srdan math

Discussion in 'Pseudoscience Archive' started by msbiljanica, Jan 5, 2012.

  1. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
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  3. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,888
    Welcome.

    And let me be the first to thank you for saving math and removing it's sinful nature.
     
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  5. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    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))
     
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  7. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  8. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  9. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  10. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  11. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  12. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  13. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  14. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  15. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    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?
     
  16. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  17. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  18. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  19. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    What use are "gap numbers"?
     
  20. msbiljanica Registered Senior Member

    Messages:
    76
    from previous exposure to guess you realize what good gap numbers , for complex shapes that are separated by gap between them
     
  21. msbiljanica Registered Senior Member

    Messages:
    76
    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)
     
  22. msbiljanica Registered Senior Member

    Messages:
    76
    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
     
  23. msbiljanica Registered Senior Member

    Messages:
    76
    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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