srdanova math

[msbiljanica]

SRDANOVA MATHEMATICS
Revision of the current mathematics

Marjanovic Srdan
M.Biljanica
16201 Manojlovce
Serbia
[email protected]
Introduction: I think that the current limited maths and sinful and should be reviewed with all new
things that I discovered. I will explain the mathematical space with two starting points ( along the
natural and real ).

Natural Base:
Natural along is what you see along the fig,1.Natural along has its beginning and its end , this
property natural long we will contact points ( fig.2).Natural length along the ground ( natural
meaning).Two more natural and longer merge points

[S1]-along nature (fig. 1-a), [Sn]-mathematical facts
[S2]-point ( natural meaning , Fig.2 -A(B))
The points will mark capital letters along the (length) small letters
Definition of - teo points A,B , the length between points AB
CM (current mathematics) - does not recognize the concept of nature along , the point is not
defined so that all l everything
__________________________________________________ ___________
Presupposition-Natural long - merge points in the direction 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
renaming of points, we get along ABC (AC)
P2-ABC (AC) .. DE ..ABCD (AD)
read it: 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 (from natural long, two or more)
Definition of the initial-and the last point, the length between the initial and final points.
CM-I do not know the connection of natural longer, not along the natural base, but the real (the line , proof)

 

[Mr. H.]

Math is sinful. Now I've heard it all.

 

[msbiljanica]

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 ),...
The 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]-numeric 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 of-numeric along a starting point, the last point at infinity
-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)
SM-I know the term but long term numeric numeric rays (line)
Natural numbers and zeros are given axiom
________________________________________________________
Presupposition-point numbers have their
The 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]-point number
CM-I do not know the item number
_________________________________________________________________
Presupposition-point numbers have opposite
The process:
P1-0 = (s.0)
P2-1 = (s.0, 1)
P3-2 = (s.0, 1.2)
P4-3 = (s.0, 1,2,3)
P5-4 = (s.0, 1,2,3,4)
...

[S7]-opposite point of
CM-I do not know the opposite point of

 

[Colin]

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 ),...
The 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, ...}
...
View attachment 17080
[S4]-numeric 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 of-numeric along a starting point, the last point at infinity
-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)
SM-I know the term but long term numeric numeric rays (line)
Natural numbers and zeros are given axiom
________________________________________________________
Presupposition-point numbers have their
The 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)
...
View attachment 17081
[S6]-point number
CM-I do not know the item number
_________________________________________________________________
Presupposition-point numbers have opposite
The process:
P1-0 = (s.0)
P2-1 = (s.0, 1)
P3-2 = (s.0, 1.2)
P4-3 = (s.0, 1,2,3)
P5-4 = (s.0, 1,2,3,4)
...
View attachment 17082
[S7]-opposite point of
CM-I do not know the opposite point of

T x 2 + W + A = TWAT

 

[msbiljanica]

Presupposition-numbers are comparable with each other
The 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 issues
CM-known two of comparability, comparability of three numbers (a number comparable with the number of the PC),comparability of the other knows.
_____________________________________________________________________________
Presupposition-number ranges number along
The process:
P1-image
P2-image
P3-image
...

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

 

[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¤

 

[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 -abc , ab
[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 -abc_d , d_abc , ab_d ,d_ab...
[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 -abcd , abc_de ,abc_def_g ,...
[S16]-two ( more) srcko
CM-[S16]-does no know

 

[Colin]

You clearly believe in making little things count. Ever considered teaching arithmetic to dwarves?

 

[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 -abcd(_e) , ab(_c_)defg , ...
[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)efg , a+(.b)c<hij ...
[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)efg , a+(.b)c>hij ...
[S20]-right inequality addition
CM-[S20]-know

 

[MeBelle]

Ahh, wait till he gets enough posts to spam the board.

 

[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+(.bcd)e=y ,a+(.bcd)e>y , a+(.bcd)e<y
a+(.bcd_e)f=y , a+(.bcd_e)f>y , a+(.bcd_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

 

[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

 

[Dr.Traveler]

This is...interesting. I'll have the time later in the month to devote to decoding your notation. Have you considered submission online to the ArXiV?

What is the larger goal? Are you trying to replace the Zermelo-Fraenkel Axiom system? I recognize trichotomy, a few Euclid's axioms, etc, but a lot of this is a bit hard to get into thanks to the notation, which is a bit non-standard.

 

[msbiljanica]

This is...interesting. I'll have the time later in the month to devote to decoding your notation. Have you considered submission online to the ArXiV?

What is the larger goal? Are you trying to replace the Zermelo-Fraenkel Axiom system? I recognize trichotomy, a few Euclid's axioms, etc, but a lot of this is a bit hard to get into thanks to the notation, which is a bit non-standard.

ArXiV?-What is it ?
notation - the same as the current math, new things that I introduce our own notation represents
My system has the following axiom
1. space (explain it with the other two axioms)
2. along the natural (from it all this evidence)
3.along the real (not even He presented)
only mathematics from other sciences is not got rid of the historical heritage (not audited mathematics) what I have realized that in fact the whole math geometry, formulas that the numbers actually refer to the object geometry (number along, plane , n-space)
___________________________________________-
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)kpg , a+(.b)c<hij ...
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)kpg¤¤e , a-(.b)c<s¤¤e+(.z)kpg¤¤e , a+(.b)c<hij¤¤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))ksg , a-(.b)c>hij ...
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)kpg¤¤e , a-(.b)c>s¤¤e-(.z)kpg¤¤e , a-(.b)c>hij¤¤e ...
[S26]-right inequality subtraction
CM-[S26]-know , forms without any gaps numbers not known
________________________
PDF - link to the pdf you will get after my 15 post

 

[Dr.Traveler]

Mathematics

This is an online publishing site for Mathematics research. I've published a few things on there that I later submitted for review. It's become a common clearing house for research.

 

[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-(.bcd)e=y ,a-(.bcd)e>y , a-(.bcd)e<y
a-(.bcd_e)f=y , a-(.bcd_e)f>y , a-(.bcd_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)

 

[msbiljanica]

3.how to solve this current knowledge of mathematics:
along a (20m) ,deleted between 10 m and 15 m (b=5m) , wet get c
20m-(.5m)=¤10m(5m)5m¤
___________________________________________________
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 5 - (.0)5=5+(s.0)11 , 5 - (.0)5<61
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)11¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤<2¤¤(2)+(.z)11¤¤(2) , ¤2(2)2¤ - (.1)¤1/1)2¤<31¤¤(2)
P2 5 - (.0)5=5+(s.0)4 , 5 - (.0)5<9
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)7¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2)¤<2¤¤(2)+(.z)7¤¤(2), ¤2(2)2¤ - (.1)¤1(1)2¤<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)kpg , a - (.b)c<hij ...
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)kpg¤¤e , a - (.b)c<s¤¤e+(.z)kpg¤¤e , a - (.b)c<hij¤¤e ...
[S29]-left inequality opposite subtraction
CM-[S29]-does no know
_________________________________
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 5 - (.0)5=5 -(s.0/s.0))115 , 5 - (.0)5>014
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)211¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)211¤¤(2) , 4¤2(2)¤ - (.1)¤1(1)2¤>011¤¤(2)
P2 5 - (.0)5=5-(s.0/s.0))4 , 5 - (.0)5>1
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)1¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)1¤¤(2) , ¤2(2)2¤ - (.1)¤1(1)2¤>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))ksg , a - (.b)c>hij ...
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)kpg¤¤e , a -(.b)c>s¤¤e-(.z)kpg¤¤e , a - (.b)c>hij¤¤e ...
[S30]-right inequality opposite subtraction
CM-[S30]-does no know

 

[msbiljanica]

Presupposition-Two ( more) opposite 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 - (.bcd)e=y ,a - (.bcd)e>y , a - (.bcd)e<y
a - (.bcd_e)f=y , a - (.bcd_e)f>y , a - (.bcd_e)f<y ,...
[S31]-function opposite subtraction
CM-[S31]-does no know
__________________________________
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
...
[S32]-multiplication
CM-[S32]-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
[S32a]-multiplication-amendment