Integland practice, cnt. ??

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

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 practice, cnt. 5

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:

  • 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

Just 1h …

… is not enough for me even to warm up, to start. And now I need to rush to another appointment. I should not complain. I was treated very nicely at Starbucks, which I am about to leave. A kind employee offered me free large cappuccino. I have no idea why. 🙂

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

Integland practice, cnt. 4

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

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.

Integland practice, cnt. 3

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

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

Integland practice, cnt. 2

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

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

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

Integland practice, cnt. 1

Element 0

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

:=  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

END of PROOF

Integland practice

Let me jot things here before a (hopefully) smoother version will go to wlod.net. Too bad I don’t know how to use css at wordpress. It’d be nice to use the same style sheets as in wlod.net.

Integers are more elegant than natural numbers but they are elusive in that it’s harder to define them. I’ve searched for an elegant, direct definition of integers but never got anything aesthetically satisfying. Even a few days ago (but never in the past), after years of dealing with the problem (not too intensively but nevertheless), I made an embarrassing error of wishful thinking, of forgetting to exclude finite (cyclic) models. My present (final :-)) definition is fine but far from breath taking, too bad.

DEFINITION 0   Integland  is an ordered triple

Z  :=  (Z – 1)

where  Z  is a set,  – : Z2 → Z  is a binary operation,  1 ∈ Z,  and the following axioms are satisfied:

  1. x-(y-z)  =  z-(y-x)
  2. x-x  =  y-y
  3. x-(x-x)  =  x
  4. 1-x  ≠ x
  5. if  1-x = x-1  then x=1
  6. ((1 ∈ A ⊆ Z) & (∀a b ∈ A  a-b ∈ A))   ⇒   A = Z

for arbitrary  x y z ∈ Z.   END of DEFINITION

(Let’s see how it looks. I’ll write a series of short posts instead of suffering editing a long one).

End of the Starbucks shift

Just a few minutes left before they close at h.23:00, it’s 22:49 at this moment according to my PC. I started with creating my wh_mth.css style sheet. It was not a smooth sailing, but I got some of it. I can quickly complete it to a first stage. It’s helpful. … Well, I better go now.

Groups (again)

An obsession?

Since I’d like to write about number theory, I can’t an urge to introduce the ring of integers from scratch. Thus first I am about to post about Abelian groups. Since I am about to post about Abelian groups, I jusrt have to introduce groups in general. This my unrealistic obsession with systematic writing is my weakness. I don’t know the secret of pragmatic writing from the middle, which is an art mixed with self-assurance or arrogance, depending on the author.

I’ve already wrote about axioms of groups, together with the first simple theorems, in the past at least twice. Once on Chimeryd site, a long time ago. That site unfortunately has dissappeared. Thus a couple years ago I wrote about groups again. That text must be still somewhere on the Internet, in one of my sites/pages, but I was unable to find it. Ok, I am not happy about it but I’ll still write about groups for the third time (the last? :)).

It’s amazing that three simple and natural axioms have at the same time so many interesting consequences (some of them deep), and they define a theory–a true profound theory, not just a loosely defined theme– which admits a whole world of models (examples), from finite to infinite, from discrete to analytic, from easy to understand to most complex.

The importance of the groups comes from their actions on other mathematical objects, thus throwing light on the structure of these objects, as well as on the group itself (we study groups by studying their actions). Thus it is good to remember about the group transformation spirit of groups. Even when studying a group alone by itself it’s good to view its elements as acting at least on itself.

A quick intro to groups (a sketch).

Let me write here a preliminary sketch, before I’ll post a more finished version on the main part of wlod.net.

Axioms

DEFINITION   A group is an ordered quadruple

Γ  :=  (G * Inv e)

where  G  is a set,  e ∈ G,  * and Inv  are a unary and binary operation in&nbsp G,  such that the following axioms hold:

  1. (x*y)*z  =  x*(y*z)
  2. e*x  =  x
  3. Inv(x)*x  =  e

It’s good to interpret these axioms in terms of the action of the group on itself, meaning that with every element  a ∈ X  we associate a (left) map

La : G → G

given by:

x ∈ G   La(x)  := a*x

First goal

Axioms 2 3 above allow symmetric versions:

  • x*e  =  x
  • x*Inv(x)  =  e

I will show below that these symmetric variations are theorems, that they follow from the axioms 1-3 of the theory group. It’s an elementary result (obviously :-)) but not trivial. (It goes without saying that this topic is classical; I don’t claim any of the results, of course not, and only the exposition is mine, at least to my limited knowledge).

Simple general properties

THEOREM 0   e*e = e

PROOF   substitute  x := e  in axiom 2 above.

THEOREM 1   If  a*x = a*y  then  x=y,  for arbitrary  a x y ∈ X. In other words, transformation  La  is an injection of  X  into itself.

PROOF   If  a*x = a*y  then:

x = Inv(a)*a*x = Inv(a)*a*y = y

END of PROOF

REMARK   More explicitly, equation  a*x = b  has at the most one solution, namely

x :=Inv(a)*b

But is this  x  actually a solution? At this moment this is still an open question, which will be answered in positive soon.

THEOREM 2   ∀x ∈ X  x*e = x

PROOF

  • Inv(x)*(x*e) = (Inv(x)*x)*e = e*e = e
  • Inv(x)*x = e

Thus, by theorem 1,  x*e = x.  END of PROOF

THEOREM 3   ∀x ∈ X  x*Inv(x) = e

PROOF By Theorem 2:

Inv(x)*e  =  Inv(x)  =  e*Inv(x)

=   (Inv(x)*x)*Inv(x) = Inv(x)*(x*Inv(x))

and by Theorem 1,  x*Inv(x) = e.   END of PROOF