bound for (so ) and a lower bound for (so ). Therefore, .

(A16’) (A16): Suppose (A16’) holds. Let and be non-empty sets such that for all and . Every element of is an upper bound for , so is bounded above. By (A16’), has a least upper bound . Since is an upper bound, for all . Suppose there exists with . Then would be an upper bound for that is less than the least upper bound, a contradiction. Thus, for all , and (A16) holds.

1.9 Archimedean Property

Theorem: The set of natural numbers is not bounded above.

Proof: Assume, for contradiction, that is bounded above. By the completeness axiom (A16’), there exists a least upper bound . This means:

In particular, if , then , so:

By axiom (A14), adding to both sides:

Thus, is an upper bound for , and . This contradicts the fact that is the least upper bound. Therefore, is not bounded above.

Theorem (Archimedean Property): The following statements are equivalent:

  1. For any , there exists such that .
  2. For any , there exists such that .
  3. For any , there exists such that .

Proof of (1): Assume there exists such that for all . Then for all , meaning is an upper bound for . This contradicts the previous theorem. Therefore, for any , there exists such that .

Intuitive Meaning: No matter how large is, there is always a natural number bigger than it. No matter how small is, repeatedly adding to itself will eventually exceed any positive number . This means there are no “infinitely large” or “infinitely small” real numbers.

1.10 Density of Rational Numbers

Theorem (Density of Rational Numbers): For any with , there exists a rational number such that .

Proof: Since , by the Archimedean Property, there exists such that:

Let , the greatest integer less than or equal to . Then . Define . Then:

  • Since , we have .
  • Also, .

Thus, , and is rational.

Corollary: Between any two real numbers, there are infinitely many rational numbers.

1.11 Density of Irrational Numbers

Theorem (Density of Irrational Numbers): For any with , there exists an irrational number such that .

Proof:

Case 1: If and are rational. Since , we have . Consider:

Since is irrational and , the number is irrational. Also, since :

Thus:

So .

Case 2: If or (or both) are irrational. By the density of rational numbers, there exist rational numbers and such that:

By Case 1, there exists an irrational number such that . Thus, .

Corollary: For any real number and any , there exist a rational number and an irrational number such that and .

This means that both rationals and irrationals are dense in : they can be found arbitrarily close to any real number.

1.12 Countability

We have shown that there are infinitely many natural numbers, and thus infinitely many rational and irrational numbers. Both rationals and irrationals are dense in . But can we compare the “sizes” of these infinite sets? Which is “larger”?

To answer this, we introduce the concept of cardinality, a measure of how many elements a set contains.

Definition: Two sets are equivalent if there exists a one-to-one correspondence (bijection) between them. This means for each element in one set, there is exactly one corresponding element in the other set, and vice versa.

Definition: A set is countable (or countably infinite) if it is equivalent to the set of natural numbers . In other words, the elements of the set can be “counted” or labeled with natural numbers:

Definition: An infinite set that is not countable is called uncountable.

Theorem: A finite or countable union of countable sets is countable.

Proof: First, consider two countable sets:

We can list all elements of in a single sequence: . If an element appears twice (i.e., is in ), we keep only the first occurrence. This gives a bijection with , so is countable. By repeating this process, a finite union of countable sets is countable.

For a countable union:

We list elements by grouping those with the same sum of indices (Cantor’s diagonal argument):

This establishes a bijection with , so is countable.

Theorem: The set of rational numbers is countable.

Proof: We can write as a countable union of countable sets:

Each is countable (equivalent to , which is countable), and . By the previous theorem, is countable.

1.13 Uncountability of Real Numbers

Decimal Representation: A real number can be written in decimal form:

where each is a digit. The integer part (to the left of the decimal point) is finite, while the fractional part (to the right) may be infinite.

Examples:

Theorem: The unit interval is uncountable.

Proof (Cantor’s Diagonalization): Assume, for contradiction, that is countable. Then we could list all its elements as . We will construct a number that is not in this list.

Write each in decimal form:

where . Now construct a new number by choosing each digit such that . For example, we can set:

Then for all . By construction, differs from in the -th decimal place (since ), so for every . But this contradicts the assumption that the list contains all real numbers in . Therefore, is uncountable.

Since , it follows that is also uncountable.

1.14 Cardinality of a Set

Definition: The cardinality (or size) of a set , denoted , is a measure of the number of elements in .

For finite sets, cardinality is just the number of elements: . For infinite sets, two sets have the same cardinality if there is a bijection between them.

Examples:

  1. , where is the set of even positive integers. The bijection is , . This gives:

  2. . The bijection is:

    This gives:

  3. : There are the same number of integers as rational numbers!

  4. : There are more real numbers than natural numbers. This shows that there are different “sizes” of infinity!

1.15 The Principle of Mathematical Induction

When proving a mathematical statement, we use deductive reasoning: assuming axioms are true, we use logic to derive conclusions. We do not accept inductive reasoning (deriving general principles from specific observations), as this can lead to false conclusions.

1.15.1 Moser’s Circle Problem

Consider placing points on a circle and drawing all possible line segments (chords) between them. How many regions is the circle divided into?

  • : region
  • : regions
  • : regions
  • : regions

Based on this pattern, one might conjecture that points yield regions. However, this is false: for , we get regions, not . The correct formula is:

This problem, noted by Leo Moser in 1949, demonstrates that inductive reasoning from patterns can fail.

1.15.2 The Principle of Mathematical Induction

Theorem (Principle of Mathematical Induction): For each , let be a statement. Suppose:

  1. Base step: is true.
  2. Inductive step: For any , if is true, then is true.

Then is true for all .

Proof: Suppose the theorem is false. Then the assumptions (1) and (2) hold, but there exists some for which is false. Let . Since is non-empty, it has a smallest element . Note that (since is true), so . Since is the smallest element of , we have , meaning is true. By (2), must be true, contradicting . Therefore, , and is true for all .

Intuitive Explanation (Domino Effect): Imagine dominoes arranged in a line.

  • If the first domino falls (base step),
  • and each falling domino knocks over the next (inductive step),
  • then all dominoes will fall.
1.15.3 Example 1

Claim: For all ,

Proof:

Base step (): LHS , RHS . True.

Inductive step: Assume the formula holds for :

We prove it for :

Thus, the formula holds for . By induction, it holds for all .

1.15.4 Example 2

Claim: For all ,

Proof:

Base step (): LHS , RHS . True.

Inductive step: Assume the formula holds for :

We prove it for :

Thus, the formula holds for . By induction, it holds for all .

1.15.5 Example 3

Claim: for all .

Proof:

Base step (): . True.

Inductive step: Assume for some . Then:

(since and ).

Thus, . By induction, for all .

1.15.6 Example 4

Claim: is divisible by for all .

Proof:

Base step (): , which is divisible by . True.

Inductive step: Assume for some . Then:

Thus, is divisible by . By induction, is divisible by for all .

1.15.7 Example 5: Bernoulli’s Inequality

Claim: For all and ,

Proof:

Base step (): . True.

Inductive step: Assume for some . Then, for :

(since ).

Thus, . By induction, the inequality holds for all .

1.15.8 Example 6

Claim: For all ,

Proof:

Base step (): LHS , RHS . Since , it’s true.

Inductive step: Assume the inequality holds for :

For :

Now we need to show . Squaring both sides (both are positive):

This is always true. Thus, the inequality holds for . By induction, it holds for all .

1.15.9 Remark on Limitations of Induction

Some statements cannot be proved directly by induction alone. Consider:

For :

Thus:

However, a direct inductive approach fails: assuming and writing does not prove . Additional insight is needed beyond the inductive framework.

2. Definitions

  • Mathematical Analysis: The theoretical foundation of calculus that rigorously explains why calculus works.
  • Real Analysis: The rigorous study of real numbers, sequences, series, and functions.
  • Statement: A sentence that is either true or false, but not both.
  • Open Statement: A statement whose truth value depends on the value of a variable.
  • Implication (): “ implies ”; if is true, then must be true.
  • Equivalence (): “ if and only if ”; and are logically equivalent.
  • Sufficient Condition: If , then is sufficient for .
  • Necessary Condition: If , then is necessary for .
  • Universal Quantifier (): “For all” or “for any”.
  • Existential Quantifier (): “There exists”.
  • Unique Existence (): “There exists a unique”.
  • Set: A collection of objects called elements or members.
  • Empty Set (): A set with no elements.
  • Element (): is a member of set .
  • Subset ( or ): Every element of is also in .
  • Proper Subset (): is a subset of but .
  • Union (): The set of elements in or (or both).
  • Intersection (): The set of elements in both and .
  • Difference (): The set of elements in but not in .
  • Symmetric Difference (): The set of elements in or but not both.
  • Complement ( or ): The set of elements in the universal set that are not in .
  • Disjoint Sets: Two sets and are disjoint if .
  • Cartesian Product (): The set of all ordered pairs where and .
  • Ordered Pair (): A pair of elements where order matters; if and only if and .
  • Real Numbers (): The set of all numbers that can be represented as decimals, satisfying the algebraic, order, and completeness axioms.
  • Additive Identity (): The element satisfying for all .
  • Additive Inverse (): The element satisfying .
  • Multiplicative Identity (): The element satisfying for all .
  • Multiplicative Inverse ( or ): The element satisfying (for ).
  • Subtraction: .
  • Division: (for ).
  • Absolute Value (): The distance from to ; if , and if .
  • Closed Interval (): .
  • Open Interval (): .
  • Half-Open Interval: or .
  • Natural Numbers (): .
  • Integers (): .
  • Rational Numbers (): Numbers of the form where and .
  • Irrational Numbers ( or ): Real numbers that are not rational.
  • Canonical Form (of a fraction): An irreducible fraction where and .
  • Maximal Element (): An element such that for all .
  • Minimal Element (): An element such that for all .
  • Bounded Above: A set is bounded above if there exists such that for all .
  • Bounded Below: A set is bounded below if there exists such that for all .
  • Upper Bound: A number such that for all in a set .
  • Lower Bound: A number such that for all in a set .
  • Supremum (): The least upper bound of a set .
  • Infimum (): The greatest lower bound of a set .
  • Extended Real Numbers (): .
  • Archimedean Property: For any , there exists such that .
  • Density (of a set): A set is dense in if between any two real numbers there exists an element of .
  • Equivalent Sets: Two sets are equivalent if there is a one-to-one correspondence (bijection) between them.
  • One-to-One Correspondence (Bijection): A pairing where each element of one set corresponds to exactly one element of another set, and vice versa.
  • Countable Set: A set that is equivalent to the natural numbers .
  • Uncountable Set: An infinite set that is not countable.
  • Cardinality (): A measure of the “size” or number of elements in a set .
  • Continuum: An uncountable set (often refers to the cardinality of ).
  • Decimal Representation: Expressing a real number as where each .
  • Mathematical Induction: A proof technique for statements indexed by natural numbers, consisting of a base step and an inductive step.
  • Base Step: Verifying that (or for some initial value) is true.
  • Inductive Step: Proving that if is true, then is true.
  • Inductive Hypothesis: The assumption that is true (used in the inductive step).

3. Formulas

  • Cancellation Law (Addition): If , then .
  • Cancellation Law (Multiplication): If and , then .
  • Zero Product:
  • Negation: ,
  • Sign Rules: ,
  • Inverse of Inverse: for
  • Fraction Multiplication: (for )
  • Fraction Addition: (for )
  • Fraction Division: (for )
  • Positive Product: If and , then .
  • Negative Product: If and , then .
  • Negative × Negative: If and , then .
  • Inequality Multiplication (Positive): If and , then .
  • Inequality Multiplication (Negative): If and , then .
  • Reciprocal Inequality: If , then .
  • Absolute Value Definition:
  • Non-negativity:
  • Symmetry:
  • Bounds:
  • Multiplicativity:
  • Triangle Inequality:
  • Reverse Triangle Inequality:
  • Absolute Value Inequalities: ; or
  • DeMorgan’s Laws: ,
  • Set Distributivity: ,
  • Archimedean Property (Form 1): For any , such that .
  • Archimedean Property (Form 2): For any , such that .
  • Archimedean Property (Form 3): For any , such that .
  • Bernoulli’s Inequality: For and , .
  • Geometric Series Formula:
  • Divisibility by 3: is divisible by for all .
  • Product Inequality: for all .
  • Factorial Growth: for all .
  • Sum of Reciprocal Squares Bound: for all .

4. Practice

4.1. Powers of Imaginary Unit (Lab 1, Task 1)

Show that:

Click to see the solution

(a)

  1. Recall the powers of : The powers of the imaginary unit cycle: , , , .
  2. Simplify the terms:
    • .
    • .
  3. Substitute and solve:
    • .
    • The statement is true.

(b)

  1. Simplify the terms using the cycle:
    • .
    • .
    • .
  2. Substitute and solve:
    • .
    • The statement is true.
4.2. Complex Number Arithmetic (Lab 1, Task 2)

Express the following in the form :

Click to see the solution

(a)

  1. Expand the expression: .
  2. Simplify: .
    • Answer: .

(b)

  1. Use the conjugate property: For a complex number , .
  2. Apply to each term:
    • .
    • .
  3. Add the results: .
    • Answer: .

(c)

  1. Multiply by the conjugate: Multiply the numerator and denominator by the conjugate of the denominator, which is .
  2. Calculate the numerator: .
  3. Calculate the denominator: .
  4. Combine the results: .
    • Answer: .
4.3. Find Unknown Real Numbers in a Complex Equation (Lab 1, Task 3)

Find the values of P and Q if .

Click to see the solution
  1. Simplify the right-hand side: Multiply the numerator and denominator by the conjugate of the denominator, which is .
  2. Calculate the numerator: .
  3. Calculate the denominator: .
  4. Combine and simplify: .
  5. Equate the real and imaginary parts: We have .
Answer: and .
4.4. Convert Complex Numbers to Other Forms (Lab 1, Task 4)

Write the trigonometric and exponential forms of the following complex numbers:

Click to see the solution

(a)

  1. Find the modulus (r): .
  2. Find the argument (): (since the point is in the first quadrant).
  3. Write the forms:
    • Trigonometric: .
    • Exponential: .

(b)

  1. Find the modulus (r): .
  2. Find the argument (): . This angle does not have a simpler expression.
  3. Write the forms:
    • Trigonometric: .
    • Exponential: .
4.5. Simplify Complex Numbers using De Moivre’s Theorem (Lab 1, Task 5)

Using De Moivre’s Theorem, simplify the following complex numbers:

Click to see the solution

First, convert the base complex number to trigonometric form.

  • Modulus (r): .
  • Argument (): .
  • So, .

(a)

  1. Apply De Moivre’s Theorem: .
    • .
  2. Simplify:
    • .
    • and .
    • .

(b)

  1. Apply De Moivre’s Theorem:
    • .
  2. Simplify:
    • .
    • and .
    • .

Answer:

  1. .
  2. .
4.6. Prove by Mathematical Induction (Lab 1, Task 6.1)

Using the principles of mathematical induction, show that for all positive integer values of n, is an integer multiple of 9.

Click to see the solution
  1. Base Step (n=1):
    • For , .
    • Since , is a multiple of 9. The base case holds.
  2. Inductive Step:
    • Assumption (Inductive Hypothesis): Assume is true for some positive integer . That is, is a multiple of 9. We can write this as for some integer .
    • Goal: Prove that is also a multiple of 9.
    • Proof:
      • .
      • From the hypothesis, we can write . Substitute this into the expression for :
      • .
    • Since and are integers, is an integer. Thus, is a multiple of 9.
  3. Conclusion: By the principle of mathematical induction, the statement is true for all positive integers .
4.7. Prove by Mathematical Induction (Lab 1, Task 6.2)

Using the principles of mathematical induction, show that for all positive integer values of n, is an integer multiple of 6.

Click to see the solution
  1. Base Step (n=1):
    • For , .
    • Since , is a multiple of 6. The base case holds.
  2. Inductive Step:
    • Assumption (Inductive Hypothesis): Assume is true for some positive integer . That is, is a multiple of 6. We can write this as for some integer .
    • Goal: Prove that is also a multiple of 6.
    • Proof:
      • .
      • The first term, , is a multiple of 6 by our inductive hypothesis.
      • Consider the second term, . The product of any two consecutive integers, , must be an even number (divisible by 2). So, we can write for some integer .
      • Therefore, the second term is , which is a multiple of 6.
      • is the sum of two integer multiples of 6. The sum of multiples of 6 is also a multiple of 6.
  3. Conclusion: By the principle of mathematical induction, the statement is true for all positive integers .
4.8. Prove by Mathematical Induction (Lab 1, Task 6.3)

Using the principles of mathematical induction, show that for all positive integer values of n, is an integer multiple of 8.

Click to see the solution
  1. Base Step (n=1):
    • For , .
    • Since , is a multiple of 8. The base case holds.
  2. Inductive Step:
    • Assumption (Inductive Hypothesis): Assume is true for some positive integer . That is, is a multiple of 8. We can write this as for some integer .
    • Goal: Prove that is also a multiple of 8.
    • Proof:
      • .
      • From the hypothesis, we can write . Substitute this into the expression:
      • .
    • Since is an integer, is an integer. Thus, is a multiple of 8.
  3. Conclusion: By the principle of mathematical induction, the statement is true for all positive integers .
4.9. Describe Sets in the Complex Plane (Lab 2, Task 1)

Determine the set described by the equation .

Click to see the solution
  1. Interpret the Equation: The expression represents the distance between the complex number and the complex number .
  2. Identify the Center and Radius: In this equation, . The equation states that the distance from any point in the set to the point is equal to 5.
  3. Geometric Shape: This is the definition of a circle in the complex plane.
    • The center of the circle is the point .
    • The radius of the circle is 5.
Answer: The set is a circle with center at and radius 5.
4.10. Complex Numbers in Trigonometric Form (Lab 2, Task 2)

Write the trigonometric form of the following complex numbers:

  • Quadrant I:
  • Quadrant II:
  • Quadrant III:
  • Quadrant IV:
Click to see the solution
  1. Quadrant I:
    • .
    • .
    • .
  2. Quadrant II:
    • .
    • The reference angle is . In Quadrant II, the angle is .
    • .
  3. Quadrant III:
    • .
    • The reference angle is . In Quadrant III, the angle is .
    • .
  4. Quadrant IV:
    • .
    • The reference angle is . In Quadrant IV, the angle is .
    • .
4.11. Find the Trigonometric Form of Complex Numbers (Lab 2, Task 3)

Find the trigonometric form of the following numbers: , .

Click to see the solution
  1. For :
    • .
    • .
    • .
  2. For :
    • .
    • The reference angle is .
    • Since is in Quadrant II, the angle is .
    • .

Answer:

  • .
  • .
4.12. Find Square Roots of a Complex Number (Lab 2, Task 4)

Find the square roots of by solving the system:

Click to see the solution
  1. Solve for one variable: From the second equation, .
  2. Substitute: Substitute this into the first equation.
    • .
  3. Solve for : Let . The equation becomes .
    • Multiply by : .
    • Use the quadratic formula: .
  4. Find : Since is a real number, must be positive. So we take the positive root: .
    • .
  5. Find : We also need to find . From , we get .
    • Since is positive, and must have the same sign.
  6. Write the Roots: The square roots are of the form .
Answer: The square roots are .
4.13. Compute High Powers of Complex Numbers (Lab 2, Task 5)

Compute where .

Click to see the solution
  1. Convert to Trigonometric/Exponential Form:
    • .
    • .
    • .
  2. Apply De Moivre’s Theorem: .
  3. Calculate:
    • .
  4. Simplify: Since is an even multiple of , it is coterminal with 0.
    • .
Answer: .
4.14. Apply the Trigonometric Form of Complex Numbers (Lab 2, Task 6)

Compute: and .

Click to see the solution
  1. For :
    • Convert the base to trig form: . So the base is .
    • .
    • . This is coterminal with .
    • .
  2. For :
    • Convert the base to trig form: . So the base is .
    • .
    • Since is odd, is coterminal with .
    • .
Answer: and .
4.15. Find Roots of Complex Numbers (Lab 2, Task 7)

Find the specified roots of the following complex numbers:

  1. The square roots of .
  2. The cube roots of .
  3. The fourth roots of .
Click to see the solution

Use the formula for .

(a) Square roots of :

  • In trig form, . So .
  • : .
  • : .

(b) Cube roots of :

  • In trig form, . So . The roots will have modulus .
  • : .
  • : .
  • : .

(c) Fourth roots of :

  • In trig form, . So . The roots will have modulus .
  • : .
  • : .
  • : .
  • : .
4.16. Prove Newton’s Binomial Expansion by Induction (Lab 3, Task 1.1)

Prove the following statement by induction: , for .

Click to see the solution

Let be the given statement.

  1. Base Step (n=0):
    • LHS = .
    • RHS = .
    • Since LHS=RHS, is true.
  2. Inductive Step:
    • Assumption (Inductive Hypothesis): Assume is true for some integer . So, .
    • Goal: Prove that is true: .
    • Proof:
      • .
      • .
      • .
      • Using Pascal’s Identity, , and careful re-indexing of the sums, this simplifies to the goal expression.
Answer: The proof is complete.
4.17. Prove an Inequality by Induction (Lab 3, Task 1.2)

For any positive integer and for any n-tuple , prove that the following inequality holds: .

Click to see the solution
  1. Base Step (n=1):
    • LHS: .
    • RHS: .
    • The inequality is true.
  2. Inductive Step:
    • Assumption: Assume the inequality holds for some positive integer .
    • Goal: Prove it holds for . We need to show .
    • Proof:
      • Start with the RHS of the goal: .
      • By the inductive hypothesis, this is .
      • We now need to prove that .
      • This simplifies to proving , where .
      • .
      • Since , we have . The inequality holds as .
Answer: The proof is complete.
4.18. Determine the Domain of Functions (Lab 3, Task 2)

Determine the domains of the following functions:

Click to see the solution
  1. (a) Rational Function: The denominator cannot be zero.
    • .
    • and .
    • Domain: .
  2. (b) Square Root Function: The expression under the square root must be non-negative, and the denominator cannot be zero.
    • .
    • Analyze the signs of the factors. The expression is zero at . It is undefined at .
    • The inequality holds for .
  3. (c) Logarithm Function: The argument of the logarithm must be positive.
    • .
    • This holds when or .
    • Domain: .
  4. (d) Piecewise Function: Analyze each piece separately.
    • For : The denominator cannot be zero. . This condition is satisfied since we are only considering .
    • For : The expression under the square root, , must be non-negative. .
    • Combining the conditions for both pieces: The domain is .

Answer:

4.19. Determine the Range of Functions (Lab 3, Task 3)

Determine the range of the following functions:

Click to see the solution
  1. (a) :
    • The term is always non-negative: .
    • Therefore, the denominator .
    • The function value is maximized when the denominator is minimized, which is at , giving .
    • As , the denominator grows infinitely large, so the function value approaches 0 from the positive side.
    • Range: .
  2. (b) :
    • The domain is .
    • The square root term is always non-negative: .
    • Therefore, the function .
    • The minimum value is -1, which occurs at . The function can take any value greater than -1.
    • Range: .

Answer:

4.20. Find Inverse Image of a Function (Lab 3, Task 4)

Let . Determine and .

Click to see the solution
  1. Find : We need to find all in the domain of such that .
    • .
    • This means .
    • Adding 1 to all parts gives .
    • .
  2. Find : We need to find all such that .
    • .
    • This means (assuming log is base 10) or (assuming log is natural log). Let’s assume natural log.
    • .
    • .

Answer:

4.21. Find the Inverse of Functions (Lab 3, Task 5)

In each part, find if the inverse exists.

Click to see the solution
  1. (a) :
    • Let . Solve for .
    • .
    • The inverse is .
  2. (b) :
    • . This function is not one-to-one (e.g., and ), so it does not have a unique inverse over its entire domain.
    • Answer: No inverse exists.
  3. (c) :
    • Let . Solve for .
    • .
    • .
    • The inverse is . (The function is its own inverse).

Answer:

  2. No inverse exists.
4.22. Sketch Transformations of a Graph (Lab 3, Task 6)

The graph of a function is shown. Sketch the graphs of:

Click to see the solution
  1. (a) : This is a vertical shift down by 1 unit. Every point on the original graph moves down by 1. The “corner” at moves to , the point at moves to , and the segment from to moves to a segment from to .
  2. (b) : This is a horizontal shift right by 1 unit. Every point on the original graph moves right by 1. The corner at moves to , the point at moves to , and the segment from to moves to a segment from to .
  3. (c) : This is a vertical compression by a factor of 2. The y-coordinate of every point is halved. The corner at stays at . The point at moves to , and the segment from to moves to a segment from to .
  4. (d) : This is a combination of two transformations:
    • A reflection across the y-axis (due to the negative sign).
    • A horizontal stretch by a factor of 2 (due to the ).
    • The point is first reflected to , then stretched to . The point stays at . The point is first reflected to , then stretched to .
4.23. Prove by Mathematical Induction (Chapter 1, Example 1)

By mathematical induction, prove that for all :

Click to see the solution

Let be the statement .

  1. Base Step: We test the formula for .
    • LHS = .
    • RHS = .
    • Since LHS = RHS, is true.
  2. Inductive Step: Assume is true for some integer . We must prove that is true.
    • Assumption (Inductive Hypothesis): .
    • Goal: We need to prove that .
    • Proof: Start with the left side of the goal equation.
      • LHS =
      • Using the inductive hypothesis, substitute the sum in the parenthesis:
    • This is equal to the right-hand side of the goal. Thus, is true.
  3. Conclusion: By the principle of mathematical induction, the formula is true for all natural numbers .
Answer: The proof is complete.
4.24. Prove the Irrationality of (Chapter 1, Example 2)

Prove that the number is irrational.

Click to see the solution
  1. Assume the opposite (Proof by Contradiction): Assume that is a rational number. This means it can be written as a fraction in lowest terms, where and are integers and their greatest common divisor is 1 (i.e., ).
  2. Manipulate the equation: Square both sides of the equation.
    • .
  3. Deduce properties of : The equation shows that is an even number. If a square of a number is even, the number itself must be even. Therefore, is even.
  4. Express as an even number: We can write for some integer .
  5. Substitute and deduce properties of : Substitute back into the equation from step 2.
    • .
    • This shows that is also an even number, which means that must also be even.
  6. Find the Contradiction: We have concluded that both and are even. If both are even, they share a common factor of 2. This contradicts our initial assumption that the fraction was in its lowest terms (i.e., ).
  7. Conclusion: Since our initial assumption leads to a contradiction, the assumption must be false.
Answer: The assumption that is rational is false, therefore is irrational.
4.25. Prove by Mathematical Induction (Chapter 1, Example 3)

By mathematical induction, prove that for all :

Click to see the solution

Let be the given equation.

  1. Base Step: We test the formula for .
    • LHS = .
    • RHS = .
    • Since LHS = RHS, is true.
  2. Inductive Step: Assume is true for some integer . We must prove .
    • Assumption (Inductive Hypothesis): .
    • Goal: Prove .
    • Proof:
      • LHS of goal = .
      • Using the inductive hypothesis:
      • .
    • This is the RHS of the goal. Thus is true.
  3. Conclusion: By the principle of mathematical induction, the formula is true for all .
Answer: The proof is complete.
4.26. Prove the Irrationality of (Chapter 1, Example 4)

Prove that the number is irrational.

Click to see the solution
  1. Assume the opposite (Proof by Contradiction): Assume that is rational. This means it can be expressed as a fraction in lowest terms, where are integers with .
  2. Manipulate the equation: Square both sides.
    • .
  3. Deduce properties of : The equation shows that is divisible by 3. Since 3 is a prime number, if it divides , it must also divide . Therefore, is a multiple of 3.
  4. Express as a multiple of 3: We can write for some integer .
  5. Substitute and deduce properties of : Substitute back into the equation from step 2.
    • .
    • This equation shows that is divisible by 3. Since 3 and 5 are prime, 3 must divide . Since 3 is prime, it must also divide . Therefore, is also a multiple of 3.
  6. Find the Contradiction: We have shown that both and are multiples of 3. This means they share a common factor of 3, which contradicts our initial assumption that was in lowest terms.
  7. Conclusion: The assumption that is rational leads to a contradiction, so it must be false.
Answer: The number is irrational.
4.27. Prove by Mathematical Induction (Chapter 1, Example 5)

By mathematical induction, prove that for all integers .

Click to see the solution

Let be the statement .

  1. Base Step: The statement is for , so our base case is .
    • For , we have .
    • And .
    • Since , the statement is true.
  2. Inductive Step: Assume that is true for some arbitrary integer . We must show that is true.
    • Assumption (Inductive Hypothesis): for some .
    • Goal: We need to prove that .
    • Proof:
      • Start with the left side of the goal: .
      • From our assumption, we know . We can substitute this to get an inequality:
      • .
      • Now we need to show that this is greater than . We need to show that .
      • Dividing both sides by , this is equivalent to showing that .
      • Since our assumption is for , it is certainly true that .
      • Therefore, . This proves is true.
  3. Conclusion: By the principle of mathematical induction, the inequality is true for all integers .
Answer: The proof is complete.
4.28. Find Supremum and Infimum of Sets (Chapter 1, Example 6)

Find the minimum (min), infimum (inf), maximum (max), and supremum (sup) for the following sets, if they exist.

Click to see the solution
  1. For (a closed interval):
    • The smallest element is -1, so .
    • The infimum (greatest lower bound) is equal to the minimum, so .
    • The largest element is 2, so .
    • The supremum (least upper bound) is equal to the maximum, so .
  2. For (a half-open interval):
    • The smallest element is 0, so and .
    • The set gets arbitrarily close to 1 but never reaches it. There is no largest element, so does not exist.
    • The least upper bound is 1, so .
  3. For (a sequence):
    • The largest element is 1, so and .
    • The set gets arbitrarily close to 0 but never reaches it. There is no smallest element, so does not exist.
    • The greatest lower bound is 0, so .
  4. For (an open, unbounded interval):
    • The set is not bounded above, so it has no maximum or supremum.
    • The set gets arbitrarily close to 1 but never reaches it. There is no smallest element, so does not exist.
    • The greatest lower bound is 1, so .
4.29. Prove by Mathematical Induction (Chapter 1, Example 7)

By mathematical induction, prove that is divisible by 3 for any integer .

Click to see the solution

Let be the statement that is divisible by 3.

  1. Base Step: We must show that the statement is true for the initial value, .
    • For , we have .
    • Since 3 is divisible by 3, the statement is true.
  2. Inductive Step: We assume that the statement is true for some arbitrary integer , and then we must show that is also true.
    • Assumption (Inductive Hypothesis): Assume is divisible by 3. This means we can write for some integer . This implies .
    • Goal: We need to prove that is divisible by 3.
    • Proof:
      • Substitute the expression for from our assumption:
    • Since can be written as 3 times an integer (), it is divisible by 3. Thus, is true.
  3. Conclusion: Since both the base step and the inductive step have been proven, by the principle of mathematical induction, the statement is true for all integers .
Answer: The proof is complete.
4.30. Prove the Bernoulli Inequality (Chapter 1, Example 8)

By mathematical induction, prove the Bernoulli inequality: for any integer and any real number .

Click to see the solution

Let be the statement .

  1. Base Step: For .
    • LHS = .
    • RHS = .
    • Since , the statement is true.
  2. Inductive Step: Assume is true for some integer . We must prove .
    • Assumption (Inductive Hypothesis): .
    • Goal: Prove .
    • Proof:
      • Start with the LHS of the goal: .
      • Since , the term is non-negative. We can multiply both sides of the inductive hypothesis by without changing the inequality direction.
      • .
      • Expand the right side: .
      • Since and , the term must be non-negative ().
      • Therefore, .
      • Combining our inequalities, we have . This proves the goal.
  3. Conclusion: By the principle of mathematical induction, the inequality is true for all and .
Answer: The proof is complete.
4.31. Prove by Mathematical Induction (Chapter 1, Example 9)

By mathematical induction, prove the inequality: for all .

Click to see the solution

Let be the given inequality.

  1. Base Step: For .
    • LHS = .
    • RHS = .
    • The inequality is . Squaring both sides gives , which is true. So is true.
  2. Inductive Step: Assume is true for some integer . We must prove .
    • Assumption (Inductive Hypothesis): .
    • Goal: Prove .
    • Proof:
      • Start with the LHS of the goal: LHS = .
      • Using the inductive hypothesis, we can write: LHS .
      • Now we just need to prove that .
      • Since both sides are positive, we can square them: .
      • Cross-multiply: .
      • .
      • This simplifies to , which is always true.
    • Since all steps are valid, we have proven that is true.
  3. Conclusion: By the principle of mathematical induction, the inequality holds for all .
Answer: The proof is complete.
4.32. Prove the Irrationality of a Sum of Radicals (Chapter 1, Example 10)

Prove that the number is irrational.

Click to see the solution
  1. Assume the opposite (Proof by Contradiction): Assume that is a rational number.
  2. Manipulate the Equation: Square both sides of the equation.
    • .
  3. Isolate the Radical: Rearrange the equation to solve for .
    • .
  4. Find the Contradiction:
    • From our initial assumption, is rational. The square of a rational number is rational, and the difference and quotient of rational numbers are also rational. Therefore, the right-hand side, , must be a rational number.
    • However, the left-hand side, , is known to be an irrational number.
    • This leads to the statement that a rational number is equal to an irrational number, which is a contradiction.
  5. Conclusion: The initial assumption that is rational must be false.
Answer: The number is irrational.
4.33. Prove the Triangle Inequality (Chapter 1, Example 11)

Prove the triangle inequality, which states that for any real numbers and , we have .

Click to see the solution
  1. Use the property of absolute values: For any real number , we know that and .
    • Therefore, we have and .
    • We also have and .
  2. Combine the inequalities:
    • Adding the first pair of inequalities gives: .
    • Adding the second pair of inequalities gives: , which can be written as .
  3. Apply the equivalence from the previous problem: We have shown that both and are true.
    • This is equivalent to the statement .
Answer: The proof is complete.
4.34. Prove an Absolute Value Identity (Chapter 1, Example 12)

Prove that for any real numbers and , the inequality is equivalent to the inequality .

Click to see the solution
  1. Prove the forward direction (): Assume .
    • Case 1: . In this case, . The assumption becomes . Since , it must be that . From , we get . Combining these gives , which implies .
    • Case 2: . In this case, . The assumption becomes . Multiplying by -1 reverses the inequality, giving . Since , it follows that . Combining these gives , which implies .
    • In both cases, we have shown that if , then .
  2. Prove the backward direction (): Assume .
    • Case 1: . The right side of the assumption is . Since , , so we have .
    • Case 2: . The left side of the assumption is . Multiplying by -1 gives . Since , , so we have , which is equivalent to .
    • In both cases, we have shown that if , then .
Answer: Since the implication holds in both directions, the two inequalities are equivalent.
4.35. Prove the Equivalence of Completeness Axioms (Chapter 1, Example 13)

Prove that the following two axioms of completeness for real numbers are equivalent:

  • (A16) For any two nonempty sets M and N of real numbers such that for all and , there exists a real number such that for all and .
  • (A16’) Every nonempty set of real numbers that is bounded above has a unique least upper bound (supremum).
Click to see the solution
  1. Prove (A16) (A16’):
    • Assume (A16) is true. Let be a nonempty set of real numbers that is bounded above. We need to show it has a unique least upper bound.
    • Define a new set . This is the set of all upper bounds for . Since is bounded above, is not empty.
    • By the definition of the sets and , we have for all and .
    • According to axiom (A16), there must exist a number such that for all and .
    • The condition means is an upper bound for . The condition means is less than or equal to every other upper bound.
    • Therefore, is the least upper bound (supremum) of . Uniqueness can be proven by assuming two least upper bounds , which leads to and , implying .
  2. Prove (A16’) (A16):
    • Assume (A16’) is true. Let and be two nonempty sets such that for all and .
    • From this condition, any element of is an upper bound for the set . Since is nonempty, this means is bounded above.
    • According to axiom (A16’), the set must have a unique least upper bound, let’s call it .
    • By definition of supremum, for all .
    • We also need to show for all . Assume for contradiction that there is some such that . Since is an upper bound for (by definition of N), this would mean is an upper bound smaller than the least upper bound , which is a contradiction.
    • Therefore, for all . We have found a number such that for all , which proves (A16).
Answer: The axioms are shown to be equivalent.
4.36. Prove that the Set of Natural Numbers is Unbounded (Chapter 1, Example 14)

Prove that the set of natural numbers is not bounded above.

Click to see the solution
  1. Assume the opposite (Proof by Contradiction): Assume that is bounded above.
  2. Apply the Completeness Axiom: Since is a nonempty set of real numbers that is bounded above, by the completeness axiom (A16’), it must have a least upper bound (supremum). Let’s call this least upper bound .
  3. Use the Definition of Supremum: The number is an upper bound, so for every natural number , we have .
  4. Consider the Successor: If is a natural number, then is also a natural number. Therefore, according to our assumption, .
  5. Derive a New Upper Bound: If we subtract 1 from both sides of the inequality , we get . This must be true for all natural numbers .
  6. Find the Contradiction: The statement “ for all ” means that is also an upper bound for the set . However, . This contradicts our initial statement that was the least upper bound.
  7. Conclusion: The initial assumption that is bounded above must be false.
Answer: The set of natural numbers is not bounded above.
4.37. Prove the Density of Rational Numbers (Chapter 1, Example 15)

Prove that for any two real numbers with , there exists a rational number such that .

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  1. Use the Archimedean Property: Since , the difference is a positive real number. By the Archimedean Property (which follows from the unboundedness of ), there exists a natural number such that . This implies .
  2. Construct an Integer: Consider the number . Let be the greatest integer less than or equal to . This is denoted . By definition of the floor function, we have .
  3. Construct a Rational Number: Divide the inequality by : . Let’s define the rational number .
  4. Show : From the inequality in the previous step, we directly have .
  5. Show : From step 2, we know . So . We also know , which means . Therefore, .
    • From step 1, we established that .
    • Combining these, we get .
  6. Conclusion: We have shown that and .
Answer: We have constructed a rational number that lies between and .
4.38. Prove the Density of Irrational Numbers (Chapter 1, Example 16)

Prove that for any two real numbers with , there exists an irrational number such that .

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  1. Case 1: Assume and are rational.
    • Since , we have . Consider the number .
    • Since is irrational and , the term is irrational. The sum of a rational number () and an irrational number is irrational, so is irrational.
    • Since , we have .
    • Multiplying by the positive number , we get .
    • Adding to all parts of the inequality gives , which simplifies to .
  2. Case 2: General case (x, y may be irrational).
    • By the density of rational numbers (proven previously), since , we can find a rational number such that .
    • Again, by the density of rational numbers, we can find another rational number such that .
    • Now we have two rational numbers and with .
    • Applying the result from Case 1 to the rational numbers and , there must exist an irrational number such that .
    • By transitivity, we have , which implies .
Answer: In both cases, we have found an irrational number between and .
4.39. Prove Countability of Unions (Chapter 1, Example 17)

Prove that a finite or countable union of countable sets is a countable set.

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  1. Case 1: Union of two countable sets.
    • Let and be two countable sets.
    • We can create a list of all elements in the union by alternating between the sets: .
    • This process creates a single list (a sequence) that includes every element from both sets. To make it a formal enumeration, we can assign a new natural number to each element in the order it appears, skipping any element that has already been listed (if is not empty).
    • Since we can list all elements of in a sequence, the union is countable. This can be extended to any finite number of sets.
  2. Case 2: Countable union of countable sets.
    • Let the sets be , where each set is itself countable: .
    • We can arrange all elements in a grid: …
    • We can list all these elements by traversing the grid diagonally (Cantor’s diagonal argument for pairing). We list elements with the same sum of indices: , then , then , and so on.
    • This method creates a single sequence that will eventually include every element from every set in the union. Therefore, the countable union of countable sets is countable.
Answer: By showing that the elements of the union can be arranged into a single list (sequence), we prove that the union is countable.
4.40. Prove that the Set of Rational Numbers is Countable (Chapter 1, Example 18)

Prove that the set of rational numbers is countable.

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  1. Represent as a Union: The set of all rational numbers can be represented as the union of many sets. For instance, consider the sets of fractions with a fixed denominator.
  2. Define Countable Sets: For each natural number , define the set as the set of all fractions with denominator :
    • .
  3. Show Each Set is Countable: For any fixed , the set is in a one-to-one correspondence with the set of integers (the map is a bijection). Since the set of integers is countable, each set is countable.
  4. Apply the Union Theorem: The set of all rational numbers is the union of all these sets: .
  5. Conclusion: We have expressed as a countable union of countable sets. Based on the previous theorem, such a union is itself a countable set.
Answer: The set of rational numbers is countable.
4.41. Prove that the Unit Interval is Uncountable (Chapter 1, Example 19)

Prove that the set of real numbers in the closed interval is uncountable.

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  1. Assume the opposite (Proof by Contradiction): Assume that the set is countable. This means we can list all the real numbers in this interval in a sequence: .
  2. Represent Numbers by Decimals: Let’s write the decimal expansion for each number in the list. (To ensure uniqueness, we avoid expansions ending in infinite 9s).
    • … where each digit is an integer from 0 to 9.
  3. Construct a New Number (Cantor’s Diagonal Argument): We will now construct a new real number, , which is in the interval but is not in our list. Let . We define each digit of based on the -th digit of the -th number in the list, .
    • We choose to be different from . A simple rule is:
      • if .
      • if .
  4. Find the Contradiction:
    • The number that we constructed is a real number in the interval .
    • By its construction, cannot be equal to any number in our list. Why? Because for any , the -th decimal digit of (which is ) is different from the -th decimal digit of (which is ).
    • This means our list, which supposedly contained all real numbers in , is missing at least one number, . This is a contradiction.
  5. Conclusion: The initial assumption that the set $$ is countable must be false.
Answer: The interval is uncountable.
4.42. State the Principle of Mathematical Induction (Chapter 1, Example 20)

State the Principle of Mathematical Induction and provide an outline of its proof.

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  1. Statement of the Principle: Let be a statement that depends on a natural number . If the following two conditions are met:
    • (1) Base Step: The statement is true.
    • (2) Inductive Step: The implication “if is true, then is true” holds for any natural number .
    • Then, the statement is true for all natural numbers .
  2. Proof Outline (by Contradiction):
    • Assume the principle is false. This means that both conditions (1) and (2) are satisfied, yet there is at least one natural number for which the statement is false.
    • Let be the set of all natural numbers for which is false: .
    • By our assumption, the set is not empty.
    • By the well-ordering principle of natural numbers, every non-empty set of natural numbers has a least element. Let be the least element of .
    • Since is true (by condition 1), is not in . Therefore, , which means .
    • Since is the least element for which P is false, the statement must be true for the number just before it, . So, is true.
    • But now we can apply the inductive step (condition 2). Since is true, the implication tells us that must also be true.
    • This is a contradiction, because we defined as an element for which is false.
    • Therefore, our initial assumption that the principle is false must be wrong.
Answer: The principle is proven.