# Ordering on Natural Numbers is Trichotomy

## Contents

## Theorem

Let $\N$ be the natural numbers.

Let $<$ be the (strict) ordering on $\N$.

Then exactly one of the following is true:

- $(1): \quad a = b$
- $(2): \quad a > b$
- $(3): \quad a < b$

That is, $<$ is a trichotomy on $\N$.

## Proof

Applying the definition of $<$, the theorem becomes:

Exactly one of the following is true:

- $(1): \quad a = b$
- $(2): \quad \exists n \in \N_{>0} : b + n = a$
- $(3): \quad \exists n \in \N_{>0} : a + n = b$

We will use the principle of Mathematical Induction.

Let $P \left({a}\right)$ be the proposition that exactly one of the above is true, for all natural numbers $b$, for fixed natural number $a$.

### Basis for the Induction

We will prove that the proposition is true for $a = 0$.

Using Proof by Cases, we divide the proposition into two cases.

#### Case 1: $b = 0$

##### $(1)$ is true

It follows trivially from the values of $a$ and $b$.

##### $(2)$ is false

Aiming for a contradiction, suppose $c$ is a non-zero natural number such that:

- $b + c = a$

Substituting the values of $a$ and $b$, we obtain:

- $0 + c = 0$

By Natural Number Addition Commutes with Zero, we can simplify the equation to:

- $c = 0$

which is a contradiction of the assumption that $c$ is non-zero.

##### $(3)$ is false

Aiming for a contradiction, suppose $c$ is an non-zero natural number such that:

- $a + c = b$

Substituting the values of $a$ and $b$, we obtain:

- $0 + c = 0$

By Natural Number Addition Commutes with Zero, we can simplify the equation to:

- $c = 0$

which is a contradiction of the assumption that $c$ is non-zero.

#### Case 2: $b \ne 0$.

##### $(1)$ is false

It follows trivially from the fact that $a = 0$.

##### $(2)$ is false

Aiming for a contradiction, suppose $c$ is an non-zero natural number such that:

- $b + c = a$

Substituting the values of $a$, we obtain:

- $b + c = 0$

By Non-Successor Element of Peano Structure is Unique, there exists a natural number $d$ such that:

- $s \left({d}\right) = c$

where $s$ denotes the successor mapping.

Therefore, we have:

- $b + s \left({d}\right) = 0$

Applying the definition of addition in Peano structure, we get:

- $s \left({b + d}\right) = 0$

which is a contradiction of $(P4)$: $0$ is not in the image of $s$.

##### $(3)$ is true

By Natural Number Addition Commutes with Zero, we have:

- $a + b = b$

The result follows.

### Inductive hypothesis

It is now assumed that the proposition is true for $a = k$, where $k$ is a natural number.

That is, for all natural numbers $b$, exactly one of the following is true:

- $(1): \quad k = b$
- $(2): \quad k > b$
- $(3): \quad k < b$

Then, it is to be proved that the proposition is true for $a = s \left({k}\right)$.

That is, for all natural numbers $b$, exactly one of the following is true:

- $(1'): \quad s \left({k}\right) = b$
- $(2'): \quad s \left({k}\right) > b$
- $(3'): \quad s \left({k}\right) < b$

### Inductive step

We have:

\(\displaystyle s \left({k}\right)\) | \(=\) | \(\displaystyle s \left({k + 0}\right)\) | Definition of Addition in Peano Structure | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k + s \left({0}\right)\) | Definition of Addition in Peano Structure | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k + 1\) | Definition of One |

#### Case 1: $k = b$

In this case:

\(\displaystyle s \left({k}\right)\) | \(=\) | \(\displaystyle k + 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b + 1\) |

##### $(1')$ is false

Aiming for a contradiction, suppose $(1')$ is true.

Then:

\(\displaystyle s \left({k}\right)\) | \(=\) | \(\displaystyle b\) | Assumption | ||||||||||

\(\displaystyle b + 1\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle 1\) | \(=\) | \(\displaystyle 0\) | Natural Number Addition is Cancellable | ||||||||||

\(\displaystyle s \left({0}\right)\) | \(=\) | \(\displaystyle 0\) | Definition of One |

which is a contradiction of $(P4)$: $0$ is not in the image of $s$.

##### $(2')$ is true

This is apparent from the definition of $>$.

##### $(3')$ is false

Aiming for a contradiction, suppose $(3')$ is true.

Let $c$ be a non-zero natural number such that:

- $s \left({k}\right) + c = b$

Then:

\(\displaystyle s \left({k}\right) + c\) | \(=\) | \(\displaystyle b\) | Assumption | ||||||||||

\(\displaystyle \left({b + 1}\right) + c\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle b + \left({c + 1}\right)\) | \(=\) | \(\displaystyle b\) | Natural Number Addition is Associative and Natural Number Addition is Commutative | ||||||||||

\(\displaystyle c + 1\) | \(=\) | \(\displaystyle 0\) | Natural Number Addition is Cancellable | ||||||||||

\(\displaystyle c + s \left({0}\right)\) | \(=\) | \(\displaystyle 0\) | Definition of One | ||||||||||

\(\displaystyle s \left({c}\right)\) | \(=\) | \(\displaystyle 0\) | Definition of Addition in Peano Structure |

which is a contradiction of $(P4)$: $0$ is not in the image of $s$.

#### Case 2: $k < b$ and $s \left({k}\right) = b$

##### $(1')$ is true

This follows from the assumption.

##### $(2')$ is false

Aiming for a contradiction, suppose $(2')$ is true.

Let $c$ be a non-zero natural number such that:

- $s \left({k}\right) = b + c$

Then:

\(\displaystyle s \left({k}\right)\) | \(=\) | \(\displaystyle b + c\) | Assumption | ||||||||||

\(\displaystyle b\) | \(=\) | \(\displaystyle b + c\) | Assumption | ||||||||||

\(\displaystyle 0\) | \(=\) | \(\displaystyle c\) | Natural Number Addition is Cancellable |

which is a contradiction of our assumption that $c$ is non-zero.

##### $(3')$ is false

Aiming for a contradiction, suppose $(3')$ is true.

Let $c$ be a non-zero natural number such that:

- $s \left({k}\right) + c = b$

Then:

\(\displaystyle s \left({k}\right) + c\) | \(=\) | \(\displaystyle b\) | Assumption | ||||||||||

\(\displaystyle b + c\) | \(=\) | \(\displaystyle b\) | Assumption | ||||||||||

\(\displaystyle c\) | \(=\) | \(\displaystyle 0\) | Natural Number Addition is Cancellable |

which is a contradiction of our assumption that $c$ is non-zero.

#### Case 3: $k < b$ and $s \left({k}\right) \ne b$

From the definition of $<$, we have the existence of a non-zero natural number $m$ such that:

- $k + m = b$

By Non-Successor Element of Peano Structure is Unique, there exists a natural number $p$ such that:

- $s \left({p}\right) = m$

Substituting:

- $k + s \left({p}\right) = b$

Applying the definition of addition in Peano structure, we get:

- $s \left({k + p}\right) = b$

##### $(1')$ is false

It is assumed in this case that:

- $s \left({k}\right) \ne b$

Therefore, $(1')$ is false.

##### $(2')$ is false

Aiming for a contradiction, suppose $(2')$ is true.

Let $c$ be a non-zero natural number such that:

- $s \left({k}\right) = b + c$

By Non-Successor Element of Peano Structure is Unique, there exists a natural number $d$ such that:

- $s \left({d}\right) = c$

Then:

\(\displaystyle s \left({k}\right)\) | \(=\) | \(\displaystyle b + c\) | Assumption | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({k + s \left({p}\right)}\right) + s \left({d}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k + \left({s\left({p}\right) + s \left({d}\right)}\right)\) | Natural Number Addition is Commutative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k + s\left({p + s \left({d}\right)}\right)\) | Definition of Addition in Peano Structure | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle s\left({k + \left({p + s \left({d}\right)}\right)}\right)\) | Definition of Addition in Peano Structure | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle s\left({k + \left({s \left({p + d}\right)}\right)}\right)\) | Definition of Addition in Peano Structure | ||||||||||

\(\displaystyle k\) | \(=\) | \(\displaystyle k + \left({s \left({p + d}\right)}\right)\) | $(P3)$$:$ $s$ is injective | ||||||||||

\(\displaystyle 0\) | \(=\) | \(\displaystyle s \left({p + d}\right)\) | Natural Number Addition is Cancellable |

which is a contradiction of $(P4)$: $0$ is not in the image of $s$.

##### $(3')$ is true

We have:

\(\displaystyle s \left({k + p}\right)\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle s \left({p + k}\right)\) | \(=\) | \(\displaystyle b\) | Natural Number Addition is Commutative | ||||||||||

\(\displaystyle p + s \left({k}\right)\) | \(=\) | \(\displaystyle b\) | Definition of Addition in Peano Structure | ||||||||||

\(\displaystyle s \left({k}\right) + p\) | \(=\) | \(\displaystyle b\) | Natural Number Addition is Commutative |

The result follows.

#### Case 4: $k > b$

From the definition of $>$, we have the existence of a non-zero natural number $m$ such that:

- $k = b + m$

By Non-Successor Element of Peano Structure is Unique, there exists a natural number $p$ such that:

- $s \left({p}\right) = m$

Substituting:

- $k = b + s \left({p}\right)$

By applying $s$ to both sides, we obtain:

- $s \left({k}\right) = s \left({b + s \left({p}\right)}\right)$

Applying the definition of addition in Peano structure, we get:

- $s \left({k}\right) = b + s \left({s \left({p}\right)}\right)$

##### $(1')$ is false

Aiming for a contradiction, suppose $(1')$ is true.

Then:

\(\displaystyle s \left({k}\right)\) | \(=\) | \(\displaystyle b\) | Assumption | ||||||||||

\(\displaystyle b + s \left({s \left({p}\right)}\right)\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle s \left({s \left({p}\right)}\right)\) | \(=\) | \(\displaystyle 0\) | Natural Number Addition is Cancellable |

which is a contradiction of $(P4)$: $0$ is not in the image of $s$.

##### $(2')$ is true

We have:

- $s \left({k}\right) = b + s \left({s \left({p}\right)}\right)$

The result follows.

##### $(3')$ is false

Aiming for a contradiction, suppose $(3')$ is true.

Let $c$ be a non-zero natural number such that:

- $s \left({k}\right) + c = b$

Then:

\(\displaystyle s \left({k}\right) + c\) | \(=\) | \(\displaystyle b\) | Assumption | ||||||||||

\(\displaystyle \left({b + s \left({s \left({p}\right)}\right)}\right) + c\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle b + \left({c + s \left({s \left({p}\right)}\right)}\right)\) | \(=\) | \(\displaystyle b\) | Natural Number Addition is Associative and Natural Number Addition is Commutative | ||||||||||

\(\displaystyle \left({c + s \left({s \left({p}\right)}\right)}\right)\) | \(=\) | \(\displaystyle 0\) | Natural Number Addition is Cancellable | ||||||||||

\(\displaystyle s \left({c + s \left({p}\right)}\right)\) | \(=\) | \(\displaystyle 0\) | Definition of Addition in Peano Structure |

which is a contradiction of $(P4)$: $0$ is not in the image of $s$.

$\blacksquare$

## Sources

- 1972: A.G. Howson:
*A Handbook of Terms used in Algebra and Analysis*... (previous) ... (next): $\S 4$: Number systems $\text{I}$: Peano's Axioms