Comprehensive Notes on Real Numbers, Natural Numbers, and Mathematical Induction

Textbook Reminder

Textbook cover may vary.
Only one edition available.
Used in the semester; good to get a copy.

Study Unit 6: Real and Complex Number Systems

6.1: Real numbers system.
6.2: Complex number system.
Start broad in 6.1.1, inspect real numbers from smallest set, adding elements.
Focus on 6.1.1 and 6.1.2.

Algebraic Structures

Studying number systems as algebraic structures.
Using real numbers in calculus.
Good definitions and properties are important.

Real Numbers

Set of elements with addition and multiplication defined.

Properties for Addition and Multiplication

Closure: Operation on any real number results in a real number.
Associativity: Holds under addition and multiplication.
Commutativity: Holds under addition and multiplication.
Zero Element: Additive identity.
Opposite: Additive inverse.
Identity Element: For multiplication.
Reciprocal: Multiplicative inverse.

Finding Opposites and Reciprocals

Examples: -2, 3/8, $(\sqrt{3})$ , 2 + e.
Opposite: Element added to a number to get zero (additive inverse).
Reciprocal: Multiplied with a number to get the multiplicative identity.

Additive and Multiplicative Identities

Additive identity: 0.
Multiplicative identity: 1.

Examples

For -2:
- Opposite: 2 (since -2 + 2 = 0).
- Reciprocal: $- \frac{1}{2}$ (since -2 * $- \frac{1}{2}$ = 1).
For 3/8:
- Opposite: $- \frac{3}{8}$ .
- Reciprocal: $\frac{8}{3}$ .
For $\sqrt{3}$ :
- Opposite: $- \sqrt{3}$ .
- Reciprocal: $\frac{1}{\sqrt{3}}$ .
For 2 + e:
- Opposite: $-2 - e$ .
- Reciprocal: $\frac{1}{2 + e}$ .

Summary of Identities and Inverses

$a + 0 = a$ (Additive identity).
$a * 1 = a$ (Multiplicative identity).
$a + (-a) = 0$ (Additive inverse).
$a * \frac{1}{a} = 1$ (Multiplicative inverse).

Distributive Law

$a(b + c) = ab + ac$ .

Difference and Quotient

Difference: Adding the opposite.
Quotient: Product with the reciprocal.

Ordering

Real numbers can be ordered; for any two real numbers a and b, either a < b, b < a, or a = b.
Complex numbers do not have this property.
Real number line is based on this ordering.

Positive Real Numbers

Inequality properties.
Multiplying by a negative number reverses inequality signs.

Properties

If a > 0, then \frac{1}{a} > 0.
If a > 0 and b > 0, then ab > 0.
If a > b, then a + c > b + c.
If a > b, then \frac{1}{a} < \frac{1}{b}.
If a > b, and c>0, then ac > bc and \frac{a}{c} > \frac{b}{c}.
If a > b, and c<0, then ac < bc and \frac{a}{c} < \frac{b}{c}.

Definition

$R^+$ denotes the set of positive real numbers.
x > 0 if and only if $x \in R^+$ .

Closure under Addition

If $x \in R^+$ and $y \in R^+$ , then $x + y \in R^+$ .

Subset Notation

$R^+ \subseteq R$ (R+ is a subset of R).

Trichotomy Property

For all $x \in R$ , either $x \in R^+$ , x = 0, or $-x \in R^+$ .

Ordering and Subtraction

If b > a, then $b - a \in R^+$ .
If a > b, then $b - a \notin R^+$ , but $a - b \in R^+$ .
If a = b, then a - b = 0.

Interval Notation

$R = (-\infty, \infty)$ .
Open interval (a, b) is contained in the closed interval [a, b].

No Smallest or Largest Element

Real numbers do not have a smallest or largest element.
Infinitely many real numbers between any two real numbers.

Theorem 46

Between any two real numbers, there is another real number.
For any real numbers a and b (a < b), there exists a $c \in R$ such that a < c < b.
Example: $c = \frac{a + b}{2}$ .

Natural Numbers

Natural numbers are the integers from 1, 2, 3, 4, etc.

Properties

No zero element.
Closed under addition and multiplication.
Not closed under subtraction or division.
No opposites.
Multiplicative identity is 1, but no reciprocals except for 1.
$n \geq 1$ for all n in the natural numbers.
If $x \in N$ , then $x + 1 \in N$ .
There does not exist a $y \in N$ such that x < y < x + 1.

Principle of Mathematical Induction (PMI)

If:
- A subset m of natural numbers contains 1.
- Whenever x is in m, x + 1 is also in m.
Then m is equal to the set of natural numbers.

Using PMI to Prove Statements

Show that the statement holds for the smallest n (usually n = 1).
Assume it holds for some k.
Prove that it holds for k + 1.

Example

Prove $1 + 3 + 5 + 7 + … + (2n - 1) = n^2$ for all natural numbers n.
For n = 3: 1 + 3 + 5 = 9 = \3^2
For n = 5: 1 + 3 + 5 + 7 + 9 = 25 = \5^2

Domino Analogy

Ensuring that one domino falls causes all subsequent dominoes to fall.
Need to make sure that:
- The first domino is standing.
- If any domino k is standing, then domino k + 1 is close enough to also fall.

Steps for PMI

Show that the statement holds for n = 1.
Assume it holds for n = k.
Prove that it holds for n = k + 1.

Detailed Example

Step 1: Show the statement holds for n = 1.
- Right-hand side: \1^2 = 1.
- Left-hand side: $2(1) - 1 = 1$ .
- Conclusion: Left-hand side = Right-hand side, thus the statement holds for n = 1.
Step 2: Assume it holds for n = k.
- Assume $1 + 3 + 5 + … + (2k - 1) = k^2$ , where k is a natural number greater than one.
Step 3: Prove that it holds for n = k + 1.
- Required to prove: $1 + 3 + 5 + … + (2(k + 1) - 1) = (k + 1)^2$ .
- Consider the left-hand side: $1 + 3 + 5 + … + (2(k + 1) - 1)$ .
- Reveal the last term: $1 + 3 + 5 + … + (2k - 1) + (2(k + 1) - 1)$ .
- Use step 2 (the assumption): $k^2 + (2(k + 1) - 1)$ .
- Simplify and expand: $k^2 + 2k + 2 - 1 = k^2 + 2k + 1 = (k + 1)^2$ .
- Conclusion: The statement holds for n = k + 1.
By the principle of mathematical induction, the statement holds for all natural numbers.

Divisibility

b is a factor of a (b divides a) if and only if there exists a q such that a = q * b.
Examples: 5 is a factor of 25, 3 is a factor of 30.
a is a multiple of b.
a is divisible by b.

Divisibility Type Example

Okay, so the statement is this thing which is in terms of n is divisible by three. So it means is divisible by three. It means it is a multiple of three or three is a factor of this

Divisibility Type Example

Statement: $(2^(2n)) - 1$ is divisible by 3.

Step 1

Show that it works for one

Let n = 1: $(2^(2*1)) - 1 = 2^2 - 1 = 4 - 1 = 3$ ..
-3 is divisible by 3 thus the statement holds for n=1
Unfortunately, I'm going to have to stop this so we have time for the class assignment. I will continue this tomorrow.