Introduction

This document presents a rigorous construction of the natural numbers from axiomatic foundations. We develop the complete theory of $\mathbb{N}$ starting only from the Peano Axioms, proving even seemingly “obvious” properties through careful logical reasoning.

Motivation

The Counting Problem

Consider ancient mathematicians attempting to answer basic questions:

“How many sheep do I have?”
“Is my flock larger than yours?”
“If I combine our flocks, how many total?”

They had available:

Physical objects (pebbles, fingers, tally marks)
The ability to match objects one-to-one
The concept of “one more”

What was difficult: How do you define “number” without using numbers? How do you ensure everyone means the same thing by “three”? How do you prove $3 + 2 = 2 + 3$ without just “seeing” it?

Historical Failures

Concrete Counting (Egyptian, ~3000 BCE)

The Egyptians used different symbols for 1, 10, 100, 1000.

Problem:

No systematic way to extend indefinitely. What comes after the largest symbol invented?

Geometric Representation (Greek, ~500 BCE)

Numbers were represented as lengths and areas.

Problem:

Cannot represent irrational numbers as ratios. The discovery that $\sqrt{2}$ exists geometrically but not arithmetically in their system led to crisis.

Intuitive Infinity (Pre-Cantor, ~1850)

Mathematicians worked with vague notions that “numbers go on forever.”

Problem:

Led to paradoxes about infinite sets. Questions like “Is $\infty + 1 = \infty$?” had no clear answer.

Peano’s Breakthrough (1889)

Giuseppe Peano asked: What is the absolute minimum we must assume to make arithmetic work?

His key insight: We don’t need to define what numbers “are”—we only need to specify how they behave.

Rejected approach:

Trying to define “three” as “the symbol after two” (circular!)
Trying to define numbers as sets of objects (conflates number with quantity)

Successful approach:

Start with one primitive object: 0 (or 1 in original formulation)
Define one primitive operation: successor $S(n)$ = “the next number”
State axioms characterizing how successor behaves
Everything else (addition, multiplication, exponentiation) derives from these axioms

This approach revolutionized mathematics by showing how to axiomatize mathematical structures. It inspired:

Zermelo-Fraenkel axioms for set theory
Hilbert’s axioms for geometry
Modern abstract algebra

Intuitive Understanding

Before seeing the formal axioms, we develop three equivalent mental models for the natural numbers.

Three Mental Models

Model 1: Chain of Beads

0 ——→ 1 ——→ 2 ——→ 3 ——→ 4 ——→ 5 ——→ ...
     S      S      S      S      S

Imagine an infinite chain of beads:

First bead is labeled 0 (starting point)
Each bead has exactly one string attached leading to the next bead (successor)
No bead has a string leading back to 0 (no cycles)
The chain continues forever (infinite)
You can reach any bead by starting at 0 and following strings

Key features:

Linear (not branching, not circular)
Infinite (no last bead)
Connected (every bead reachable from 0)
Directed (arrows go one way)

Model 2: Stepladder

        ...
         ║
        ║ 3
         ║
        ║ 2
         ║
        ║ 1
         ║
Ground ▓▓▓ 0

Imagine an infinite ladder reaching upward:

Ground level = 0 (you start here)
Each rung is one step up (successor function)
You can always climb one more rung (no highest step)
You can’t skip rungs (must go $0 \to 1 \to 2 \to 3$, not $0 \to 5$ directly)
You can never climb below ground level (no negative rungs in $\mathbb{N}$)

This model emphasizes:

Well-ordered: Clear notion of “before” and “after”
Discrete steps: No “between” 3 and 4 (unlike real numbers)
Grounded: Everything builds from 0

Model 3: Construction Process

Stage 0: {0}
Stage 1: {0, 1}           where 1 = S(0)
Stage 2: {0, 1, 2}        where 2 = S(1) = S(S(0))
Stage 3: {0, 1, 2, 3}     where 3 = S(2) = S(S(S(0)))
...
Stage ω: ℕ = {0, 1, 2, 3, ...}

Think of natural numbers as iterative construction:

We start with just 0
At each stage, we add the successor of the largest element so far
Continue this process infinitely
$\mathbb{N}$ is exactly what this process generates

This model emphasizes:

Generative: Numbers are built from simpler numbers
Minimal: $\mathbb{N}$ contains only what this process generates (nothing extra)
Recursive: Each stage uses the previous stage

The Core Principle

All three models reveal the same fundamental insight: recursive structure.

To define what happens at stage $n$, we refer to stage $n-1$. This recursive pattern manifests as:

Inductive definition (define complex things from simpler things)
Structural recursion (build on previous structure)
Mathematical induction (prove for $n$ by assuming true for $n-1$)

This recursive structure is why the Peano axioms work and why mathematical induction is the natural proof technique for $\mathbb{N}$.

Edge Cases and Counterexamples

These mental models help us predict what properties the formal axioms must satisfy.

What if we had cycles?

Consider clock arithmetic modulo 5:

Where $S(4) = 0$ (loops back). This violates the chain/ladder/construction models:

Chain: String leads back to first bead
Ladder: Can climb back down to ground
Construction: Process doesn’t generate anything new after 5 steps

This structure satisfies some natural properties but lacks others. Distinguishing $\mathbb{N}$ from finite cycles requires an explicit axiom.

What if elements were unreachable from 0?

Consider $\mathbb{N} \cup {\infty}$ where $\infty$ is isolated:

0 → 1 → 2 → 3 → ...  (reachable chain)

        ∞             (unreachable element)

This violates the construction model: $\infty$ is not generated by repeatedly applying successor to 0.

An axiom ensuring minimality—that $\mathbb{N}$ contains exactly what’s reachable from 0—is necessary.

The Peano Axioms

We now make our intuitive understanding precise. The natural numbers are characterized by:

A set $\mathbb{N}$
A distinguished element $0 \in \mathbb{N}$
A function $S: \mathbb{N} \to \mathbb{N}$ (the successor function)

These must satisfy five axioms.

Axiom 1: Existence

Axiom 2: Closure

Axiom 3: Injectivity

$$\forall n, m \in \mathbb{N}, \quad S(n) = S(m) \implies n = m$$

Contrapositive form (equivalent):

$$\forall n, m \in \mathbb{N}, \quad n \neq m \implies S(n) \neq S(m)$$

In words:

Different natural numbers have different successors.

This prevents “collisions”—situations where two different numbers have the same successor.

Distinction from function property:

Function (Axiom 2): Each input has exactly one output
Injective (Axiom 3): Different inputs have different outputs

All injective functions are functions, but not all functions are injective.

Visual representation:

Not injective:
3 → 5    (collision - both map to 5)
4 ↗

Injective:
3 → 5
4 → 6    (different outputs)

Axiom 4: Zero is Not a Successor

Axiom 5: Induction Principle

Set-theoretic form:

Let $P \subseteq \mathbb{N}$. If:

$0 \in P$, and
$\forall n \in \mathbb{N}, , n \in P \implies S(n) \in P$

then $P = \mathbb{N}$.

Predicate form:

Let $P(n)$ be a property of natural numbers. If:

$P(0)$ is true, and
$\forall n \in \mathbb{N}, , P(n) \implies P(S(n))$

then $P(n)$ is true for all $n \in \mathbb{N}$.

In words:

If a set $P$ contains 0, and whenever $P$ contains $n$ it also contains $S(n)$, then $P$ contains all natural numbers.

Two purposes:

Proof technique: Gives us mathematical induction as a method
Minimality (more fundamental): Ensures $\mathbb{N}$ is the minimal set satisfying Axioms 1-4

The induction axiom prevents “ghost elements” that are not reachable from 0 by repeatedly applying successor. It ensures:

$$\mathbb{N} = {0, S(0), S(S(0)), S(S(S(0))), \ldots}$$

and nothing more.

Domino analogy:

Condition (i): First domino falls
Condition (ii): If domino $n$ falls, then domino $S(n)$ falls
Conclusion: All dominoes fall

Understanding the Axioms

Why Each Axiom is Necessary

We examine what breaks when each axiom is removed or violated.

Without Axiom 1 (Existence)

Violating structure:

Empty set $\emptyset$

What breaks:

No starting point. Cannot begin building anything.

Without Axiom 2 (Closure)

Violating structure:

$S(3)$ is undefined, or $S(3) \notin \mathbb{N}$

What breaks:

Cannot always compute successor
Induction fails (cannot proceed from $n$ to $S(n)$ if $S(n)$ doesn’t exist in $\mathbb{N}$)
Arithmetic becomes impossible: What is $3 + 1$ if $S(3)$ is undefined?

Example:

0 → 1 → 2 → 3
             ↓
           (dead end, S(3) undefined)

Without Axiom 3 (Injectivity)

Violating structure:

Colliding paths

0 → 1 → 2
         ↓
3 → 4 → 5 → 6 → ...
    ↑

Where $S(2) = 5$ and $S(4) = 5$ (collision at 5)

What breaks:

Addition ill-defined: If we define $n + 1 = S(n)$, then $2 + 1 = S(2) = 5$, but also $4 + 1 = S(4) = 5$. This implies $2 + 1 = 4 + 1$, so $2 = 4$. Contradiction!
No unique representation: The number 5 can be written as $S(2)$ or $S(4)$. Which predecessor is “real”?
Cannot invert $S$ uniquely: If $S(n) = 7$, is $n = 6$ or something else? Ambiguous!

Without Axiom 4 (Zero Not Successor)

Violating structure:

Clock arithmetic $\mathbb{Z}/5\mathbb{Z}$

Where $S(4) = 0$ (loops back)

What breaks:

No first element: Every element has a predecessor. Cannot identify a unique starting point for induction.
Cannot distinguish finite from infinite: This structure is finite (5 elements), but without Axiom 4, we cannot rule it out as a model of $\mathbb{N}$.
Arithmetic differs: In $\mathbb{Z}/5\mathbb{Z}$, we have $3 + 2 = 0$ (modular arithmetic), whereas in $\mathbb{N}$, we have $3 + 2 = 5$ (standard arithmetic).
Well-ordering fails: In $\mathbb{N}$, every non-empty subset has a least element. In circular structures, no minimum exists.

Without Axiom 5 (Induction)

Violating structure:

$\mathbb{N} \cup {\infty}$ where $S(\infty) = \infty$

0 → 1 → 2 → 3 → ...  (reachable from 0)

        ∞             (isolated, unreachable)

Check Axioms 1-4:

✓ Axiom 1: $0$ exists
✓ Axiom 2: $S(\infty) = \infty \in \mathbb{N} \cup {\infty}$
✓ Axiom 3: $S$ is injective (different elements map to different successors)
✓ Axiom 4: No element maps to 0

Axiom 5 fails:

Let $P = {0, 1, 2, 3, \ldots}$ (standard naturals, excluding $\infty$)

(i) $0 \in P$ ✓
(ii) If $n \in P$, then $S(n) \in P$ ✓

But $P \neq \mathbb{N} \cup {\infty}$ because $\infty \notin P$!

What breaks:

Non-uniqueness: Which structure is THE natural numbers? $\mathbb{N}$ alone? $\mathbb{N} \cup {\infty}$? $\mathbb{N} \cup {\omega, \omega+1, \omega+2, \ldots}$?
Induction as proof technique fails: We can prove properties for ${0, 1, 2, \ldots}$ but not for ghost elements.
Recursive definitions don’t work on ghost elements: If we recursively define operations, they only apply to reachable elements.

Connecting Formal to Intuitive

Return to our three mental models:

Chain of Beads:

Axiom 1: First bead exists (0)
Axiom 2: Every bead has a string to next bead (closure)
Axiom 3: No two beads connect to same next bead (injective)
Axiom 4: No string connects back to first bead (no cycle)
Axiom 5: Chain contains only beads reachable from first (minimality)

Ladder:

Axiom 1: Ground level exists
Axiom 2: Every rung has a rung above it
Axiom 3: No two rungs lead to same upper rung
Axiom 4: No rung leads down to ground
Axiom 5: Ladder contains only rungs reachable from ground

Construction:

Axiom 1: Process starts with {0}
Axiom 2: Process can always add successor
Axiom 3: New element is distinct from all previous
Axiom 4: Process never returns to 0
Axiom 5: $\mathbb{N}$ is exactly what process generates

The formal axioms precisely capture these intuitive pictures.

Key Properties

We now derive fundamental properties from the axioms alone. Even “obvious” facts require proof.

Property 1: Zero is Unique

Proof:

We know from Axiom 4 that 0 is not a successor.

Uniqueness: Suppose $x$ is another element that’s not a successor, with $x \neq 0$.

Consider the set $P = {0, S(0), S(S(0)), S(S(S(0))), \ldots}$—all elements reachable from 0.

Check induction conditions:

(i) $0 \in P$ ✓ (by definition of $P$)

(ii) If $n \in P$, then $S(n) \in P$ ✓ (by construction of $P$)

By Axiom 5 (induction): $P = \mathbb{N}$

But $x \notin P$ (since $x$ is not a successor and $x \neq 0$, so $x$ is not reachable from 0).

This contradicts $P = \mathbb{N}$.

Therefore, no such $x$ exists. Zero is the unique minimal element. ∎

Property 2: Every Non-Zero Element is a Successor

Proof by Induction:

Let $P = {0} \cup {n \in \mathbb{N} : n = S(m) \text{ for some } m}$

In words: $P$ consists of 0 and all successors.

Claim: $P = \mathbb{N}$

Base case: $0 \in P$ ✓ (by definition of $P$)

Inductive step: Suppose $n \in P$. Then $S(n) \in P$ because $S(n) = S(n)$ (it’s a successor) ✓

By Axiom 5 (induction): $P = \mathbb{N}$

Therefore, every element is either 0 or a successor. ∎

Property 3: Successor is Bijection onto $\mathbb{N} \setminus {0}$

Proof:

Injective: Axiom 3 directly states $S$ is injective. ✓

Surjective onto $\mathbb{N} \setminus {0}$: Property 2 states every non-zero element is a successor. ✓

Therefore $S$ is a bijection from $\mathbb{N}$ to $\mathbb{N} \setminus {0}$. ∎

Consequence:

We can define a predecessor function $P: \mathbb{N} \setminus {0} \to \mathbb{N}$:

$$P(n) = \text{the unique } m \text{ such that } S(m) = n$$

This exists and is unique by the bijection property.

Property 4: $\mathbb{N}$ is Infinite

Proof by Contradiction:

Assume, for contradiction, that $\mathbb{N}$ has a largest element $k$.

Then $\mathbb{N} = {0, 1, 2, \ldots, k}$ for some $k \in \mathbb{N}$.

Apply Axiom 2 (Closure): Since $k \in \mathbb{N}$, we have $S(k) \in \mathbb{N}$ by Axiom 2.

Derive contradiction:

We show $S(k)$ cannot equal any element of ${0, 1, 2, \ldots, k}$:

$S(k) \neq 0$ by Axiom 4 ✓

$S(k) \neq 1, 2, \ldots, k$ because these are $S(0), S(1), \ldots, S(k-1)$ respectively, and $S$ is injective (Axiom 3). If $S(k) = S(i)$ for $i < k$, then $k = i$ by injectivity, contradicting $k > i$. ✓

Therefore $S(k) \notin {0, 1, 2, \ldots, k} = \mathbb{N}$.

But this contradicts Axiom 2, which states $S(k) \in \mathbb{N}$.

Conclusion: Our assumption was false. $\mathbb{N}$ has no largest element. ∎

Alternative formulation:

By induction, the first $k+1$ elements generated from 0 are distinct:

$$0, S(0), S^2(0), \ldots, S^k(0)$$

are all different (by injectivity and Axiom 4). If $\mathbb{N}$ had exactly $k+1$ elements, then:

$$\mathbb{N} = {0, S(0), S^2(0), \ldots, S^k(0)}$$

But then $S(S^k(0)) = S^{k+1}(0)$ must equal one of these elements. It cannot equal 0 (Axiom 4), and it cannot equal $S^i(0)$ for $i < k$ (by injectivity). Contradiction!

Property 5: Successor Has No Fixed Points

Proof by Induction:

Let $P(n)$ be the statement “$S(n) \neq n$”.

Base case ($n = 0$):

By Axiom 4, $S(n) \neq 0$ for all $n \in \mathbb{N}$.

In particular, $S(0) \neq 0$. ✓

So $P(0)$ is true.

Inductive step:

Inductive Hypothesis (IH): Assume $S(k) \neq k$ for some $k \in \mathbb{N}$.

Goal: Prove $S(S(k)) \neq S(k)$.

Proof by contradiction:

Suppose $S(S(k)) = S(k)$.

By Axiom 3 (injectivity): If $S(a) = S(b)$, then $a = b$.

Applying this with $a = S(k)$ and $b = k$:

$$S(S(k)) = S(k) \implies S(k) = k$$

But this contradicts our inductive hypothesis that $S(k) \neq k$!

Therefore our assumption was false: $S(S(k)) \neq S(k)$. ✓

Conclusion:

By the principle of mathematical induction (Axiom 5):

$P(0)$ is true

For all $k$, $P(k) \implies P(S(k))$

Therefore $P(n)$ is true for all $n \in \mathbb{N}$.

That is, $S(n) \neq n$ for all $n \in \mathbb{N}$. ∎

Defining Addition

We now construct addition from first principles.

The Problem

We have:

$\mathbb{N}$ (set of natural numbers)
$0$ (the starting element)
$S$ (successor function)

We want:

“$+$” (addition operation)

But “$+$” doesn’t exist yet!

We must create it using only what the axioms give us.

Recursive Definition Strategy

Key insight:

We cannot define all additions at once (infinitely many pairs!).

Solution:

Define addition recursively:

Give a rule for the simplest case (base case: adding 0)
Give a rule for reducing complex cases to simpler cases (recursive case: adding $S(m)$ in terms of adding $m$)

This is analogous to:

Defining what happens when you add 0 (base)
Defining what happens when you add “one more,” given you know how to add the previous amount (recursive)

Definition of Addition

For any $n, m \in \mathbb{N}$, we define $n + m$ by recursion on $m$:

Base Case

$$n + 0 = n$$

Meaning:

Adding zero to anything gives that thing back.

Why this choice?

Makes 0 the “additive identity”
Matches intuition: “3 apples plus 0 apples = 3 apples”
Provides the base for our recursion

Important:

This is a definition, not a theorem from any axiom. We choose to define addition this way.

Recursive Case

$$n + S(m) = S(n + m)$$

Meaning:

To add $n$ to the successor of $m$, take the successor of $(n + m)$.

In words:

“Adding to one-more-than-$m$ is the same as doing one-more-than (adding to $m$)”

Why this choice?

Breaks down $n + S(m)$ into a simpler problem: $n + m$
We “peel off” the $S$ from the right, wrap it around the result
Matches intuition: $3 + 5 = 3 + (4+1) = (3+4) + 1$

How the Definition Works

Example 1: Computing $3 + 2$

Notation:

$1 = S(0)$
$2 = S(S(0)) = S(1)$
$3 = S(S(S(0))) = S(2)$

Computation:

$$ \begin{aligned} 3 + 2 &= 3 + S(1) && \text{[because } 2 = S(1)\text{]} \ &= S(3 + 1) && \text{[recursive rule: } n + S(m) = S(n+m)\text{]} \ &= S(3 + S(0)) && \text{[because } 1 = S(0)\text{]} \ &= S(S(3 + 0)) && \text{[recursive rule again]} \ &= S(S(3)) && \text{[base case: } n + 0 = n\text{]} \ &= S(S(S(S(S(0))))) && \text{[because } 3 = S(S(S(0)))\text{]} \ &= 5 && \text{[by definition of 5]} \end{aligned} $$

What happened:

We applied the recursive rule repeatedly
Each time, we “peeled off” one $S$ from the right
We “wrapped” that $S$ around the outside
Eventually we reached $+ 0$, where we used the base case
We ended up with the right number of $S$’s wrapped around 0

Example 2: Computing $0 + 3$

Is this the same as $3 + 0$?

$$ \begin{aligned} 0 + 3 &= 0 + S(2) && \text{[because } 3 = S(2)\text{]} \ &= S(0 + 2) && \text{[recursive rule]} \ &= S(0 + S(1)) && \text{[because } 2 = S(1)\text{]} \ &= S(S(0 + 1)) && \text{[recursive rule]} \ &= S(S(0 + S(0))) && \text{[because } 1 = S(0)\text{]} \ &= S(S(S(0 + 0))) && \text{[recursive rule]} \ &= S(S(S(0))) && \text{[base case]} \ &= 3 && \text{[by definition]} \end{aligned} $$

So $0 + 3 = 3$ ✓

Notice:

We needed to use the recursive rule 3 times, then the base case.

Compare with $3 + 0$:

$$3 + 0 = 3 \quad \text{[IMMEDIATELY by base case!]}$$

This is much simpler!

A Shocking Truth

From the definition alone:

✓ We know $n + 0 = n$ (that’s the base case—part of the definition)
✗ We don’t immediately know $0 + n = n$ (must prove as a theorem!)
✗ We don’t immediately know $S(n) + m = S(n + m)$ (must prove!)
✗ We don’t immediately know $m + n = n + m$ (must prove!)

What’s Next

Part 2 picks up from here and proves the remaining structure:

Left identity: $0 + n = n$ — surprisingly non-trivial!
Left successor property: $S(m) + n = S(m + n)$ — harder!
Commutativity: $m + n = n + m$ — very hard, requires previous results!
Associativity: $(a + b) + c = a + (b + c)$
Multiplication, defined recursively from addition
Left identity, left successor, commutativity, distributivity, and associativity for multiplication

Each theorem builds on previous results, forming a logical hierarchy from the simple Peano Axioms to the full arithmetic structure.

Continue to Operations on Natural Numbers (Part 2).