Integland 0

Integland 1

Integland 2

Integland 3

Integland 4

# Natural numbers

## Doubling domains and natural numbers

I’ll call a set of integers A ⊆ **Z** a doubling domain ⇐:⇒ the following two conditions are satisfied:

- 1 ∈ A
- ∀
_{x ∈ A} ( x+x and x+x+1 ∈ A)

For instance the whole set **Z** is a doubling domain. Now the set of natural numbers **N** is defined as the intersection of all doubling domains. Obviously **N** itself is a doubling domain. Thus it is the smallest doubling domain, meaning that **N** is contained in every other doubling domain.

**THEOREM 11** 0 is not a natural number, i.e. 0 ∉ **N**.

**PROOF** Define

N’ **:=** **N** \ {0}

Since 1 ≠ 0, we have 1 ∈ N’. Next, consider arbitrary x ∈ N’, so that x ≠ 0. Then integers x+x and x+x+1 both belong to **N**, and–by parity Theorem 10–both these integers are different from 0 = 0+0. It follows that both integers x+x and x+x+1 belong to N’. We have proved that N’ is a doubling domain. Obviously N’ is a subset of **N**, and vice versa. Thus **N** = N’, which means that 0 is not natural.

**END of PROOF**

**THEOREM 12** Every natural number z ∈ **N** is either 1 or of the form z = x+x+e, where x is natural, and e = 0 or 1.

**PROOF** Consider

N’ **:=** {1} ∪ { x+x+e : x ∈ **N** and e = 0 or 1 }

Obviously 1 ∈ N’ ⊆ **N**. Thus it follows that:

∀_{x ∈ N’} x+x x+x+1 ∈ N’

In other words, N’ is a doubling domain. Since N’ is contained in **N**, the two are identical:

**N** = N’

**END of PROOF**

## Natural domains

I’ll call a set of integers A ⊆ **Z** a natural domain ⇐:⇒ the following two conditions are satisfied:

For instance the whole set **Z** is a natural domain.

**THEOREM 13** The set of natural numbers **N** is a natural domain.

**PROOF** Let

N’ **:=** { x ∈ **N** : n+1 ∈ **N** }

Then 1 ∈ N’. Furthermore, if x ∈ N’ then x and x+1 belong to **N**, hence

x+x x+x+1 (x+1)+(x+1) (x+1)+(x+1)+1 ∈ **N**

The first and second element above show that x+x ∈ N’. The second and then third element above, since

(x+1)+(x+1) = (x+x+1) + 1

show that x+x+1 ∈ N’. Thus N’ is a doubling domain, contained in **N**, hence **N** = N’. Thus **N** is also a natural domain.

**END of PROOF**

## Properties of natural domains

First notation:

- X + a
**:=** { x+a : x ∈ X }
- A
_{X} **:=** { a ∈ **Z** : X + a ⊆ X }

Of course X ⊆ **Z** is a natural domain ⇔ the following two conditions are satisfied:

Furthermore,

A_{X} + 1 ⊆ X

for every natural domain X.

Now let’s formulate a general and straightforward theorem:

**THEOREM 14** Let X ⊆ **Z**. Then A_{X} is an additive monoid, meaning that:

- 0 ∈ A
_{X}
- ∀
_{a b ∈ AX a+b ∈ AX}

Now I’ll formulate one of the main results of this note:

**THEOREM 15** The set of natural numbers **N** is contained in every natural domain, i.e. it is, in this sense, the smallest natural domain.

This theorem follows immediately from the following more detailed theorem:

**THEOREM 16** Let X be an arbitrary natural domain. Let

A’ **:=** A_{X} + 1

Then

- A’ is a natural domain
- A’ is a doubling domain
**N** ⊆ A’ ⊆ X

**PROOF** Of course 1 ∈ A’. Additionally, for the first part of our theorem, let

a’ **:=** a+1 ∈ A’, i.e. a ∈ A_{X}.

We need to show that a’+1 ∈ A’.

**Indeed**, we already know that

1 a ∈ A

hence, by theorem 14, a’ **:=** a+1 ∈ A_{X}, thus A’ is a natural domain.

Now, let again a’ **:=** a+1 ∈ A’, i.e. a ∈ A_{X}. Then

- a’+a’ = ((a+1)+a) + 1 ∈ A
_{X} + 1 = A’, and now:
- a’+a’+1 ∈ A’ since A’ is a natural domain.

Thus A’ is a doubling domain, while **N** is the smallest doubling domain. It follows that **N** ⊆A’. We already saw earlier that A’ ⊆X. This completes the proof.

**END of PROOF**