Beiträge von ms.srki

Theorem - The contact number is sorted vertically only be a natural straight line that gives a natural gap
-When has two (more) solutions they are connected
Proof -[tex]1\rightarrow \underline1(\underline{s})[/tex]

[tex]4{+_{12}^{0}}2=\underline2[/tex] or [tex]4{+_{12}^{\underline4}}2= \underline2[/tex]

[tex]4{+_{12}^{1}}2=\underline2[/tex] or [tex]4{+_{12}^{\underline3}}2= \underline2[/tex]

[tex]4{+_{12}^{2}}2=\underline2[/tex] or [tex]4{+_{12}^{\underline2}}2= \underline2[/tex]

[tex]4{+_{12}^{3}}2=\underline4[/tex] or [tex]4{+_{12}^{\underline1}}2= \underline4[/tex]

[tex]4{+_{12}^{4}}2=\underline6[/tex] or [tex]4{+_{12}^{\underline0}}2= \underline6[/tex]

[tex]+_{12}[/tex]- addition rule 12
(CM.) - no addition rule 12

ERROR - in the above posts should be like this
[tex]4{+_{10}^{0}}2=\underline2[/tex] or [tex]4{+_{10}^{\underline4}}2=\underline2[/tex]

[tex]4{+_{10}^{1}}2=('\underline1,\underline1)[/tex] or [tex]4{+_{10}^{\underline3}}2=('\underline1,\underline1)[/tex]

[tex]4{+_{10}^{2}}2=\underline2[/tex] or [tex]4{+_{10}^{\underline2}}2=\underline2[/tex]

[tex]4{+_{10}^{3}}2=('\underline3,\underline1)[/tex] or [tex][4{+_{10}^{\underline1}}2=('\underline3,\underline1)[/tex]

[tex]4{+_{10}^{4}}2=\underline6[/tex] or [tex]4{+_{10}^{\underline0}}2=\underline6[/tex]

Theorem - The contact number is sorted vertically only be a natural straight line that gives a natural gap
-When has two (more) solution between them becomes straights lines

PROOF - [tex]1\rightarrow \underline1(1)[/tex]

[tex]4{+_{11}^{0}}2=\underline2[/tex] or [tex]4{+_{11}^{\underline4}}2=\underline2[/tex]

[tex]4{+_{11}^{1}}2=('\underline12\underline1)[/tex] or [tex]4{+_{11}^{\underline3}}2=('\underline12\underline1)[/tex]

[tex]4{+_{11}^{2}}2=\underline2[/tex] or [tex]4{+_{11}^{\underline2}}2=\underline2[/tex]

[tex]4{+_{11}^{3}}2=('\underline31\underline1)[/tex] or [tex]4{+_{11}^{\underline1}}2=('\underline31\underline1)[/tex]

[tex]4{+_{11}^{4}}2=\underline6[/tex] or [tex]4{+_{11}^{\underline0}}2=\underline6[/tex]

[tex]+_{11} [/tex]- addition rule 11

(CM.) - no addition rule 11

Solution trisecting (n-sections) and n-proper construction of the polygon using divider and straightedge.

The ball shall be awarded in two equal parts, obtained half contains two areas, the circle (represents a planar geometry) and a semi-sphere (represents spherical geometry), the circle is the boundary between the circuit and semi-sphere
View photo (below)
CIRCLE
within a given arbitrary angle BAC,
straightedge (straightedge is flexible, it can draw on a sphere) along the BA to extend the circle to give the point D
SPHERE
straightedge connect points B and D, you get the curve BD
straightedge and divider - a procedure divisions curve into two equal parts is the same as the process of division shall exceed the level in two equal parts, we get the point E
straightedge - connect points C and E and get the curve CE

Proportion longer exists in plane geometry, I found that the process can be applied to the sphere
choose point G
divider EC, from point G we get a point H
divider EC, from point H we get a point I
divider EC, from point E twe get a point J
divider EC, from point J get the point K
divider EC, from point K get the point L
straightedge point L and point I and connect, we get curve LI
EG divider, from the point L we get a point P
EG divider, from the point P and get the point O
straightedge join the dots E and P and proceed to the circle, we get the point Q
straightedge merge points E and O and proceed to the circle, we get the point R
CIRCLE
straightedge connect point A and point Q, we get along AQ
straightedge connect point A and point R, we get along AR

these have carried out the production of a given trisection angle, is obtained from the rest of the (n-section, a regular n-polygon) ...

Now proclaim everywhere that I decided 2- millennium math problems

Theorem - The contact number is sorted vertically only one natural straight line gives a natural gaps

PROOF [tex] 1\rightarrow \underline1[/tex]

[tex]4{+_1^{0}}2=\underline2[/tex] or [tex] 4{+_1^{\underline4}}2=\underline2[/tex]

[tex]4{+_1^{1}}2=('\underline1,\underline1)[/tex] or [tex]4{+_1^{\underline3}}2=('\underline1,\underline1)[/tex]

[tex]4{+_1^{2}}2=\underline2[/tex] or [tex]4{+_1^{\underline2}}2=\underline2[/tex]

[tex]4{+_1^{3}}2=('\underline3,\underline1)[/tex] ior [tex]4{+_1^{\underline1}}2=('\underline3,\underline1)[/tex]

[tex]4{+_1^{4}}2=\underline6[/tex] or [tex]4{+_1^{\underline0}}2=\underline6[/tex]

[tex]+_{10} [/tex]- addition rule10

(CM.) - no addition rule 10

[tex]+_9[/tex]
Theorem - The contact number is sorted vertically:
- Only one natural straight line gives a natural straight line
- Two natural straight line that gives a natural straight line
- when there are two (more) solutions between them, are connected
PROOF - [tex](1,11)\rightarrow1(\underline{s})[/tex]
[Blockierte Grafik: http://217.26.67.168/uploads/3/5/3576551/y4.png]
[tex]2\underline{2 }2+_9^02\underline{2 }2=4[/tex] or [tex]2\underline{2 }2+_9^{\underline6}2\underline{2 }2=4[/tex]

[tex]2\underline{2 }2+_9^12\underline{2 }2=6[/tex] or [tex]2\underline{2 }2+_9^{\underline5}2\underline{2 }2=6[/tex]

[tex]2\underline{2 }2+_9^22\underline{2 }2=8[/tex] or [tex]2\underline{2 }2+_9^{\underline4}2\underline{2 }2=8[/tex]

[tex]2\underline{2 }2+_9^32\underline{2 }2=7[/tex] or [tex]2\underline{2 }2+_9^{\underline3}2\underline{2 }2=7[/tex]

[tex]2\underline{2 }2+_9^42\underline{2 }2=6[/tex] or [tex]2\underline{2 }2+_9^{\underline2}2\underline{2 }2=6[/tex]

[tex]2\underline{2 }2+_9^52\underline{2 }2=7 [/tex] or[tex]2\underline{2 }2+_9^{\underline1}2\underline{2 }2=7[/tex]

[tex]2\underline{2 }2+_9^62\underline{2 }2=8[/tex] or [tex]2\underline{2 }2+_9^{\underline0}2\underline{2 }2=8[/tex]

[tex]+_9[/tex] - addition rule 9

(CM.) - no addition rule 9

https://onedrive.live.com/redir?resid=70…hint=file%2cpdf

[tex]+_8[/tex]
Theorem - The contact number is sorted vertically:
- Only one natural straight line gives a natural straight line
- Two natural straight line that gives a natural straight line
- When there are two (more) solution between them becomes gaps
PROOF - [tex](1,11)\rightarrow1(\underline1)[/tex]

[tex]2\underline{2 }2+_8^02\underline{2 }2=2\underline22 [/tex] or [tex]2\underline{2 }2+_8^{\underline6}2\underline{2 }2=2\underline22[/tex]

[tex]2\underline{2 }2+_8^12\underline{2 }2=3\underline13[/tex] or [tex]2\underline{2 }2+_8^{\underline5}2\underline{2 }2=3\underline13[/tex]

[tex]2\underline{2 }2+_8^22\underline{2 }2=8[/tex] or [tex]2\underline{2 }2+_8^{\underline4}2\underline{2 }2=8[/tex]

[tex]2\underline{2 }2+_8^32\underline{2 }2=2\underline13\underline12[/tex] or [tex]2\underline{2 }2+_8^{\underline3}2\underline{2 }2=2\underline13\underline12[/tex]

[tex]2\underline{2 }2+_8^42\underline{2 }2=2\underline22\underline22[/tex] or [tex]2\underline{2 }2+_8^{\underline2}2\underline{2 }2=2\underline22\underline22[/tex]

[tex]2\underline{2 }2+_8^52\underline{2 }2=2\underline23\underline22[/tex] or [tex]2\underline{2 }2+_8^{\underline1}2\underline{2 }2=2\underline23\underline22[/tex]

[tex]2\underline{2 }2+_8^62\underline{2 }2=2\underline24\underline22[/tex] or [tex]2\underline{2 }2+_8^{\underline0}2\underline{2 }2=2\underline24\underline22[/tex]

[tex]+_8 [/tex]- addition rule 8

(CM.) - no addition rule 8

error corrected in PDF, in Serbian language, soon will be in English

https://onedrive.live.com/redir?resid=70…hint=file%2cpdf

Theorem - contact number is sorted horizontally, two natural straight line provide a natural straight line

Proof - [tex]11\rightarrow1[/tex]
[attachment=19640]

[tex]2\underline{2 }2+_4^{\underline0}2\underline{2 }2=(2,2)[/tex]

[tex]2\underline{2 }2+_4^{\underline1}2\underline{2 }2=(1,1)[/tex]

[tex]2\underline{2 }2+_4^{\underline2}2\underline{2 }2=0[/tex]

[tex]2\underline{2 }2+_4^{\underline3}2\underline{2 }2=1[/tex]

[tex]2\underline{2 }2+_4^{\underline4}2\underline{2 }2=2[/tex]

[tex]2\underline{2 }2+_4^{\underline5}2\underline{2 }2=1[/tex]

[tex]2\underline{2 }2+_4^{\underline6}2\underline{2 }2=0[/tex]

[tex]+_4[/tex] - addition rule 4

(CM.) - No "addition rule 4"

Theorem - The contact number is sorted horizontally only be a natural straight line that gives a natural straight line , when there are two (more) results merge

Proof - [tex]1\rightarrow 1 (\underline{s})[/tex]

[tex]4{+_3^{\underline0}}2=2[/tex]

[tex]4{+_3^{\underline1}}2=2[/tex]

[tex]4{+_3^{\underline2}}2=2[/tex]

[tex]4{+_3^{\underline3}}2=4[/tex]

[tex]4{+_3^{\underline4}}2=6[/tex]

[tex]+_3 [/tex]- addition rule 3

(SM.) - no "addition rule 3 "

Theorem - The contact numbers is sorted horizontally to be the only one natural straight line that gives a natural straight line , when there are two (more) results between them becomes a gap.

PROOF -[tex]1\rightarrow1(\underline{1})[/tex]

[tex]4{+_2^{\underline0}}2=2[/tex]
[tex]4{+_2^{\underline1}}2=1\underline{2}1[/tex]
[tex]4{+_2^{\underline2}}2=2[/tex]
[tex]4{+_2^{\underline3}}2=3\underline{1}1[/tex]
[tex]4{+_2^{\underline4}}2=6[/tex]

+₂ - addition rule 2

(CM.) - No "addition rule 2"

Theorem - The contact numbers is sorted horizontally to be the only one natural straight line that gives a natural straight line

PROOF - [tex]1\rightarrow 1[/tex]

4[tex]{+_1^{\underline0}[/tex]2=2

4[tex]+_1^{\underline1}[/tex]2=(1,1)

4[tex]{+_1^{\underline2}[/tex]2=2

4[tex]{+_1^{\underline3}}[/tex]2=(3,1)

4[tex]{+_1^{\underline4}}[/tex]2=6 or 4+2=6

+₁ - addition rule 1

(CM.) - There are no "addition rule 1" only when the contact point number, the axiom

Theorem - Two numbers have contact, their condition is described counts of first number

PROOF - number 3 and number 2 have a contact at point 0
[tex]3^{\underline{0}} 2[/tex]

- number 3 and number 2 have a contact at point 1
[tex]3^{\underline{1}}2[/tex]

- number 3 and number 2 have a contact at point 2
[tex]3^{\underline{2}}2[/tex]

- number 3 and number 2 have a contact at point 3
[tex]3^{\underline{3}}2[/tex]

(CM.) - Knows no contact numbers

Theorem - natural numbers and natural numbers gaps can be connected in the direction AB (0.1)

PROOF - Number 1 and number [tex]\underline {1} [/tex]receives the combined number of [tex]1\underline {1}[/tex] or dup (duž , praznina )

-Number [tex]\underline {1}[/tex] and number 1 receives the combined number of [tex]\underline {1}1[/tex] or dup 

-Number 1 and number [tex]\underline {2}[/tex] receives the combined number of [tex]1\underline {2}[/tex] or dup

[size=24pt]...[/size]
- A basic set of combined natural numbers [tex]K^o=(a_n,\underline{b}_n,a_n\in{N^o},\underline{b}_n\in{N_p^o},(a_n,\underline{b}_n)>0)[/tex]

[tex]a_1\underline{b}_1[/tex]
[tex]\underline{b}_1a_1[/tex]
[tex]a_1\underline{b}_1a_2[/tex]
[tex]\underline{b}_1a_1\underline{b}_2[/tex]
...

(CM.) - Dup do not know, not know the combined numbers (there is this form, but not numbers [tex]\{0,a\}\cup\{c,c\},\{0,0\}\cup\{a,b\},\{0,b\}\cup\{c,d\},\{0,0\}\cup\{a,b\}\cup\{c,c\},...[/tex] )

Theorem - there is a relationship between the points 0 and all points one-way infinite straight line(one-way infinite gaps) including points 0

PROOF - relationship points 0 points 0 and the number 0

-relationship points 0 points 1 and the number 1( [tex]\underline{1}[/tex])

-relationship points 0 points 2 and the number 2 ([tex]\underline{2}[/tex])

...

basic set of natural numbers [tex]N^o=\{0 , 1 , 2 ,3 ,4 ,5 ,...\}[/tex]
basic set of natural numbers gaps [tex]N_p^o=\{0 , \underline{1} ,\underline{2} ,\underline{3} ,\underline{4} ,\underline{5} ,...\}[/tex]

(CM.) - natural numbers are given as an axiom, there is no natural gaps numbers (there is this form, but do not call numbers [tex](\{0,0\}\cup\{a,a\} a\in N)[/tex]

Theorem - Natural straight line (natural gap) are connected in the direction of the points AB (0.1)
PROOF - straight line (gaps) b ([tex]\underline{b}[/tex]) -defined AC (0,2)

- straight line (gaps) c ([tex]\underline{c}[/tex]) -defined AD (0,3)

...
infinite one way straight line (oneway infinite gaps) ∞ ([tex]\underline{\infty}[/tex]) defined A∞ (0, ∞)

(CM.) - straight line (not from the natural basis), there is gaps, a one-way infinite straight line the (semi-line (not from natural base)), one-way infinite gaps does not exist

I will present you a math composed of only two basis (natural and realistic basis)

Current mathematics (CM.)

Natural Base
-natural straight line the main axiom, its beginning or end point and natural straight line a defined length and with two points
NOTATION - natural straight line (lower case), points (capital letters or numbers (when specified point uploads metric (such as the number line)))

-natural gaps negation natural straight line , natural emptiness and emptiness is defined with two points
NOTATION - natural gaps (small underlined letter)

-basic rule merger - natural straight line and natural gaps are connected only points
-basic set - all possibilities defined theorem
(CM.)does not know the natural straight line , point is not defined, knows no natural gap, is not defined by basic set