This should be very easy on paper (those who play such games can do it most likely in their head). But how will it go in real time on-line? It feels to me awkward but let me try for awhile.

First try

Axioms

Let’s consider system  (X -),  where  –  is a binary operation in  X,  which satisfies the following axioms:

1. x – (y – z)  =  z – (y – x)
2. x-x = y-y
3. x – (y-y)  =  x

for every  x y z ∈ X.

Consequences

From axioms  1. and 3.  (let z:=y)  we get:

Let’s define:

• 0  :=  x-x     (the choice of  x  is irrelevant)
• Neg(x)   :=  0 – x
• x+y  :=  x – Neg(y)

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Wen can rewrite axioms 3 as:

THEOREM 0

Neg(Neg(x)) = x

PROOF

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

END of PROOF

THEOREM 1

x+y = y+x

PROOF

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

END of PROOF

THEOREM 2

x+0 = x

PROOF

x+0  =  x – (0 – 0)  =  x – 0  =  x

END of PROOF

THEOREM 3

(x+y)+z  =  x+(y+z)

PROOF

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

z – ((0 – y) – (x – 0))  =  z – ((0 – y) – x)

By the same token (swap  x  and  z):

(z+y)+x  =  x – ((0 – y) – z)

But

x – ((0 – y) – z)  =  z – ((0 – y) – x)

hence

((x+y)+z  =  ((z+y)+x

and we already showed that operation + is commutative–therefore

((x+y)+z  =  (x+(y+z)

END of PROOF

THEOREM 4

x-y  =  x + Neg(y)

PROOF

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

END of PROOF

Ooooph

Oooph, I got lucky :-). It was hard for me to type, copy/paste, and manage html, and at the same time to think. Fortunately, I’ve selected the axioms well, and then there was very little left to think about, lucky me.

Now the road to integland is wide open–hurray!

Should I start another elementary mathematical topic (and which) or should I stop writing, and spend some time now on learning CSS (Cascading Style Sheets), and then JavaScript (in this order, because they are presented in my textbook in this order, and I’d like to be systematic, at least on this occasion)?

One of the topics is the beginning Number Theory. It’s rather clear that One should not start with Peano axioms, followed by the inductive definitions of the operations. It’d take too long and be too boring to get to the first juicy results. The start point can be a strong algebraic system, which includes the ordering of the natural or integer numbers. It’s a better option. But the list of axioms has to be quite extensive, which means back to boring. One would like to believe that elementary mathematics is simple but it’s not. And neither is the task of presenting elementary mathematics.

I could leave unclear what is assumed about natural or integer numbers. That’s what usually is done. But I would feel uncomfortable all the time.

Hey, I have a promising idea. In general, of the equivalent systems of axioms some seemingly weak simple ones and some seemingly very strong systems can be highly elegant, for different reasons of course. If you want to prove that a model satisfies the axioms then the weak system of axioms is useful. But when you want to derive new theorems then the strond system gives you a head start. In the theory of smooth manifolds John Milnor superbly selects axioms and definitions which are in between, neither weak nor very strong. It’s a true art. Beautiful and pragmatic at the same time. Let me attempt something similar for the beginning of the Number Theory. I’ll sketch my idea below (in the actual note the axioms will be spelled out in a more detailed way, and will be nicer–no Neg nor + will be mentioned). The axioms will describe only the group of rational integers. The rest–multiplication and ordering–will be defined afterward. Thus indeed this approach is stronger than Peano axioms, but avoids immediate definition of integers as an ordered ring.

DEFINITION   Integland  is an ordered triple   Z  :=  (Z  –  1), which satisfies the following axioms (this are not the ultimate axioms yet, but what the axioms should achieve):

1. Z  is a set;
2. 1 ∈ Z;
3. symbol  –  stands for a binary operation  – : Z² → Z;
4. 1-x ≠ x
5. if  1-x = x-1  then  x=1;
6. (Z + Neg 0)  is an abelian group, where  0 := 1-1,  and the unary operation  Neg,  and binary operation +:

   Neg : Z → Z   and   + : Z² → Z

   are defined as follows:

   Neg(x) := 0-x   and   x+y := x-Neg(y)

   for every  x y ∈ Z;

7. If  A ⊆ Z  is such that:
   • 1 ∈ A
   • ∀_{x y ∈ A}  x-y ∈ A

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END of DEFINITION

Now  *  is defined as the unique binary operation such that

• 1*x = x*1 = x
• (a-b)*x = a*x – b*x
• a*x = x*a

for every  a b x ∈ Z.

This is actually an inductive definition of *.

in order to get–withing a link–from here to the index page of wlod.net I need to go up twice. Otherwidse, until now, I was not able within this blog to create a working link to wlod.net. It was so frustrating!!!

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And I still am not able to write an absolute link from this blog to wlod.net. Am I hopelessly embarrassing myself? (Of course! What a stupid question :-)).

Tomorrow is a special day. Or taking into account time zones, it is already around.

I’ve checked once again on LaTeX on these pages. I am poor at such things. Nevertheless now I know, as expected, that one kind of latex works on the wordpress of this site. My knowledge and control over this important to me (to this site) issue is still almost non-existent.

At wlod.net I was so far using only html, even for mathematics. This was a step back. Should I get more ambitious? However I feel like developing the main part of this site outside the wordpress. I don’t feel free under wordpress. I’d like to limit wordpress on wlod.net just to my blog Blastog.

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I closed the topic of counting unicolor triangles. Getting the examples which showed that the bound of the main theorem (word main sounds so impossibly serious :-)) was rather trivial. The whole thing is very simple, just one simple idea, quite natural. Can it be new? Hard to believe. But then, why doesn’t it appear in presentations of bicoloring the complete 6-vertex graph  K₆?

Now it’s time for me to write about the probabilistic approach. I should also add a comment or to to what I’ve already written, especially about the relation of the 3 mod 4 case complication to the theorem 0 of combinatorics.

\[\LaTeX\] page:

$3^2-2^3=1$

Now a different mode:

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\[1+3+\ldots+(2\cdot n-1) = n^2\]

\[1^3+2^3+…+n^3 = (1+2+\ldots+n)^2\]

How was it?

Are adjectives unicolor and bicolor acceptable to native Anglo-Saxons? Should I use uni-colored and bi-colored instead? Can I use all four of them?

Anyway, in the past week or ten days I was writing about the count of unicolor triangles when the set of pairs (edges) is painted, each edge in one of the two colors, say blue and red. I don’t know if the result is new (perhaps not) or the method (I hope so). Most of this writing repeats my older notes. But I was happy that two days ago or so I got also something new–a very tiny something but it still makes me feel better. I have provided (simple!) extreme examples also in the

ξ ≡ 1 mod 4

case. Case  ξ ≡ 3 mod 4  is still open (to me).
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My counting idea is adequate for triangles of colored edges, but not for tetrahedra of colored triangles, not by itself–I need an additional fresh idea. One day I will get it.

I am in Starbucks. I forgot to bring my mouse, how dumb! and frustrating! Somehow today I am doing not bad without the mouse, using just the touch pad. I mean typing. Otherwise I am not doing much. Thus I better write some emails. Will I?

In less than 15m Starbucks gets closed but it’s almost full, everybody is comfortable, noboedy seems to be going home. It feels strange, surrealistic. A few minutes ago I looked for any employee, to ask about the closing time, just in case. I couldn’t find any. They happened to hide in the back in their room. TThey have cleared the shelves of the food, so it does look like the end of the day here. And still…

Somehow time goes by. Can I do anything about it? Do I have any ability to do anything about getting more intensive?

It is strange today. Ordinarily, they would already be cleaning the place for the night, sweeping the floor, but not yet, not tonight.

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Finally, a single guy a couple minutes ago, and a couple right now, have left. But then, there are only 7m left!

Yes, they are closing in a few minutes. It’s just a bit different today. What a pity! I’d love to stay for another hour, at least.

I am at Barnes & Noble for a change. My neighborhood Starbucks was full around 11am, and again in the early afternoon, except for one sit near an obnoxious guy (I had a displeasure of suffering his presence at Starbucks a week or so ago). Hence the bookstore.

Only now I see that pressing “+ New” at the top menu line of WordPress opens a drop-down menu, which offers an option of creating a new page or a new post (plus 3 other options).

It’s my 4th time in Michigan. This time I moved from California. Michigan climate is known to be brutal, and Californian to be the best. But these days it’s beautiful here, and I like it better for its authentic feel, while California climate (at its best) feels sweet like a candy bar.

There is so little time, and I am so slow, slower than in the past. Time is taken also by some annoying task which should never be needed but they are. Not to mention the necessary evil (:-)) of dental and medical appointments.

I am writing the mathematical part of this site in a flexible way. Simple ad hoc notes will be loosely placed straight in the Mathematics folder. More systematic notes will be organized into subfolders of Mathematics.
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DIGRESSION   If this were the regular part of my site, not WordPress, I’d make a relative link to folder Mathematics. But here, under WordPress, I don’t know where is what, and I don’t want to devote time to correct creating a relative link, I made an absolute one–relative link is preferable!

This blog is for me to relax, to jot some plans and items so that I will have a place to check on them later, will remember about them this way. We will see.

Oh, WordPress editor displays the word count of the edited text just underneath, how nice! 🙂 There were 312 words, not counting this very sentence :-).

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