srdan math

M

msbiljanica

Registered Senior Member
SRDANOVA MATHEMATICS
Revision of the current mathematics
View attachment 4479
Marjanovic Srdan
M.Biljanica
16201 Manojlovce
Serbia
ms.biljanica@gmail.com
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
View attachment 4480

[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)
...

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)
...
View attachment 4481
[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)
________________________________________________________________
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 4482
[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)
CM-I know the term but long term numeric numeric rays (line)
Natural numbers and zeros are given axiom
 
Welcome.

And let me be the first to thank you for saving math and removing it's sinful nature.
 
msbiljanica, your math seems far more limited than the mathematics of set theory, category theory or Euclidean geometry in that those subjects are capable of being communicated to other human beings.

I think what you are saying is this:

Definition 1:
A segment has two end points.
Definition 2:
A segment has a length.
Definition 3:
A new segment may be constructed, given any two segments.
Axiom 1:
A segment of length 1 exists.
Axiom 2:
A new segment constructed from a given segment and a segment of length 1 has a length distinct from the given segment.

Theorem:
A unbounded sequence of segments with distinct lengths exists, constructed by adding a segment of length 1 to itself an arbitrary number of times.

Definition 4:
The length of a segment created by adding a segment of length 1 to itself N times is given as N+1.

--
This seems less useful than the Peano axioms of arithmetic.

  • 0 is a number
  • if x is number, x = x
  • if x and y are numbers, x = y means y = x
  • if x, y and z are numbers, x = y and y = z means x = z
  • if x is a number and x = y then you can be assured that y is a number
  • if x is a number, S(x) is a number
  • there is no number x such that S(x) = 0
  • if x and y are numbers, S(x) = S(y) means x = y
  • if C is a collection of numbers and 0 is in C and if x is a number in C that means S(x) is in C then C contains every natural number

http://en.wikipedia.org/wiki/Peano_axioms#The_axioms

Given this, we can define P(,) so that
P(x,0) = x
P(x,S(y)) = S(P(x,y))

And we can define M(,) so that
M(x,0) = 0
M(x,S(y)) = P(x,M(x,y))

Then we can prove:
P(x,S(0)) = S(P(x,0)) = S(x)
M(x,S(0)) = P(x,M(x,0)) = P(x,0) = x
M(x,S(S(0))) = P(x,M(x,S(0))) = P(x,x)
P(S(0),S(0)) = S(S(0))
P(S(S(0)),S(0)) = S(S(S(0)))
M(x,S(S(S(0)))) = P(x, M(x,S(S(0))) = P(x, P(x, x))
Which justifies the definitions
1 = S(0)
2 = S(1)
3 = S(2)
Reexpressing these theorems in terms of these definitions:
P(x,1) = S(x)
M(x,1) = x
M(x,2) = P(x,x)
P(1,1) = 2
P(2,1) = 3
M(x,3) = P(x, P(x, x))
 
rpenner-What's the score Z÷(10^n)=?
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 4488
[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 4489
[S7]-opposite point of number
SM-I do not know the opposite point of
 
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
View attachment 4492
P3-four numbers (a, b, c, d) are comparable with each other
View attachment 4493
...
[S8]-comparability issues
CM knows the comparability of two numbers, the comparability of three numbers (a number comparable with the numbers b and c)
comparability of the other knows.
__________________________________________________ _________
Presupposition-number ranges number along
The process:
P1-image
P2-image
P3-image
...
View attachment 4494
[S9]-mobility of number
CM does not know the number of mobility
 
Presupposition-number and mobile number of a contact
The process:
P1-3 + (.0) 2 = 3
P2-3 + (.1) 2 = 3
P3-3 + (.2) 2 = 4
P4-3 +2 = 5, same as the current sum
View attachment 4495
[S10]-addition of
CM knows only one type of addition is given as an axiom
____________________________________________________________
Presupposition-No I do not have a mobile contact number, except to point
The process:
P1-¤3(0)2¤
P2-¤3(1)2¤
P3-¤3(2)2¤
...
View attachment 4496
Next-gap number and mobile number have no contact, except to point
...
[S11]-gap numbers Gn = {¤a(b)c¤,...,¤ (b )...( d) e ¤}
[S12]-gap along the
CM-gap does not know the numbers, gaps along
 
Presupposition-gap number is comparable with the gap number and
The process:
P1-¤a(b)c¤ , a+c=z
P2-¤a(b)c(d)e¤ , a+c+e=z
P3-¤a(b)c(d)e(f)g¤ , a+c+e+g=z
...
Bringing the number of gaps in the number of z, as it compares the number of
[S13]-comparability gap of
CM-comparability gap does not know the number of
___________________________________________________________________________
Presupposition-two or more of the same numbers can be written in abbreviated form
The process:
P1-{a, a} = af2
P2-{a, a, a} = af3
P3-{a, a, a, a} = af4
...
[S14]-the same number of frequency
CM does not know the frequency of the same number
_____________________________________________________________________-
Presupposition-this can be written in abbreviated form
-growing (a, a + b, a + b + b,..., a + b + b +...+ b)
-descending (a + b + b +...+ b, a + b + b, a + b, a)
P1-abc, c = a + b, c = a + b +b, ..., c=a + b + b +...+ b-final
P2-ab - infinity
[S15]-srcko
CM-does not know srcko
 
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
 
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
 
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¤
View attachment 4502
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
 
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:

What use is your new number theory?

What new results have you discovered using your theory?

What are the advantages of your theory over conventional number theory?

Where does your work fit into the general body of mathematical knowledge?
 
James R said:
msbiljanica:

1.What use is your new number theory?

2.What new results have you discovered using your theory?

3.What are the advantages of your theory over conventional number theory?

4.Where does your work fit into the general body of mathematical knowledge?
Click to expand...
1.relationship and the existence of new types of numbers
2.existence of gap numbers
3.current numbers are set axiom, my numbers arise from something (along, the numbers are different name for the longer (except for the number 0, which is the point))
4.when you compare my math (somewhere around May this year, when I show a lot of my math) and current mathematics get an answer to this question
 
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
 
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
 
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:
View attachment 4504
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)
 
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))1[sub]1[/sub] or d+(s.0)number (more) from 1[sub]1[/sub]
and d+(.z)1[sub]1[/sub]¤¤e or d+(.z) number ( more ) from 1[sub]1[/sub]¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 5 - (.0)5=5+(s.0)1[sub]1[/sub] , 5 - (.0)5<6[sub]1[/sub]
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)1[sub]1[/sub]¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤<2¤¤(2)+(.z)1[sub]1[/sub]¤¤(2) , ¤2(2)2¤ - (.1)¤1/1)2¤<3[sub]1[/sub]¤¤(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)1[sub]1[/sub] ,a - (s.b)c<g[sub]1[/sub]
a - (.b)c=d+(s.0)g , a - (.b)c<l
a - (.b)c=d+(s.0)k[sub]p[/sub]g , a - (.b)c<h[sub]i[/sub]j ...
a -(.b)c=d+(.z)1[sub]1[/sub]¤¤e , a - (.b)c<s¤¤e+(.z)1[sub]1[/sub]¤¤e , a - (s.b)c<g[sub]1[/sub]¤¤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[sub]p[/sub]g¤¤e , a - (.b)c<s¤¤e+(.z)k[sub]p[/sub]g¤¤e , a - (.b)c<h[sub]i[/sub]j¤¤e ...
[S29]-left inequality opposite subtraction
CM-[S29]-does no know
_________________________________
Presupposition-In the expression a - (.b)c=d , d-(s.0/s.0))1[sub]1[/sub]p or d+(s.0/s.0)number (more) from
1[sub]1[/sub]p and d-(.z)1[sub]1[/sub]p¤¤e or d-(.z) number ( more ) from 1[sub]1[/sub]¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 5 - (.0)5=5 -(s.0/s.0))1[sub]1[/sub]5 , 5 - (.0)5>0[sub]1[/sub]4
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)2[sub]1[/sub]1¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)2[sub]1[/sub]1¤¤(2) , ¤2(2)2¤ - (.1)¤1(1)2¤>0[sub]1[/sub]1¤¤(2)
P2 5 - (.0)5=5-(s.0/s.0))1 , 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)1[sub]1[/sub]p ,a - (s.b)c>g[sub]1[/sub]l
a - (.b)c=d-(s.0/s.0))g , a - (.b)c>l
a - (.b)c=d-(s.0/s.0))k[sub]s[/sub]g , a - (.b)c>h[sub]i[/sub]j ...
a - (.b)c=d-(.z)1[sub]1[/sub]p¤¤e , a - (.b)c>s¤¤e-(.z)1[sub]1[/sub]p¤¤e , a - (s.b)c>g[sub]1[/sub]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[sub]p[/sub]g¤¤e , a - (.b)c>s¤¤e-(.z)k[sub]p[/sub]g¤¤e , a - (.b)c>h[sub]i[/sub]j¤¤e ...
[S30]-right inequality opposite subtraction
CM-[S30]-does no know
 
Presupposition-Two ( more) opposite subtraction (left and right inequalities) can be short to write
Process:
P1 7 - (.0[sub]1[/sub]3)4=y , 7 - (.0[sub]1[/sub]3)4>y ,7 - (.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 ,...
[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
 
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