[msbiljanica]

questions
1.Z÷(10^n)=?,Z-integers
2.write in abbreviated form (if the function can be final and natural)
2+5=7 , 2+10=12 , 2+15=17, 2+20=22 , 2+25=27 , 2+30=32 , 2+35=37 , 2+38=40,
2+40=42, 2+41=43 , 2+44=46 , 2+45=47, 2+47=49 , 2+50=52 ,2+57=59 , 2+60=62 ,
2+64=66, 2+70=72, 2+71=73 , 2+78=80 , 2+80=82 , 2+85=87 , 2+90=92 ,2+92=94
3.how to solve this current knowledge of mathematics:
along a (20m) ,deleted between 10 m and 15 m (b=5m) , wet get c (image)

Can mathematics explain the only two axiom that the rest are just evidence (experiments), if you think so join me show you Srdanova math, see you
_______________________________________________________________________________________________________________-
I figure this way, from education school has 12 years, I was always the subject of mathematics and physics had the best grades, math deal amateur, studying mathematics I came to know that mathematics can be simplified and be connected (to be explained only with two axiom) and extend the mathematics that can solve math problems that present no solution.

Marjanovic Srdan
M.Biljanica
16201 Manojlovce
Serbia
ms.biljanica@gmail.com
natural axiom
What is " nature along "?
-nature along in figure 1
What is "point"?
-start (end) natural long in figure 2

What is the " basic rule "?
-basic rule is determined that the two ( more) longer only have to connect the points
[Sn]-mathematical facts
[S₁]-nature along
[S₂]-point (natural meaning of)
Definition[natural along]-two points , distance between two points
CM (current mathematics)-[S₁]-does not know , [S₂]-point is not defined , so anything and everything

[msbiljanica]

NATURAL MATHEMATICS



Presupposition-natural long merge points in the direction of the first natural along AB
Process:
P1-AB..CD..ABC(AC)
to read- natural along AB to point B, is connected to the natural long CD to point C, shall be
P2-ABC(AC)..DE..ABCD(AD)
read- along the ABC(AC) to point C , connecting with the natural long DE to point D is done
renaming of points , we get along ABCD(AD)
P3-ABCD(AD)..EF..ABCDE(AE)
...


[S₃]-along (natural basis)
Definition[along]-the first and last point and the distance between points
CM-[S₃]-does not know
_________________________________________________________________________________
Presupposition - All points of a longer (the infinite form) can be replaced with labels: (0), (0,1 ),...,
(0,1,2,3,4,5,6,7,8,9 ),...
Process:
P₁-N (₀) = {0,00,000,0000,...}
P₂-N (_0,1) = {0,1,10,11,100,...}
...
P₁₀-N (_{0,1,2,3,4,5,6,7,8,9}) = {0,1,2,3,4,5,6,7,8,9,10,11, ...}
...

[S₄]-number along
[S₅]-set of natural numbers N
We will use N (_{0,1,2,3,4,5,6,7,8,9}) = {0,1,2,3,4,5,6,7,8,9,10,11,12,...}
Definition[number along]- a starting point (0), the last point at infinity
[number N]-The number 0 is the point 0
-Other numbers are longer, the first item is 0, the last point is the point of the name (number)
CM-[S₄].does not know , [S₅]-axiom

[sigurdV]

I dont understand, please make a step by step repeat.

[msbiljanica]

sigurdV, on 23 January 2012 - 08:57 AM, said:

I dont understand, please make a step by step repeat.

This presentation is a step by step, each new piece of evidence derives from "natural along" or previous evidence, it can be easier than this
Presupposition-Numbers have their points
Process:
P₁ 0=(.0)
P₂ 1={(.0),(.1)}
P₃ 2={(.0),(.1),(.2)}
P₄ 3={(.0),(.1),(.2),(.3)}
P₅ 4={(.0),(.1),(.2),(.3),(.4)}
...

[S₆]-number points
CM-[S₆]does not know
___________________________________________________________
Presupposition-numbers have opposite points
Process:
P₁ 0=(s.0)
P₂ 1={(s.0),(s.1)}
P₃ 2={(s.0),(s.1),(s.2)}
P₄ 3={(s.0),(s.1),(s.2),(s.3)}
P₅ 4={(s.0),(s.1),(s.2),(s.3),(s.4)}
...

...

[S₇]-number opposite points
CM-[S₇]does not know

[sigurdV]

msbiljanica, on 23 January 2012 - 09:09 AM, said:

This presentation is a step by step, each new piece of evidence derives from "natural along" or previous evidence, it can be easier than this
Presupposition-Numbers have their points
Process:
P₁ 0=(.0)
P₂ 1={(.0),(.1)}
P₃ 2={(.0),(.1),(.2)}
P₄ 3={(.0),(.1),(.2),(.3)}
P₅ 4={(.0),(.1),(.2),(.3),(.4)}
...
a5.png
[S₆]-number points
CM-[S₆]does not know
___________________________________________________________
Presupposition-numbers have opposite points
Process:
P₁ 0=(s.0)
P₂ 1={(s.0),(s.1)}
P₃ 2={(s.0),(s.1),(s.2)}
P₄ 3={(s.0),(s.1),(s.2),(s.3)}
P₅ 4={(s.0),(s.1),(s.2),(s.3),(s.4)}
...
a6.png
...

[S₇]-number opposite points
CM-[S₇]does not know

I still dont understand it all,
but now i begin to think there might be something to understand.

Since you have an interest in maths, perhaps you also care for logic?
If so, I invite you to the thread "The Final Solution of the Liar".(#1)

[msbiljanica]

Presupposition-numbers are comparable with each other
Process:
P₁-two numbers (a, b ) are comparable with each other - a> b, a =b, a <b, ).(=(>,=,<)
P₂-three numbers (a, b, c) are comparable with each other

P₃-four numbers (a, b, c, d) are comparable with each other

...
[S₈]-comparability numbers
CM-[S₈]known two of comparability, comparability of three numbers(a number comparable with the numbers b and c),
comparability of the other knows.
___________________________________________________________________________________________
Presupposition-number ranges for number along
Process:
P₁-image
P₂-image
P₃-image

[S₉]-mobility of number
CM-[S₉]-does not know

[msbiljanica]

Presupposition-Number (a) and mobile number (b ) of a contat ,merge
Proces:
P₁ 3+(.0/.0)2=3
P₂ 3+(.1/.0)2=3
P₃ 3+(.2/.0)2=4 - image
P₄ 3+(.3/.0)2=5

designation means the more we write (.0/.0) , unless specifically indicated to be
P₁ 3+(.0)2=3
P₂ 3+(.1)2=3
P₃ 3+(.2)2=4
P₄ 3+(.3)2=5
General form
a+(.0)b=c
a+(.1)b=c
...
a+(.d)b=c
[S₁₀]-addition
CM-only form a+(s.0/.0)b=c , others do not know , axiom
______________________________________
Presupposition-Number (a) and mobile number (b ) but have no contact with the item
Process:
P₁ ¤3(0)2¤
P₂ ¤3(1)2¤
P₃ ¤3(2)2¤
...

Next - gap number and mobile number have no contact , except to point
...
[S₁₁]-gap number GN={¤a(b)c¤,...,¤g(f)...(d)e¤}
[S₁₂]-gap along
Definition[gap along] a>0-¤0(0)0¤-point
-¤0(a)0¤-two point ,separated by a gap
-¤a(0)0¤,¤0(0)a¤,¤a(0)a¤-along , two points
-¤a(a)a¤-two along , 4 points
...
CM-[S₁₁],[S₁₂] -does no know
______________________________________________
Third Now we can describe ( c ) the number of gaps ,¤10m(5m)5m¤
PDF - http://www.mediafire...qaqp63/srdanova math.pdf

[msbiljanica]

Presupposition-Gap number is comparable with the gap number and number
Process:
P₁ ¤a(b)c¤ , a+(s.0)c=z
P₂ ¤a(b)c(d)e¤ , a+(s.0)c+(s.0)e=z
P₃ ¤a(b)c(d)e(f)g¤ ,a+(s.0)c+(s.0)e+(s.0)g=z
...
number z as it compares as a number of
[S₁₃]-comparability gap numbers
CM-[S₁₃]-does no know
__________________________________________________
Presupposition-Adding the result can be written in short form:
a) a+(s.0)0 , a+(s.0)b , a+(s.0)b+(s.0)b , ...,a+(s.0)b+...+(s.0)b
b ) a+(s.0)b+...+(s.0)b,...,a+(s.0)b+(s.0)b, a+(s.0)b , a+(s.0)0
Process:
P₁ - 3+(s.0)0=3 , 3+(s.0)4=7 , 3₄7
3+(s.0)0=3 , 3+(s.0)4=7 , 3+(s.0)4+(s.0)4=11 , 3₄11
3+(s.0)0=3 , 3+(s.0)4=7 , 3+(s.0)4+(s.0)4=11 , 3+(s.0)4+(s.0)4+(s.0)4=15 , 3₄15
...
3+(s.0)0=3 , ... , 3+(s.0)4+...+(s.0)4=d , 3₄

P₂ - 3+(s.0)4+(s.0)4+(s.0)4=15 , 3+(s.0)4+(s.0)4=11 , 3+(s.0)4=7 , 3+(s.0)0=3 , 15₄3
...
3+(s.0)4=7 , 3+(s.0)0=3 , 7₄3
General form -a_bc , a_b
[S₁₄]-srcko
CM-[S₁₄]-does no know

[msbiljanica]

Presupposition-Srcko can join a number not that can not be in the structure srcko
Process:
P₁ 10₁₀70 and 5 , 5_10₁₀70
P₂ 5₅20 and 22 ,5₅20_22
P₃ 7₅ and 25 , 7₅_25
P₄ 6₈ and 2 ,2_6₈
...
General form -a_bc_d , d_a_bc , a_b_d ,d_a_b...
[S₁₅]-pendant srcko
CM-[S₁₅]-does no know
Note-only one number can be pendand , number two goes into a complex srcko
__________________________________________________________
Presupposition-Two ( more ) srcko (pendand srcko) are combined into one unit
Process:
P₁ 10₆ and 11₈ , 10₆11₈
P₂ 10₅65 and 70₃ ,10₅65_70₃
P₃ 30₃60 and 45₂77_78 ,30₃60_45₂77_78
...
General form -a_bc_d , a_bc_d_e ,a_bc_d_ef_g ,...
[S₁₆]-two ( more) srcko
CM-[S₁₆]-does no know

[msbiljanica]

Presupposition-Two ( more ) srcko have the first ( last) common number
Process:
P₁ 10₅30 and 3₃30 , 10₅3₃(_30)
P2 4₄44 and 44₁₀94 and 44₂56 , 4₄(_44_)₁₀94₂56
...
General form -a_bc_d(_e) , a_b(_c_)_de_fg , ...
[S₁₇]-two ( more) first-last srcko
CM-[S₁₇]-does no know
______________________________________________________
Presupposition-In the expression a+(.b)c=d , d+(s.0)1₁ or d+(s.0)number (more) from 1₁
Process:
P₁ 3+(.s.0)5=8+(s.0)1₁ , 3+(s.0)5<9₁
P₂ 5+(.0)5=5+(s.0)2₂4 , 5+(.0)5<7₂9
...
General form - a+(.b)c=d+(s.0)1₁ ,a+(s.b)c<e₁
a+(.b)c=d+(s.0)e , a+(.b)c<f
a+(.b)c=d+(s.0)e_fg , a+(.b)c<h_ij ...
[S₁₈]-left inequality
CM-[S₁₈]-know
_______________________________________
2+5=7 , 2+10=12 , 2+15=17, 2+20=22 , 2+25=27 , 2+30=32 , 2+35=37 , 2+38=40,
2+40=42, 2+41=43 , 2+44=46 , 2+45=47, 2+47=49 , 2+50=52 ,2+57=59 , 2+60=62 ,
2+64=66, 2+70=72, 2+71=73 , 2+78=80 , 2+80=82 , 2+85=87 , 2+90=92 ,2+92=94
srcko
5₅50={5,10,15,20,25,30,35,40,45,50}
38₃50={38,41,44,47,50}
50₁₀90={50,60,70,80,90}
50₇92={50,57,64,71,78,85,92}
two(more) first-last srcko
5₅38₃(_50_)₁₀90₇92
remains part of the function, when we come to it

[msbiljanica]

Presupposition-Parts number (a) and mobile number (b ) have a contact , the contact is delete
Process:
P₁ 4-(.0)2=2
P₂ 4-(.1)2=¤1(2)1¤ image
P₃ 4-(.2)2=2
P₄ 4-(.3)2=¤3(1)1¤
P₅ 4-(.4)2=¤4(0)2¤

General form
a-(.0)b=c
a-(.1)b=c
...
a-(.d)b=c
[S₁₉]-subtraction
CM-only form a-(s.0/s.0)b=c , others do not know , axiom
_______________________________________________________________________
Presupposition-In the expression a+(.b)c=d , d-(s.0/s.0))1₁f or d-(s.0/s.0))number (more) from 1₁f
Process:
P₁ 3+(.s.0)5=8-(s.0/s.0))1₁8 , 3+(s.0)5>0₁7
P₂ 5+(.0)5=5-(s.0/s.0))2₂4 , 5+(.0)5>3₂1
...
General form - a+(.b)c=d-(s.0/s.0))1₁f ,a+(s.b)c>0₁e
a+(.b)c=d-(s.0/s.0)e , a+(.b)c>f
a+(.b)c=d-(s.0/s.0)e_fg , a+(.b)c>h_ij ...
[S₂₀]-right inequality addition
CM-[S₂₀]-know

[msbiljanica]

Presupposition-Two ( more) addition (left and right inequalities) can be short to write
Process:
P1 3+(.0₁3)4=y , 3+(.0₁3)4>y ,3+(.0₁3)4<y
P2 8+(.2₂8)5=y , 8+(.2₂8)5>y ,8+(.2₂8)5<y
...
General form - a+(.b_cd)e=y ,a+(.b_cd)e>y , a+(.b_cd)e<y
a+(.b_cd_e)f=y , a+(.b_cd_e)f>y , a+(.b_cd_e)f<y ,...
[S21]-function addition
CM-[S21]-does no know
______________________________________________________
Presupposition-Gap number ( value z) is a variable (same value) with the gaps that is constant
Process:
P1 ¤5(2)0¤,¤4(2)1¤,3(2)2¤,¤2(2)3¤,¤1(2)4¤,¤0(2)5¤ --5¤¤(2)

P2 ¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤
¤2(2)0¤,¤1(2)1¤.¤0(2)2¤
¤1(2)0¤,¤0(2)1¤
¤0(2)0¤ --0₁3¤¤(2)


¤2(2)0¤,¤1(2)1¤,¤0(2)2¤
¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤
...
2₁¤¤(2)

[S22]-variability of z number
CM-[S22]-does no know
___________________________
PDF - http://www.mediafire...jzheej0qu93jjkm

[msbiljanica]

Presupposition-Translation of gap number in the variability z ( with constant gap ) can addition
(s.0)
Process:
P1 ¤3(2)3¤+(.z)¤4(2)4=6¤¤(2)+(.z)8¤¤(2)=14¤¤(2)
P2 ¤1(6)1(9)1¤+(.z)¤3(6)2(9)1¤=3¤¤(6)(9)+(.z)6¤¤(6)(9)=9¤¤(6)(9)
...
General form ¤a(b)c¤+(.z)¤d(b)e¤=f¤¤(b )+(.z)g¤¤(b )=h¤¤(b ) ...
[S23]-z addition
CM-[S23]-does no know
__________________________________________
Presupposition-Translation of gap number in the variability z ( with constant gap ) can subtraction
(s.0/s.0)
Process:
P1 ¤3(2)3¤-(.z)¤1(2)1=6¤¤(2)-(.z)2¤¤(2)=4¤¤(2)
P2 ¤3(6)2(9)1¤-(.z)¤1(6)1(9)1¤=6¤¤(6)(9)-(.z)3¤¤(6)(9)=3¤¤(6)(9)
...
General form ¤a(b)c¤-(.z)¤d(b)e¤=f¤¤(b )-(.z)g¤¤(b )=h¤¤(b ) ...
[S24]-z subtraction
CM-[S24]-does no know

[msbiljanica]

Presupposition-In the expression a-(.b)c=d , d+(s.0))1₁ or d+(s.0)number (more) from 1₁
and d+(.z)1₁¤¤e or d+(.z) number ( more ) from 1₁¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 4-(.2)2=2+(s.0)1₁ , 4-(.2)2<3₁
4-(.3)2=¤1(2)1¤+(.z)1₁¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)1₁¤¤(2) , 4-(.3)2<3₁¤¤(2)
P2 4-(.2)2=2+(s.0)4 , 4-(.2)2<6
4-(.3)2=¤1(2)1¤+(.z)7¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)7¤¤(2), 4-(.3)<9¤¤(2)
...
General form - a-(.b)c=d-(s.0)1₁ ,a+(s.b)c<g₁
a-(.b)c=d+(s.0)g , a+(.b)c<l
a-(.b)c=d+(s.0)k_pg , a+(.b)c<h_ij ...
a-(.b)c=d+(.z)1₁¤¤e , a-(.b)c<s¤¤e+(.z)1₁¤¤e , a+(s.b)c<g₁¤¤e
a-(.b)c=d+(.z)g¤¤e , a-(.b)c<s¤¤¤e+(.z)g¤¤e , a+(.b)c<l¤¤e
a-(.b)c=d+(.z)k_pg¤¤e , a-(.b)c<s¤¤e+(.z)k_pg¤¤e , a+(.b)c<h_ij¤¤e ...
[S25]-left inequality subtraction
CM-[S25]-know , forms without any gaps numbers not known
____________________________________________________________________
Presupposition-In the expression a-(.b)c=d , d-(s.0/s.0))1₁p or d+(s.0/s.0)number (more) from 1₁p
and d-(.z)1₁p¤¤e or d-(.z) number ( more ) from 1₁¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 4-(.2)2=2-(s.0/s.0))1₁2 , 4-(.2)2>0₁1
4-(.3)2=¤1(2)1¤-(.z)2₁1¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)2₁1¤¤(2) , 4-(.3)2>0₁1¤¤(2)
P2 4-(.2)2=2-(s.0/s.0))1 , 4-(.2)2>1
4-(.3)2=¤1(2)1¤-(.z)1¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)1¤¤(2) , 4-(.3)>1¤¤(2)
...
General form - a-(.b)c=d-(s.0/s.0)1₁p ,a-(s.b)c>g₁l
a-(.b)c=d-(s.0/s.0))g , a-(.b)c>l
a-(.b)c=d-(s.0/s.0))k_sg , a-(.b)c>h_ij ...
a-(.b)c=d-(.z)1₁p¤¤e , a-(.b)c>s¤¤e-(.z)1₁p¤¤e , a-(s.b)c>g1k¤¤e
a-(.b)c=d-(.z)g¤¤e , a-(.b)c>s¤¤e-(.z)g¤¤e , a-(.b)c>l¤¤e
a-(.b)c=d-(.z)k_pg¤¤e , a-(.b)c>s¤¤e-(.z)k_pg¤¤e , a-(.b)c>h_ij¤¤e ...
[S26]-right inequality subtraction
CM-[S26]-know , forms without any gaps numbers not known

[msbiljanica]

Presupposition-Two ( more) subtraction (left and right inequalities) can be short to write
Process:
P1 7-(.0₁3)4=y , 7-(.0₁3)4>y ,7-(.0₁3)4<y
P2 8-(.2₂8)5=y , 8-(.2₂8)5>y ,8-(.2₂8)5<y
...
General form a-(.b_cd)e=y ,a-(.b_cd)e>y , a-(.b_cd)e<y
a-(.b_cd_e)f=y , a-(.b_cd_e)f>y , a-(.b_cd_e)f<y ,...
[S27]-function subtraction
CM-[S27]-does no know
_____________________________________
Presupposition-Parts number (a) and mobile number (b ) have a contact , contact remains , the
rest is delete
Process:

P1 4 - (.0)2=2
P2 4 - (.1)2=2 image
P3 4 - (.2)2=2
P4 4 - (.3)2=1
P5 4 - (.4)2=1
General form
a - (.0)b=c
a - (.1)b=c
...
a - (.d)b=c

[S28]-opposite subtraction
CM-[S28]-does no know
-sign for. opposite subtraction (when you download the following PDF you will see how it looks)