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#1 2012-02-13 22:57:37

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

srdanova math

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)
--yy--
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.
--a1--
Marjanovic Srdan
[ contact information removed by moderator ]
natural axiom
What is " nature along "?
-nature along in figure 1
What is "point"?
-start (end) natural long in figure 2
--a2--
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

View Image: a2.png View Image: yy.png

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#2 2012-02-14 00:41:17

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,574

Re: srdanova math

Hi;

Can mathematics explain the only two axiom that the rest are just evidence (experiments),

This is hardly true. Most modern mathematicians do not even know how to experiment so how can mathematics be full of experiments? Go to D Zeilberger's page and listen to his lectures
online. Or read Experimental mathematics by the Borweins. Coming from their camp, even 2 axioms is 3 too many.

A question to you, in the last month or so you have posted this message to about 75
forums. Those forums are mostly about philosophy and a smattering of other subjects. But no maths forums
except this one. Why?

I also think you have been here before and we already had this discussion so I know where you are going with this. I also urged you back then to forget your singular ideas and get on the bandwagon of the Discretists. They are already attempting to overhaul mathematics. "A purge now and then is a good thing" - Marko Rahmius. Too many purges creates chaos.

Welcome to the forum.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#3 2012-02-15 01:05:48

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

bobbym wrote:

A question to you, in the last month or so you have posted this message to about 75
forums. Those forums are mostly about philosophy and a smattering of other subjects. But no maths forums
except this one. Why?



Welcome to the forum.

who wants me to discuss and learn new things in mathematics is welcome who wants me to listen to his will
__________
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)
...
--image--

[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, ...}
...
--image--
[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
________________________________________________
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)}
...
--image--
[S6]-number points
CM-[S6]does not know

View Image: a3.png View Image: a4.png View Image: a5.png

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#4 2012-02-15 23:09:21

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

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)}
...
--image--
...

[S7]-number opposite points
CM-[S7]does not know
------
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
--image--
P3-four numbers (a, b, c, d) are comparable with each other
--image--
...
[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.

View Image: a6.png View Image: a7.png View Image: a8.png

Last edited by msbiljanica (2012-02-15 23:10:32)

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#5 2012-02-17 20:21:07

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-number ranges for number along
Process:
P1-image
P2-image
P3-image
--image--
[S9]-mobility of number
CM-[S9]-does not know
______________
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¤
...
--image--
Next - gap number and mobile number have no contact , except to point
...
[S10]-gap number GN={¤a(b)c¤,...,¤g(f)...(d)e¤}
[S11]-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
________________
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
form (.a/0) and (s.a/0) we’ll continue this write (.a) and (s.a)
--image--
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
[S12]-addition
CM-only form a+(s.0/.0)b=c , others do not know , axiom

View Image: a10.png View Image: a11.png View Image: a9.png

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#6 2012-02-19 01:22:40

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-Gap number is comparable with the gap number and number
Process:
P[sub]1[/sub] ¤a(b)c¤ , a+(s.0)c=z
P[sub]2[/sub] ¤a(b)c(d)e¤ , a+(s.0)c+(s.0)e=z
P[sub]3[/sub] ¤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
[S[sub]13[/sub]]-comparability gap numbers
CM-[S[sub]13[/sub]]-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:
P[sub]1[/sub] - 3+(s.0)0=3 , 3+(s.0)4=7 , 3[sub]4[/sub]7
       3+(s.0)0=3 , 3+(s.0)4=7 , 3+(s.0)4+(s.0)4=11 , 3[sub]4[/sub]11
       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 , 3[sub]4[/sub]15
       ...
       3+(s.0)0=3 , ... , 3+(s.0)4+...+(s.0)4=d , 3[sub]4[/sub]
   
P[sub]2[/sub] - 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 , 15[sub]4[/sub]3
...
3+(s.0)4=7 , 3+(s.0)0=3 , 7[sub]4[/sub]3
General form -a[sub]b[/sub]c , a[sub]b[/sub]
[S[sub]14[/sub]]-srcko
CM-[S[sub]14[/sub]]-does no know
____________________________________________
Presupposition-Srcko can join a number not that can not be in the structure srcko
Process:
P[sub]1[/sub] 10[sub]10[/sub]70 and 5 , 5_10[sub]10[/sub]70
P[sub]2[/sub] 5[sub]5[/sub]20 and 22 ,5[sub]5[/sub]20_22
P[sub]3[/sub] 7[sub]5[/sub] and 25 , 7[sub]5[/sub]_25
P[sub]4[/sub] 6[sub]8[/sub] and 2 ,2_6[sub]8[/sub]
...
General form -a[sub]b[/sub]c_d , d_a[sub]b[/sub]c , a[sub]b[/sub]_d ,d_a[sub]b[/sub]...
[S[sub]15[/sub]]-pendant srcko
CM-[S[sub]15[/sub]]-does no know
Note-only one number can be pendand , number two goes into a complex srcko

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#7 2012-02-20 05:09:56

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-Two ( more ) srcko (pendand srcko) are combined into one unit
Process:
P[sub]1[/sub] 10[sub]6 [/sub]and 11[sub]8[/sub] , 10[sub]6[/sub]11[sub]8[/sub]
P[sub]2[/sub] 10[sub]5[/sub]65 and 70[sub]3[/sub] ,10[sub]5[/sub]65_70[sub]3[/sub]
P[sub]3[/sub] 30[sub]3[/sub]60 and 45[sub]2[/sub]77_78 ,30[sub]3[/sub]60_45[sub]2[/sub]77_78
...
General form -a[sub]b[/sub]c[sub]d[/sub] , a[sub]b[/sub]c_d[sub]e[/sub] ,a[sub]b[/sub]c_d[sub]e[/sub]f_g ,...
[S[sub]16[/sub]]-two ( more) srcko
CM-[S[sub]16[/sub]]-does no know
_________________-
Presupposition-Two ( more ) srcko have the first ( last) common number
Process:
P[sub]1[/sub] 10[sub]5[/sub]30 and 3[sub]3[/sub]30 , 10[sub]5[/sub]3[sub]3[/sub](_30)
P2 4[sub]4[/sub]44 and 44[sub]10[/sub]94 and 44[sub]2[/sub]56 , 4[sub]4[/sub](_44_)[sub]10[/sub]94[sub]2[/sub]56
...
General form -a[sub]b[/sub]c[sub]d[/sub](_e) , a[sub]b[/sub](_c_)[sub]d[/sub]e[sub]f[/sub]g , ...
[S[sub]17[/sub]]-two ( more)  first-last srcko
CM-[S[sub]17[/sub]]-does no know
______________________________________________________
Presupposition-In the expression a+(.b)c=d , d+(s.0)1[sub]1[/sub] or d+(s.0)number (more) from 1[sub]1[/sub]
Process:
P[sub]1[/sub] 3+(.s.0)5=8+(s.0)1[sub]1[/sub] , 3+(s.0)5<9[sub]1[/sub]
P[sub]2[/sub] 5+(.0)5=5+(s.0)2[sub]2[/sub]4 , 5+(.0)5<7[sub]2[/sub]9
...
General form - a+(.b)c=d+(s.0)1[sub]1[/sub] ,a+(s.b)c<e[sub]1[/sub]
a+(.b)c=d+(s.0)e , a+(.b)c<f
a+(.b)c=d+(s.0)e[sub]f[/sub]g , a+(.b)c<h[sub]i[/sub]j ...
[S[sub]18[/sub]]-left inequality
CM-[S[sub]18[/sub]]-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
5[sub]5[/sub]50={5,10,15,20,25,30,35,40,45,50}
38[sub]3[/sub]50={38,41,44,47,50}
50[sub]10[/sub]90={50,60,70,80,90}
50[sub]7[/sub]92={50,57,64,71,78,85,92}
two(more) first-last srcko
5[sub]5[/sub]38[sub]3[/sub](_50_)[sub]10[/sub]90[sub]7[/sub]92
remains part of the function, when we come to it

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#8 2012-02-21 04:37:46

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-Parts number (a) and mobile number (b ) have a contact , the contact is delete
Process:
P[sub]1[/sub] 4-(.0)2=2
P[sub]2[/sub] 4-(.1)2=¤1(2)1¤ image
P[sub]3[/sub] 4-(.2)2=2
P[sub]4[/sub] 4-(.3)2=¤3(1)1¤
P[sub]5[/sub] 4-(.4)2=¤4(0)2¤
--image--
General form
a-(.0)b=c
a-(.1)b=c
...
a-(.d)b=c
[S[sub]19[/sub]]-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))1[sub]1[/sub]f or d-(s.0/s.0))number (more) from 1[sub]1[/sub]f
Process:
P[sub]1[/sub] 3+(.s.0)5=8-(s.0/s.0))1[sub]1[/sub]8 , 3+(s.0)5>0[sub]1[/sub]7
P[sub]2[/sub] 5+(.0)5=5-(s.0/s.0))2[sub]2[/sub]4 , 5+(.0)5>3[sub]2[/sub]1
...
General form - a+(.b)c=d-(s.0/s.0))1[sub]1[/sub]f ,a+(s.b)c>0[sub]1[/sub]e
a+(.b)c=d-(s.0/s.0)e , a+(.b)c>f
a+(.b)c=d-(s.0/s.0)e[sub]f[/sub]g , a+(.b)c>h[sub]i[/sub]j ...
[S[sub]20[/sub]]-right inequality addition
CM-[S[sub]20[/sub]]-know
____________________
Presupposition-Two ( more) addition (left and right inequalities) can be short to write
Process:
P1 3+(.0[sub]1[/sub]3)4=y , 3+(.0[sub]1[/sub]3)4>y ,3+(.0[sub]1[/sub]3)4<y
P2 8+(.2[sub]2[/sub]8)5=y , 8+(.2[sub]2[/sub]8)5>y ,8+(.2[sub]2[/sub]8)5<y
...
General form - a+(.b[sub]c[/sub]d)e=y ,a+(.b[sub]c[/sub]d)e>y , a+(.b[sub]c[/sub]d)e<y
a+(.b[sub]c[/sub]d_e)f=y , a+(.b[sub]c[/sub]d_e)f>y , a+(.b[sub]c[/sub]d_e)f<y ,...
[S21]-function addition
CM-[S21]-does no know

View Image: a12.png

Last edited by msbiljanica (2012-02-21 04:38:15)

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#9 2012-02-23 04:23:11

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

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¤  --0[sub]1[/sub]3¤¤(2)

   
     ¤2(2)0¤,¤1(2)1¤,¤0(2)2¤
     ¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤
                       ...
      2[sub]1[/sub]¤¤(2)

[S22]-variability of z number
CM-[S22]-does no know
_______________
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

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#10 2012-02-24 03:25:20

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-In the expression a-(.b)c=d , d+(.z)1[SIZE="1"]1[/SIZE]¤¤e or d+(.z) number ( more ) from
1[SIZE="1"]1[/SIZE]¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:

P1 4-(.3)2=¤1(2)1¤+(.z)1[SIZE="1"]1[/SIZE]¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)1[SIZE="1"]1[/SIZE]¤¤(2) , 4-(.3)2<3[SIZE="1"]1[/SIZE]¤¤(2)
P2 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+(.z)1[SIZE="1"]1[/SIZE]¤¤e , a#(.b)c<s¤¤e+(.z)1[SIZE="1"]1[/SIZE]¤¤e , a#(s.b)c<g[SIZE="1"]1[/SIZE]¤¤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[SIZE="1"]p[/SIZE]g¤¤e , a#(.b)c<s¤¤e+(.z)k[SIZE="1"]p[/SIZE]g¤¤e , a#(.b)c<h[SIZE="1"]i[/SIZE]j¤¤e ..., #-calculation operations
(+,-,×,..)
[S25]-left inequality gap number
CM-[S25]-does no know
_____________________________________
Presupposition-In the expression a-(.b)c=d , d-(.z)1[SIZE="1"]1[/SIZE]p¤¤e or d-(.z) number ( more ) from
11p¤¤e ,e={(f),(f)(f),(f)(f)(f),...}
Process:
P1  4-(.3)2=¤1(2)1¤-(.z)2[SIZE="1"]1[/SIZE]1¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)2[SIZE="1"]1[/SIZE]1¤¤(2) , 4-(.3)2>0[SIZE="1"]1[/SIZE]1¤¤(2)
P2  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-(.z)1[SIZE="1"]1[/SIZE]p¤¤e , a-(.b)c>s¤¤e-(.z)1[SIZE="1"]1[/SIZE]p¤¤e , a-(s.b)c>g[SIZE="1"]1[/SIZE]k¤¤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[SIZE="1"]p[/SIZE]g¤¤e , a-(.b)c>s¤¤e-(.z)k[SIZE="1"]p[/SIZE]g¤¤e , a-(.b)c>h[SIZE="1"]i[/SIZE]j¤¤e .. ,#-calculation operations
(+,-,×,...)
[S26]-right inequality gap number
CM-[S26]-does no know
_______________________________
Presupposition-The numbers are added , contact remains the rest is deleted
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=0
--image--
P1 ¤1(1)2¤ - (.0)2=1
P2 ¤1(1)2¤ - (.1)2=1 image
P3 ¤1(1)2¤ - (.2)2=2
P4 ¤1(1)2¤ - (.3)2=1
P5 ¤1(1)2¤ - (.4)2=0
General form
a - (.0)b=c
a - (.1)b=c
...
a - (.d)b=c

[S27]-opposite subtraction
CM-[S27]-does no know

View Image: a12.png

Last edited by msbiljanica (2012-02-24 03:27:56)

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#11 2012-02-24 22:41:40

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-Two ( more) addition the same number can be written in shorted form
Process:
P1 a+(s.c)a=a×(s.c)2 , a×(s.c)b
P2 a+(s.c)a+(s.c)a=a×(s.c)3 , a×(s.c)b
P3 a+(s.c)a+(s.c)a+(s.c)a=a×(s.c)4 , a×(s.c)b
...
[S28]-multiplication
CM-[S28]-know , axiom
_____
Presupposition-In terms a×(s.c)b , b can be number 0 (1)
Process:
P1 a×(s.c)0
P2 a×(s.c)1
[S28a]-multiplication-amendment
_____________
Presupposition-In terms a×(s.c)b , multiplication
Process:
P1 4×(s.0)4=16
P2 4×(s.1)4=13 image
P3 4×(s.2)4=10
P4 4×(s.3)4=7
P5 4×(s.4)4=4

--image--
P1 ¤1(2)1¤×(s.0)4=¤1(2)2(2)2(2)2(2)1¤
P2 ¤1(2)1¤×(s.1)4=¤1(2)1(2)1(2)1(2)1¤ image
P3 ¤1(2)1¤×(s.2)4=¤1(1)6(1)1¤
P4 ¤1(2)1¤×(s.3)4=7
P5 ¤1(2)1¤×(s.4)4=¤1(2)1¤
--image--
General form
a×(s.0)b=c
a×(s.1)b=c
...
a×(c.d)b=c

[S28b]-multiplication
CM-[S28b]-know a×(s.0)b other do not know , forms withuut any gaps numbers not known

___________________________________________
Presupposition-The numbers are multiplication , where a contact is deleted
Process:
P1 4 × (s.0)4=¤4(0)4(0)4(0)4¤
P2 4 × (s.1)4=¤3(1)2(1)2(1)3¤ image
P3 4 × (s.2)4=¤2(6)2¤
P4 4 × (s.3)4=¤1(5)1¤
P5 4 × (s.4)4=0
--image--
General form
a × (s.0)b=c
a × (s.1)b=c
...
a × (c.d)b=c

[S29]-multiplication subtraction
CM-[S29]-does no know
×You will see a sign in the PDF

View Image: a14.png View Image: a15.png View Image: a16.png

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#12 2012-02-27 04:14:02

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-The numbers are multiplication , contact remains the rest is deleted
Process:
P1 4 × (s.0)4=0
P2 4 × (s.1)4=¤1(2)1(2)1¤ image
P3 4 × (s.2)4=6
P4 4 × (s.3)4=5
P5 4 × (s.4)4=4
--image--
General form
a × (s.0)b=c
a ×(s.1)b=c
...
a × (s.d)b=c

[S30]-multiplication opposite subtraction
CM-[S30]-does no know
________________________
Presupposition-Three (more) merger (multiplication )
Process:
P1 5×(s.3ß3)5=¤1(1)1¤ - image
P2 5×(s.4ß3)5=¤1(3)1¤ - image
     5×(s.4ß4)5=¤1(1)1¤ -image
    5×(s.4ß5)5=1- image
P3 5×(s.5ß5)=5
--image--
General form
a#(s.1ß3)b=c
a#(s.2ßd)b=c
...
a#(s.eßf)b=c , #- calculation operations (×,...)

[S31]-srki
CM-[S31]-does no know
___________________
Presupposition-Three (more) gap merger (multiplication )
Process:
P1 ¤1(4)1¤×(s.5¤ß3)5=¤1(2)1¤ - image
     ¤1(4)1¤×(s.5¤ß4)5=2-image
P2 ¤1(4)1¤×(s.6¤ß4)5=4
--image--
General form
a#(s.1¤ß3)b=c
a#(s.2¤ßd)b=c
...
a#(s.e¤ßf)b=c , #- calculation operations (×,...)

[S32]-gap srki
CM-[S32]-does no know

View Image: a17.png View Image: a18.png View Image: a19.png

Last edited by msbiljanica (2012-02-27 04:16:30)

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#13 2012-02-28 03:16:03

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-Srki and  gap srki merger (multiplication )
Process:
P1 ¤1(7)1¤=¤1(1)1(1)1(1)1¤×(s.5¤|ß3)6=¤1(1)1(1)1(1)1¤
     ¤1(1)1(1)1¤=¤1(1)1(1)1(1)¤×(s.5¤|ß4)6=0
P2 4=¤1(1)1(1)1(1)1¤×(s.6¤|ß3)6=2 image
P3  ¤1(1)1(1)1(1)1¤=¤1(1)1(1)1(1)1¤×(s.7¤|ß6)6=¤1(1)1(1)1¤
   [attachment=2682:a20.png]
General form
w=a#(s.1¤|ß3)b=c
w=a#(s.2¤|ßd)b=c
...
w=a#(s.e¤|ßf)b=c , #- calculation operations (×,...)

[S33]-two srki
CM-[S33]-does no know
[size="4"]Note - this is a two-way calculation, and therefore has two equals signs left(srki) right(gap srki)[/size]
__________
Presupposition-Subtraction form a-(s.0/s.0)b=0,a-(s.0/s.0)b-(s.0/s.0)b=0 , ...,a-(s.0/s.0)b-
...-(s.0/s.0)b=0 can be written differently counting the subtraction
Process:
P1 12-(..0)4=12    (..c) -counting the subtraction
    12-(..1)4=8
    12-(..2)4=4
    12-(..3)4=0
    12:4=3
General form
a-(..0)b=a
a-(..1)b=c
...
a-(..d)b=0
a:b=d

[S34]-counting the subtraction
[S35]-divide
CM-[S34]-does no know ,[S35]-know, axiom
_____________________________
Presupposition-Subtraction form a-(s.0/s.0)b=0,a-(s.0/s.0)b-(s.0/s.0)b=0 , ...,a-(s.0/s.0)b-
...-(s.0/s.0)b=0, number b can replaced b= ((s.0/s.0)c-(s./s.0)d) , b= ((s.0/s.0)c-(s.0/s.0)d-
(s./s.0)e),...,b= ((s.0/s.0)...(s.0/s.0)w) ,  can be written differently counting the subtraction
Process:
P1 12-(..0)1[SUB]1[/SUB]2=12
12-(..1)1[SUB]1[/SUB]2=11
12-(..2)1[SUB]1[/SUB]2=9
12-(..3)1[SUB]1[/SUB]2=8
12-(..4)1[SUB]1[/SUB]2=6
12-(..5)1[SUB]1[/SUB]2=5
12-(..6)1[SUB]1[/SUB]2=3
12-(..7)1[SUB]1[/SUB]2=2
12-(..8)1[SUB]1[/SUB]2=0
12:1[SUB]1[/SUB]2=8
General form
a-(..0)b[SUB]e[/SUB]f=a
a-(..1)b[SUB]e[/SUB]f=c
...
a-(..d)b[SUB]e[/SUB]f=0
a:b[SUB]e[/SUB]f=d  ...
[S36]-inequality divide
CM-[S36]-does no know

View Image: a20.png

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#14 2012-02-28 03:54:25

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,574

Re: srdanova math

Hi;

Can you pause and explain your ideas before moving on? If you are just intending to publish your book then vixra is a much better place to do that.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#15 2012-02-28 07:18:21

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

bobbym wrote:

Hi;

Can you pause and explain your ideas before moving on? If you are just intending to publish your book then vixra is a much better place to do that.

what I've realized that the whole thing in math geometry, and that everything is a ratio of two (more) geometrical object in a line, plane, n-volume, I am totally rejected the historical legacy that has a current math, I started from the first geometry object (along natural) and the concept of point and a basic rule to connect the two (more) along the natural area this is the first axiom and it does not prove,each new object (concept, calculation, ...) occurs along the natural object or arising out along natural ...
If something is unclear ask

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#16 2012-03-03 01:09:23

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

Presupposition-Two ( more) multiplying the same numbers can be for short write
Process:
P1 a×(s.c)a=

,

P2 a×(s.c)a×(s.c)a=
,

P3 a×(s.c)a×(s.c)a×(s.c)a=
,

...
General form

[S37]-graduation
CM-[S37]-know,axiom
__________________________________
Presupposition-In terms a^(s.a)b , b can be number 0 (1)
Process:
P1

P2

[S37a]-graduation -amendment
_______________
Presupposition-Graduation
Process:
P1-
=1024
P2-
=769
P3-
=514
P4-
=259
P5-
=4
General form
=c
=d
...
=a
[S37b]-graduation
CM-[S37b]-know a(s.0)b=c , other do not know
______________
Presupposition-Two ( more) same number are abbreviated
Process:
P1- {4,4}=

P2-{4,4,4}=

P3-{4,4,4,4}=

....
General form
{a,a}=

{a,a,a}=

...
{a,...,a}?=

[S38]-the same numbers
CM-[S38]- do not know

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#17 2012-09-21 05:50:49

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

because I do not get it, here's the work of the occurrence arranged by the sheer number of mathematical notation, mathematical journals refused to publish it, see if it's worth.
https://docs.google.com/file/d/0BzkWG0xdRpPYVTZxVThoUkJlRWs/edit

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#18 2012-09-21 11:20:55

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,391

Re: srdanova math

Hello msbiljanica

Should it be that you are from the Balkan area?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#19 2012-09-22 02:15:19

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

anonimnystefy wrote:

Hello msbiljanica

Should it be that you are from the Balkan area?

I am from Serbia - South Eastern Europe

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#20 2012-09-22 02:22:56

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,391

Re: srdanova math

Well, you needn't explain where Serbia is. I live there too! smile


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#21 2013-04-08 06:18:02

msbiljanica
Member
Registered: 2012-01-18
Posts: 17

Re: srdanova math

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