Number Systems

Natural Numbers

1. There exists an element $0 \in \mathbb{N}$.

2. There exists a function

   $$\begin{equation*} S : \mathbb{N} \to \mathbb{N} \end{equation*}$$

   called the successor function, such that for every $n \in \mathbb{N}$, $S(n)$ represents the next natural number after $n$.

3. Zero is not the successor of any number

   $$\begin{equation*} \forall n \in \mathbb{N},\; S(n) \neq 0 \end{equation*}$$

4. Successor function is injective (one-to-one)

   $$\begin{equation*} \forall m, n \in \mathbb{N},\; S(m) = S(n) \implies m = n \end{equation*}$$

5. Principle of mathematical induction. For any property $P(n)$ defined on $\mathbb{N}$: If

   $$\begin{equation*} P(0) \text{ is true, and } \forall n \in \mathbb{N},\; P(n) \Rightarrow P(S(n)), \end{equation*}$$

​ then

$$\begin{equation*} \forall n \in \mathbb{N},\; P(n) \text{ is true.} \end{equation*}$$

Addition

The addition $+ : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ is defined recursively as follows:

$$\begin{equation*} a + 0 = a \end{equation*}$$ $$\begin{equation*} a + S(b) = S(a + b) \end{equation*}$$

Multiplication

The multiplication $\times : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ is defined recursively as follows:

$$\begin{equation*} a \times 0 = 0 \end{equation*}$$ $$\begin{equation*} a \times S(b) = a \times b + a \end{equation*}$$

Order Relation

For all $m, n \in \mathbb{N}$:

$$\begin{equation*} m < S(n) \iff (m < n) \text{ or } (m = n). \end{equation*}$$

This defines $<$ recursively based on the successor structure.

Theorems

1. $a + (b + c) = (a + b) + c$.

2. $a + 0 = 0 + a = a$.

3. $a + b = b + a$.

4. $(a + b)\times c = a \times c + b \times c$.

5. $a \times 1 = 1 \times a = a$.

6. $a \times b = b \times a$.

7. $a \times (b \times c) = (a \times b) \times c$.

Integers

Integers can be defined from natural numbers as equivalence classes of ordered pairs of natural numbers:

$$\begin{equation*} \mathbb{Z} = \{ (a, b) \mid a, b \in \mathbb{N} \} \end{equation*}$$

We define an equivalence relation:

$$\begin{equation*} (a, b) \sim (c, d) \iff a + d = b + c \end{equation*}$$

Each equivalence class $[(a, b)]$ represents the difference between $a$ and $b$. Thus we have:

$$\begin{equation*} \begin{aligned} 0 &= [(0, 0)], \\ 1 &= [(1, 0)], \\ -1 &= [(0, 1)], \\ a - b &= [(a, b)]. \end{aligned} \end{equation*}$$

Addition

$$\begin{equation*} [(a, b)] + [(c, d)] = [(a + c,\, b + d)] \end{equation*}$$

Multiplication

$$\begin{equation*} [(a, b)] \times [(c, d)] = [(ac + bd,\, ad + bc)] \end{equation*}$$

These definitions are well-defined, meaning the result does not depend on the choice of representatives of the equivalence classes.

Order Relation

$$\begin{equation*} [(a,b)] < [(c,d)] \iff a + d < b + c. \end{equation*}$$

Theorems

1. $a + b = b + a, \qquad a \times b = b \times a$.

2. $a + (b + c) = (a + b) + c, \qquad a \times (b \times c) = (a \times b) \times c$.

3. $a + 0 = a, \qquad a \times 1 = a$.

4. $\forall a \in \mathbb{Z}, \; \exists (-a) \in \mathbb{Z} \text{ such that } a + (-a) = 0$.

5. $a \times (b + c) = a \times b + a \times c$.

Rational Numbers

The set of rational numbers $\mathbb{Q}$ is constructed from the integers $\mathbb{Z}$ as the set of equivalence classes of ordered pairs of integers:

$$\begin{equation*} \mathbb{Q} = \{ (a, b) \mid a \in \mathbb{Z},\; b \in \mathbb{Z} \setminus \{0\} \} \end{equation*}$$

Each pair $(a,b)$ intuitively represents the fraction $\dfrac{a}{b}$.

We define an equivalence relation:

$$\begin{equation*} (a,b) \sim (c,d) \iff ad = bc \end{equation*}$$

Each equivalence class $[(a,b)]$ corresponds to a rational number.

We write:

$$\begin{equation*} \dfrac{a}{b} = [(a,b)], \quad \text{where } b \neq 0 \end{equation*}$$

Addition

$$\begin{equation*} \dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd} \end{equation*}$$

Multiplication

$$\begin{equation*} \dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd} \end{equation*}$$

Additive Inverse

$$\begin{equation*} -\dfrac{a}{b} = \dfrac{-a}{b} \end{equation*}$$

Multiplicative Inverse

$$\begin{equation*} \left(\dfrac{a}{b}\right)^{-1} = \dfrac{b}{a}, \quad a \neq 0 \end{equation*}$$

Order Relation

To compare two rational numbers, let

$$\begin{equation*} \dfrac{a}{b}, \; \dfrac{c}{d} \in \mathbb{Q}, \quad b > 0, \; d > 0. \end{equation*}$$

We define the order relation $<$ on $\mathbb{Q}$ by:

$$\begin{equation*} \dfrac{a}{b} < \dfrac{c}{d} \iff ad < bc. \end{equation*}$$

Theorems

1. $a+b = b+a, \qquad a\times b = b\times a$.

2. $a+(b+c) = (a+b)+c, \qquad a\times(b\times c) = (a\times b)\times c$.

3. $a+0 = a, \qquad a\times1 = a$.

4. $\forall a \in \mathbb{Q},\; \exists (-a) \text{ such that } a+(-a)=0$.

5. $\forall a \in \mathbb{Q}\setminus\{0\},\; \exists a^{-1} \text{ such that } a\times a^{-1}=1$.

6. $a\times(b+c) = a\times b + a\times c$.

Real Numbers

A Dedekind cut in $\mathbb{Q}$ is a partition $(A, B)$ of $\mathbb{Q}$ satisfying:

$$\begin{equation*} \begin{aligned} 1.\;& A, B \subset \mathbb{Q}, \; A \cup B = \mathbb{Q}, \; A \cap B = \varnothing, \\[4pt] 2.\;& \forall a \in A,\; \forall b \in B,\; a < b, \\[4pt] 3.\;& A \text{ has no greatest element.} \end{aligned} \end{equation*}$$

Each real number is identified with one such cut $A$, which intuitively represents “all rationals less than that real number.”

For example:

$$\begin{equation*} \sqrt{2} = \{ q \in \mathbb{Q} \mid q < 0 \text{ or } q^2 < 2 \}. \end{equation*}$$

The set of all Dedekind cuts is denoted by:

$$\begin{equation*} \mathbb{R} = \{ A \subset \mathbb{Q} \mid A \text{ is a Dedekind cut} \}. \end{equation*}$$

Addition

$$\begin{equation*} A + B = \{ a + b \mid a \in A,\, b \in B \}. \end{equation*}$$

Multiplication

(for positive cuts)

$$\begin{equation*} A \times B = \{ a \times b \mid a \in A,\, b \in B,\; a,b > 0 \}. \end{equation*}$$

Negatives and general cases are defined symmetrically by extending signs.

Order Relation

Define:

$$\begin{equation*} A < B \iff A \subsetneq B. \end{equation*}$$

This relation extends the usual order on $\mathbb{Q}$ and makes $(\mathbb{R}, <)$ a totally ordered set.

Theorems

1. $a+b = b+a, \qquad a\times b = b\times a$.

2. $a+(b+c) = (a+b)+c, \qquad a\times(b\times c) = (a\times b)\times c$.

3. $a+0 = a, \qquad a\times1 = a$.

4. $\forall a \in \mathbb{Q},\; \exists (-a) \text{ such that } a+(-a)=0$.

5. $\forall a \in \mathbb{Q}\setminus\{0\},\; \exists a^{-1} \text{ such that } a\times a^{-1}=1$.

6. $a\times(b+c) = a\times b + a\times c$.

7. The set of rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$:

$$\begin{equation*} \forall a,b \in \mathbb{R},\; a < b \Rightarrow \exists r \in \mathbb{Q} \text{ such that } a < r < b. \end{equation*}$$

8. Every nonempty subset $S \subset \mathbb{R}$ that is bounded above has a least upper bound (supremum) in $\mathbb{R}$:

$$\begin{equation*} \forall S \subset \mathbb{R},\; S \neq \varnothing,\; \text{if } S \text{ is bounded above, then } \sup S \in \mathbb{R}. \end{equation*}$$

​ This property makes $\mathbb{R}$ a complete ordered field.