The integers came from God, everything else is man-made.

• $Leopold$ $Kronecker$

ℕ⁰ = {0,1,2,3,4,5,6,7,8,9,10...}
ℕ⁺ =   {1,2,3,4,5,6,7,8,9,10...}

∀𝚗{∃𝑠(𝚗)}
   𝑠(𝚗): ℕ⟶ℕ  -- Mapping for natural numbers
                  (ℕ)<ℕ>   Input (ℕ); Output <ℕ> 
   
   𝚗≠𝚖 ⇒ 𝑠(𝚗)≠𝑠(𝚖)    -- Injective property

𝚖+1 = 𝑠(m)
𝚖+𝚗 = 𝑠(𝑠(𝑠(𝑠(𝑠(...𝑠(𝚖)｠⃨｠= 𝑠ⁿ(𝚖)
      ︸𝚗 iterations
𝚖+𝑠(𝚗) = 𝑠(𝚖+𝚗)

function addition(n,m : Natural)return Natural is
   begin
   if m = 1 then return natural'succ(n);
     else return natural'succ(addition(n,m-1));
     end if;
   end addition;

0  1  2  3  4  5  6   
┠──╂──╂──╂──╂──╂──╂╌╌╌
𝑠(𝚗)
↓  ↓  ↓  ↓  ↓  ↓  ↓
1  2  3  4  5  6  7
┠──╂──╂──╂──╂──╂──╂╌╌╌

Commutative

Associative

Cancellation

The product of two natural numbers is defined recursively.

 𝚖·𝟷＝𝚖
 𝚖·𝑠(𝚗)＝𝚖·𝚗﹢𝚖

function product(n,m : Natural) return Natural is
   begin
   if n = 1 then return m;
     else return product(n-1,m)+m;
     end if;
   end product;

0   1   2   3   4   5   6   
┠───╂───╂───╂───╂───╂───╂╌╌╌
⨉𝚖 ↓
0   𝚖   𝟸𝚖  𝟹𝚖  𝟺𝚖  𝟻𝚖   𝟼𝚖 
┠───╂───╂───╂───╂───╂───╂╌╌╌

0   1   2   3   4   5   6   
┠───╂───╂───╂───╂───╂───╂╌╌╌
⬛️  🟦  🟩   🟨  🟧  🟥   🟪
⨉2
0   1   2   3   4   5   6   7   8   9   10  11  12 
┠───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂─╌╌╌
⬛️      🟦      🟩       🟨      🟧       🟥      🟪

Distributive

Associative

Commutative

Cancellation

function power(base, exponent : Natural) return Natural is
   begin
   if exponent = 1 
      then return base;
      else return power(base, exponent - 1) * base;
      end if;
   end power;

0   1   2   3   4   5   6   
┠───╂───╂───╂───╂───╂───╂╌╌╌
⬛️  🟦  🟩   🟨  🟧  🟥   🟪
╱╲2
0   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16 
┠───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂───╂╌╌╌
⬛️  🟦          🟩                   🟨                          🟧    

For any two natural numbers $m$ and $n$:

A crucial property of natural numbers is that they are well-ordered.

This property might seem obvious for simple subsets, but it applies even to complex sets defined indirectly. For example, the set of all numbers that can be expressed as $(12x + 28y)$ (where $(x)$ and $(y)$ are integers) has a smallest natural number, even though its structure is not immediately clear.

Mathematical induction is a powerful proof technique that relies on the well-ordering principle.

𝚗 ∊ 𝔸 ⊂ ℕ
𝑃(𝚗) : Injection Property  -- 𝑃(𝚗): Congruent with the property 𝑃 on 𝚗
𝚗₀: 𝑃(𝚗)		   -- Base case, typically 0 or 1
                   
𝚗:𝑃(𝚗) → 𝚗+1:𝑃(𝚗)	   -- Inductive step
                           -- If 𝚗 was congruent to the property
                           -- then it implies 𝚗+1 must also be congruent

∴∀𝚗 ∶ 𝑃(𝚗)		   -- Then all 𝚗 must be congruent with 𝑃

1. Base Case: Show that the smallest element (often $(1)$ or $( 0 )$ is in $( X )$.
2. Inductive Step (Bootstrap): Show that if the property holds for $( k < n )$, it must also hold for $( n )$.

To prove by induction that a finite set with ( n ) elements has exactly ( 2^n ) subsets:

• Base Case: For ( n = 1 ), a set with one element has two subsets (itself and the empty set).
• Inductive Step: For a set with ( n ) elements, divide subsets into those containing a chosen element and those not containing it. This gives ( 2^{n-1} + 2^{n-1} = 2^n ) subsets.

The sum of the first $( n )$ positive odd integers is $( n^2 )$:

• Well Ordering Axiom
  • Every nonempty subset of the set of nonnegative integers contains a smallest element.

These axioms provide a framework for defining the structure of $ℕ$.

𝚗∊ℕ
𝑠: ℕ⟶ℕ	
| ∀𝚗( ∃!𝚖 := 𝑠(𝚗)	--- Each n has a unique successor
| 𝚗₀∊ℕ | ∀𝚗[ 𝑠(𝚗)≠𝚗₀ ]	
| 𝕌⊂ℕ:
  | 𝚗₀∊𝕌
  | 𝚗∊𝕌⟶𝑠(𝚗)∊𝕌

∴ 𝕌＝ℕ

𝚖≤𝚗 : { 𝚖<𝚗 ∨ 𝚖=𝚗}
𝚖≥𝚗 : { 𝚖>𝚗 ∨ 𝚖=𝚗}

𝚖≤𝚗 ⟹ 
   𝚖+𝚙 ≤ 𝚗+𝚙
   𝚖·𝚙 ≤ 𝚗·𝚙   𝚖,𝚗,𝚙∊ℕ

• Reflexiva
  • 𝚖≤𝚖
• Antisimétrica
  • { m≤n ∧ n≤m } ⟹ m=n
• Transitiva
  • { m≤n ∧ n≤𝚙 } ⟹ m≤𝚙

El conjunto ℕ con la relación ≤ es un conjunto bien ordenado.

El conjunto ℕ de los números naturales es un semianillo ordenado conmutativo y con unidad.

• ≤
• ＋
  • Asociativa
  • Conmutativa
• ×
  • Asociativo
  • Distributivo en ＋
  • Conmutativo
• 𝟷
  • 𝟷×n = n

In mathematics, we use axioms to define fundamental truths without proof to avoid circular reasoning or infinite regress. Here, we accept:

• The Well-Ordering Principle as an axiom about $( \mathbb{N} )$.
• The elementary arithmetic operations on $( \mathbb{Z} )$, such as addition, subtraction, and multiplication, as foundational knowledge.

[!NoOTE]

Historical Note

The axiomatic approach has roots in ancient Greek mathematics, particularly in Euclid's Elements, where geometry was derived from a small set of axioms. The power of this method lies in deriving vast mathematical truths from a minimal set of assumptions.

In this chapter, we’ve laid the groundwork for exploring the structure and properties of $( \mathbb{N} )$. Here’s a summary of key concepts:

• $( \mathbb{N} )$ is closed under addition and multiplication.
• Well-Ordering Principle: Every non-empty subset of $( \mathbb{N} )$ has a least element.
• Principle of Mathematical Induction: Enables proofs that certain properties hold for all elements in $( \mathbb{N} )$.
• Introduction of $( \mathbb{Z} )$ as the smallest set containing $( \mathbb{N} )$ and closed under subtraction.

This structured understanding of $( \mathbb{N} )$ will serve as a foundation for more complex topics in algebra and other mathematical fields.

Let’s prove by mathematical induction that for all positive integers $( n )$:

$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(2n + 1)(n + 1)}{6}.$

We start by verifying the formula for $( n = 1 )$.

• For $( n = 1 )$, the left side of the equation is simply $( 1^2 = 1 )$.
• The right side of the equation, with $( n = 1 )$, becomes:

$\frac{1(2 \cdot 1 + 1)(1 + 1)}{6}$ $= \frac{1 \cdot 3 \cdot 2}{6}$ $= \frac{6}{6}$ $= 1$

• Since both sides are equal, the formula holds for $( n = 1 )$.

Thus, the base case is true.

Assume that the formula holds for some positive integer $( k )$; that is, assume:

This assumption is called the inductive hypothesis.

We need to prove that if the formula holds for $( n = k )$, then it also holds for $( n = k + 1 )$. In other words, we need to show:

Starting with the left side of the equation for $( n = k + 1 )$, we have:

By the inductive hypothesis, we know that:

So, we can substitute this into our expression:

Now, we’ll simplify this expression by combining terms.

1. Rewrite $( (k + 1)^2 )$ with a common denominator:

2. Factor out $( (k + 1) )$ from the numerator:

3. Expand the terms inside the brackets:

4. Combine like terms:

5. Factor the quadratic expression in the brackets:

This matches the right side of the expression we wanted to prove for $( n = k + 1 )$:

Since we have shown that:

1. The formula holds for the base case $( n = 1 )$,
2. If the formula holds for $( n = k )$, then it also holds for $( n = k + 1 )$,

we conclude by the principle of mathematical induction that the formula is true for all positive integers $( n )$.

To prove the generalized triangle inequality, we’ll use mathematical induction on $( n )$. We want to show that for any real numbers $( x_1, x_2, \dots, x_n )$:

For $( n = 1 )$, the statement is trivially true:

This is clearly correct, so the base case holds.

Assume that the statement holds for some positive integer $( k )$; that is, assume:

This assumption is called the inductive hypothesis.

We need to prove that if the statement holds for $( n = k )$, then it also holds for $( n = k + 1 )$. In other words, we need to show:

Consider the expression $( |x_1 + x_2 + \dots + x_k + x_{k+1}| )$. We can rewrite this as:

Now, we apply the triangle inequality for two terms $( A = x_1 + x_2 + \dots + x_k )$ and $( B = x_{k+1} )$, which states that $( |A + B| \leq |A| + |B| )$. Using this inequality, we get:

By the inductive hypothesis, we know that:

Substituting this into our inequality, we obtain:

This completes the inductive step.

Since we have shown that:

1. The statement holds for the base case $( n = 1 )$,
2. If the statement holds for $( n = k )$, then it also holds for $( n = k + 1 )$,

we conclude by the principle of mathematical induction that the inequality is true for all positive integers $( n )$: