Observable sets

It is important to know the normal world. Arts and sciences are complicated, but we need to perceive the common sense. I will describe it mathematically.

1=objects, 2=devices or machines (eyes), 3=observers (brain)

This case is very complicated, and it is almost artificial intelligence, according to John von Neumann.
At first, our eyes are not devices like Google Glass.
When you see the objects, the information is sent to your brain. If your brain were broken like me, this must be disorder.

Math isn't chaos.

However, your eyes and brain are connected, so your perception may be sometimes different from others. You are the subject, not objects like devices. You may say that reality is abstract. You are also egocentric. This is our existential problem. You may be the transcendence like Buddha, and you are myself and there is no boundary in this world. I see and you also see. This is integration, no duality.

Math isn't religion.

We share common knowledge. This is commutative.

	A	B	C	D	E	F	G
A	○	×	○	○	○	○	○
B	×	○	×	○	○	○	○
C	○	×	○	○	○	○	○
D	○	○	○	○	○	○	○
E	○	○	○	○	○	×	○
F	○	○	○	○	×	○	○
G	○	○	○	○	○	○	○

[X,Y]=XY-YX≠0→×=noncommutative

[X,Y]=XY-YX=0→○=commutative

A and B are noncommutative, and B and C are also noncommutative. They are linked.

S={A,B,C,D}

S and S' are commutative. A-B-C is noncommutative, so it isn't S'. D and G are commutative, so they are S'. E-F is noncommutative, but D and G are commutative. Therefore, it is S'.

S'={D,E,F,G}

S'' is double commutative, so it includes D and G.

D is the center, but G connect S and S'.

D and G are commutative, so S and S'' are commutative.

S=S''

Quaternion

Prime numbers include the imaginary space in two dimensions. William Rowan Hamilton has found the quaternions. It extends complex numbers.

H is the sum of quaternions.
R is space of numerical vectors.

H={(a,b,c,d)|a,b,c,d∈R}
H=a+bi+cj+dk

i^2=j^2=k^2=ijk=-1

-1=ijk
-i=i^2jk
-i=-jk
∴i=jk

×	1	i	j	k
1	1	i	j	k
i	i	-1	k	-j
j	j	-k	-1	i
k	k	j	-i	-1

∴
ij=k, ji=-k, jk=i, kj=-i, ki=j, ik=-j

ij≠ji

H=a+bi+cj+dk (i=jk,j=ki,k=ij)

You are dazing.

Hausdorff space

It is hard to define emptiness. Hausdorff space is known as empty set Φ.



X is a topological space.

X=(x,y)

x and y are dots.

U∩V=Φ

U is vicinities of x.
V is vicinities of y.

X×X→Δ={(x,x)|x∈X} is closed.

Busy Beaver

Gregory Chaitin has written about mathematical evolution. If you handled more numbers, your information would be reliable. This is expansion and power law.

It is also called "Busy beaver".

N≦N+N,N×N,N^N,N^N^N^N^N

When you added more numbers, uncertainty would be decreased.

Ω mean the stability or the end.

Ω=Σ2^-P
Ωk=Σ2^-P
Ωk≦Ωk+1

Goldbach's weak conjecture

Every odd number greater than 5 can be expressed as the sum of three primes. (A prime may be used more than once in the same sum.) This is called Goldbach's weak conjecture. I think that it is almost the same as Goldbach's conjecture.

a＞2
2a+1=P1+P2+P3
2a=(P1+P2+P3)-1

P is prime numbers.

a＞2
2a+1=O1+O2+O3
2a=(O1+O2+O3)-1

O is odd numbers.



For example,

2a=6 (a=3)

6=(3+2+2)-1 This is prime numbers.

6=(3+3+1)-1=(5+1+1)-1 This is odd numbers.


2a=8 (a=4)

8=(5+2+2)-1=(3+3+3)-1 This is prime numbers.

8=(7+1+1)-1=(5+3+1)-1=(3+3+3)-1 This is odd numbers.


2a=10 (a=5)

10=(7+2+2)-1=(5+3+3)-1 This is prime numbers.

10=(5+5+1)-1=(7+3+1)-1=(9+1+1)-1=(5+3+3)-1 This is odd numbers.

∴ (P1+P2+P3)-1⊂(O1+O2+O3)-1⊂2a

Therefore, Goldbach's weak conjecture is correct.

Goldbach's conjecture

Every even integer greater than 2 can be expressed as the sum of two primes. This is called Goldbach's conjecture.

2a=P1+P2

when a is 2, 4=2+2.

In the case of a>2, P1 and P2 must be odd numbers.


P is prime numbers. O is odd numbers.

∴ P1+P2⊂O1+O2⊂2a

I have found the pattern of prime numbers which is based on odd numbers.

P1+P2⊂O1+O2⊂2a is never broken.
It is quite natural that you can keep finding P1.
P2=2a-P1



Therefore, Goldbach's conjecture must be correct.

Landau symbol

When you keep adding integers infinitely, O (not Zero) is usable to express this difficulty. It is called Landau symbol.

O(n), O(n^2), O(nlogn)

This is expansion anyway.

Traveling salesman problem is well known as Landau symbol.





In this case, there are 4 places. The salesman starts from A and comes back to the same place. He must choose the fastest. This is easy one. However, when you keep adding places, the salesman would be confused.

It is said that there are (n-1)!/2 choices.

If you have 10 places, (10-1)!/2=(9*8*7*6*5*4*3*2*1)/2=181440. There are 181440 choices.
If you have 14 places, (14-1)!/2=3113510400.

This is expansion.

Max=T(1)

k=2
T(k)＞T(1)
Max=T(k)

k＜n and T(k+1)=T(n)＞T(k)
Max=T(k+1)=T(n)


∴