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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)
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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:
- 1 ∈ A
- ∀x ∈ A x+1 ∈ A
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 }
- AX := { a ∈ Z : X + a ⊆ X }
Of course X ⊆ Z is a natural domain ⇔ the following two conditions are satisfied:
- 1 ∈ X
- 1 ∈ AX
Furthermore,
AX + 1 ⊆ X
for every natural domain X.
Now let’s formulate a general and straightforward theorem:
THEOREM 14 Let X ⊆ Z. Then AX is an additive monoid, meaning that:
- 0 ∈ AX
- ∀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’ := AX + 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 ∈ AX.
We need to show that a’+1 ∈ A’.
Indeed, we already know that
1 a ∈ A
hence, by theorem 14, a’ := a+1 ∈ AX, thus A’ is a natural domain.
Now, let again a’ := a+1 ∈ A’, i.e. a ∈ AX. Then
- a’+a’ = ((a+1)+a) + 1 ∈ AX + 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