I just can’t work away from my place. I am wasting a lot of time at Starbucks. I’ll move to a new place in one month, and I hope to do better then.

Let me describe a sequence of fields isomorphic to R via iterations of exp. The multiplication of each of them is the sum of the next one. I could vary the base, make it dependent of the index of the consecutive iteration but first let me keep it simple.

Let:

• e_-2  :=  -∞
• e_-1  :=  0   (thus  e_-1 = e^e_-2)
• e_n  :=  e^e_n-1   for every  n=0 1 …

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so that  e_n = eⁿ  for every  n=0 1 …. We have the standard addition and multiplication operations in

R = (e_-2; ∞)

namely

• #₀  :=  + : (e_-2; ∞)² → (e_-2; ∞)
• #₁  :=  * : (e_-2; ∞)² → (e_-2; ∞)

followed by more advanced admun (add-multiplication) operations:

#_n : (e_n-3; ∞)² → (e_n-3; ∞)

defined by:

exp(x) #_n exp(y)  :=  exp(x #_n-1 y)

for every  n = 2 3 …

Let  R_n  :=  (e_n-2); ∞)  for each  n=0 1…  Then

(R_n  #_n  #_n+1  e_n-1  e_n)

is a field, where:

• R_n  is the set of elements of the field;
• #_n  is the field’s addition operation (it’s commutative and associative);
• #_n+1  is the field’s multiplication operation (it’s commutative and associative);
• e_n-1  is the field addition’s neutral element (“zero”, even if it looks like something else )
• e_n  is the field multiplication’s neutral element (“one”, even if it looks like something else)

The Starbucks is about to close, I need to go (while Starbucks at Arborland is open 24h–should I write there before I move?).

Yes, I went home then, but now I am editing my entry (cheating? :-)).

They’ll close Starbucks in 13 minutes. I’ve spent here half a day and accomplished a big… NOTHING. I hope to be more efficient once I move to a new apartment.

I pushed Integland further at home, on my Eec. It’s good that it still works despite the battery going down the drain. I met the mild challenge presented by Integland (IntegLand?–either way it is a bit ugly). Now it’s just a chore, and I will have to force myself to continue, to finish.

I feel like writing about my old ideas related with the iteration of the isomorphism

exp : R → (0;∞)

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a#b := a^log(b)

Operation # is of course commutative and associative, it’s the multiplication in a disguise.

I should really go now, it’s 22:57.

Negative domains. Negative integers, Z^#

A set  A ⊆ Z  is called a negative domain   ⇐:⇒   the following two conditions are satisfied:0

• 1 ∉ A
• ∀_{x ∈ A}   x-1 ∈ A

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The union of any family of negative domains is a negative domain. In particular the union  Z^#  of all negative domains is called the set of negative integers. It contains every other negative domain.

The following theorem is straightforward:

THEOREM ??   A set of integers  B ⊆ Z  is a negative domain   ⇔   A := Z \ B  is a natural domain.

Since  N  is a natural domain, and  0 ∉ N, it follows that  0 ∈ Z^# — indeed,

0  ∈  Z \ N  ⊆  Z^#

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)

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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 }
• A_X  :=  { a ∈ Z : X + a ⊆ X }

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

• 1 ∈ X
• 1 ∈ A_X

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

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I am glad about my efforts related to my Integland. (Only “gland” in this name bothers me, it’s a minor nuisance).

Integland 0

Integland 1

Integland 2

Integland 3

Parity

Axioms 4 5 (see Integland 0) are concerned with inequalities:

• 1-x  ≠  x
• if  1-x = x-1  then x=1

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for all  x ∈ Z.

The latter axiom above admits the following equivalent version:

if  Neg(x) = x   then  x=0.

On the other hand the former axiom implies another two equivalent versions of itself:

• x+x  ≠  1
• x+(x-1)  ≠  0

Thus, as we already have seen it in Integland 1:

0  ≠  1

Our goal in this note is to get a complete picture with respect to some such inequalities. It is described by the following special case of the classical Euclid theorem about the integer division and remainder:

THEOREM 10 (Parity)   For every integer  z ∈ Z  there exists exactly one integer  x ∈ Z,  and exactly one value  e = 0 or 1,  such that:

z  =  x + x + e

PROOF   We will need several steps (observations).

Observation 0   Let  e f ∈ {0 1}.  Then

e-f  ∈  {Neg(1) 0 1}

Observation 1   Let  t ∈ Z.  Integer  t + t + Neg(1)  is of the form  x+x+e  (as in the theorem).

Indeed:   t + t + Neg(1)  =  (t-1) + (t-1) + 1

Observation 2   Integer  1  is of the form  x+x+e.

Indeed:   1 = 0+0+1

Observation 3   The difference of arbitrary two integers of the form  x+x+e  is of the same form.

Indeed:   Let  x+x+e  and  y+y+f  be two integers of the said form. Then:

(x+x+e) – (y+y+f)  =  (x-y) + (x-y) + (e-f)

If  e-f  is  0 or 1  then the claim holds.  If  e-f = Neg(1)  then apply observations 0 1.

Observation 4 (the first half of the theorem)   The set of all integers of the form  x+x+f  coincides with  Z.

Indeed:   Apply axiom 6 to observations 2 3.

Observation 5   Let  x y ∈ Z  be such that

x+x = y+y

Then  x = y

Indeed:   if  x+x = y+y  then:

x-y = y-x

hence

x-y = Neg(x-y)

Thus  x-y = 0,  i.e. x=y.

Observation 6   Let  x y ∈ Z  be such that

x+x+1 = y+y+1

Then  x = y

Observation 7   There does not exist any   x y ∈ Z  such that

x+x+1 = y+y     (not possible!)

Indeed:   if  x+x+1 = y+y   then

1 – (y-x)  =  y-x

in a contradiction to axiom 4 (substitute  x  of axiom 4 by  y-x).

Observation 8 (uniqueness–the second and last part of the theorem)   Let

x y e f  ∈  Z

be such that  e f  ∈  {0 1},  and

x + x + e  =  y + y + f

Then   x = y   e = f.

Indeed   This is what observations 5 6 7 tell us.

Addition  +

Define addition  + : Z² → Z  as follows:

∀_{x y ∈ Z}   x+y  :=  x – Neg(y)

It follows that:

THEOREM 4   ∀_{x y ∈ Z}   x-y  =  x + Neg(y)

PROOF

x + Neg(y)  =  x – Neg(Neg(y))  =  x-y

END of PROOF

Let  x=y  in the above theorem. Then we get:

∀_{x ∈Z}   x + Neg(x)  =  0

THEOREM 5   ∀_{x y ∈ Z}   x+y = y+x

PROOF

x+y  =  x – Neg(y)  =  x – (0 – y)  =  y – (0 – x)  =  y – Neg(x) =   y+x

END of PROOF

THEOREM 6   ∀_{x y ∈ Z}   Neg(x-y)  =  Neg(x) – Neg(y)

Neg(x-y)  =  y-x  =  y + Neg(x)  =  Neg(x) + y  =  Neg(x) – Neg(y)

END of PROOF
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THEOREM 7   ∀_{x y ∈ Z}   Neg(x+y)  =  Neg(x) + Neg(y)

PROOF

Neg(x+y)   =   Neg(x – Neg(y))

=   Neg(x) – Neg(Neg(y)) =   Neg(x) + Neg(y)

END of PROOF

THEOREM 8   ∀_{x y z∈ Z}   (x-y)-z  =  (x-z)-y  =  x – (y+z)

PROOF

x – (y+z)   =   x – (y – Neg(z))

=   Neg(z) – (y – x)   =   Neg(z) – Neg(x-y)

=  (x-y) + Neg(z)  =   (x-y) – Neg(Neg(z))   =   (x-y)-z

Now the rest is obvious.   END of PROOF

THEOREM 9   ∀_{x y z ∈ Z}   (x+y)+z  =  x+(y+z)

PROOF

    (a+b)+c  =   (a – Neg(b)) – Neg(c)   =   a – ((Neg(b) + Neg(c))

    =   a – Neg(b+c)   =   a + (b + c)

END of PROOF

Negative operation Neg(x)

Let’s define unary operation  Neg : Z → Z  as follows:

∀_{x ∈ Z}   Neg(x) := 0 – x

Thus

Neg(0)  =  0

THEOREM 2   ∀_{x y ∈ Z}   Neg(x-y) = y-x

PROOF

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END of PROOF

As a corollary we get:
THEOREM 3   ∀_{x ∈ Z}   Neg(Neg(x)) = x

PROOF

Neg(Neg(x))  =  Neg(0-x)  =  x-0  =  x

END of PROOF

Element 0

Consider integland  Z := (Z  –  1).  Let’s define

0  :=  1-1

so that (by axiom 2)

∀_{x ∈ Z}   x-x = 0

It also follows (see axiom 3) that:

∀_{x ∈ Z}   x-0 = x

and, by substituting  x  :=  1  in axiom 4:

0 ≠ 1

DEFINITION 1   Set  A ⊆ Z  is called a subminusop  ⇐:⇒ 

∀_{a b ∈ A}   a-b ∈ A

Instead of the long and cumbersome term subminusop I’ll be using its abbreviation submop.

THEOREM 0   Element  x := 0  is the only  x ∈ Z  such that the one-element set  {x}  is a submop.

Also

THEOREM 1   x = y   ⇔   x-y = 0

PROOF

• If  x=y  then  x-y = x-x = 0.   That’s the first implication.
• If  x-y = 0  then

  x  =7nbsp; x-0  =  x – (x – y)  =  y – (x – x)  =  y-0  =  y

  That’s the second implication

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END of PROOF