srdanova math

Discussion in 'Math' started by msbiljanica, Feb 14, 2012.

  1. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    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.png
    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.png
    Marjanovic Srdan
    M.Biljanica
    16201 Manojlovce
    Serbia
    <SNIP>
    natural axiom
    What is " nature along "?
    -nature along in figure 1
    What is "point"?
    -start (end) natural long in figure 2
    a2.png
    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
     
    Last edited by a moderator: Feb 14, 2012
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Hello msbiljanica and welcome to All About Circuits.

    I am not quite sure what points you are making here.

    Could you separate them and post them one at a time?
     
  3. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    Thank you for the welcome, this is like a like Euclid's Elements, in the simplest form, as geometric objects (the first natural geometrical object along the notion of point, the basic rule), all other evidence (slightly different than the current mathematics), which arise as a ratio of two (more) geometrical object formed from natural or along previous geometrical object. the point is that I think all of their math geometry, and that the numbers are a calculation function is actually another name for the related geometrical object
    ___________________
    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)
    ...

    a3.png
    [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, ...}
    ...
    a4.png
    [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)}
    ...
    a5.png
    [S6]-number points
    CM-[S6]does not know
     
  4. studiot

    AAC Fanatic!

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    In addition to the axioms of Euclid's Geometry, which were stated without proof there were a number of other supporting statements, also without proof.

    In all there were

    23 Definitions

    5 Postulates

    5 Common Notions or Axioms

    There were also axioms added after Euclid, eg

    "two straight lines do not enclose or contain a space."

    Any new system will need to provide equivalents to all of these, which I suspect will add up to rather more than two statements made without proof.

    You seem to be putting points on a line into one to one correspondence with the integers.
    It is easy to show that there are point on a continuous line for which this cannot be done.
     
    Last edited: Feb 16, 2012
  5. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    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
    _______________________________
    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
    a7.png
    P3-four numbers (a, b, c, d) are comparable with each other
    a8.png
    ...
    [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.
     
  6. studiot

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    Please clear up something for me.

    By "does not know" are you referring to what we call an open or closed set or interval?

    An interval is part of a line or the set of all numbers in that part of a line.

    An open interval does not contain its end points

    A closed interval does

    So for instance the open interval 0, 1 contains all the numbers between 0 and 1 but not 0 or 1 themselves.

    So a closed interval contains all the numbers between 0 and 1 and 0 and 1 themselves.

    We write the open inteval 0,1 as ] 0 , 1 [

    and the closed interval 0,1 as [ 0 , 1 ]

    to distinguish them.
     
    Last edited: Feb 16, 2012
  7. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    calculation, or the name of a geometry object, and the rest, which does not exist in the current mathematics
     
  8. studiot

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    Thanks for your last comments.
    Reading your other posts again is helping a little but

    I am still not sure why you are trying to identify natural numbers with lines or is it points on a line?

    As I pointed out in post#6 although all the numbers in N have equal significance, there are two types of points on a line and these types do not have equal significance.
     
  9. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    numbers are another name for the point (0) and longer (1,2,3,4, ....) of my math
    ____________________________

    Presupposition-number ranges for number along
    Process:
    P1-image
    P2-image
    P3-image
    a9.png
    [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¤
    ...
    a11.png
    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)
    a10.png
    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
     
  10. studiot

    AAC Fanatic!

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    What I understand you are saying is

    Take a stick of length 4 units.
    Place another stick of length 2 units on one end or altenatively
    cut the first stick into two pieces and place the second stick between the pieces of the first.

    Either way the result will be a length of 6 units?

    Please correct this if I have missed something.

    As regards the motivation for this,

    You have used (assumed) several underlying properties from set theory about the natural numbers.

    1)
    If a and b are members of the set N then a+b = c is also a member of the set.
    Furthermore c is unique. There is no other member, say d, for which a + b = d.

    2)
    N contains a member 0 such that a + 0 = a for every member of N
    There is only one member 0 is unique.

    3)
    N contains a member (-a) such that a + (-a) = 0 for every member of N
    Again (-a) is unique
     
  11. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    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
    _______________________________________
    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
     
  12. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    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
     
  13. studiot

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    I'm sorry but you have lost me somewhere along the way.

    It would be very helpful if you could answer one my basic questions rather than ploughing on with more of your maths.

    Otherwise I fear you are doing a lot of work for nothing.

    Please state what these two axioms are and how far they extend.

    Please also state what problems conmventional maths cannot solve, but your system can, so that a comparison can be made.

    That is pose a conventional maths problem in a conventional way.
     
    Last edited: Feb 20, 2012
  14. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    You know the first axiom is called the natural along ,term point, the basic rule for linking the two (more) natural along.
    second axiom when we come to it
    all I ever show starts along the natural, or what became of natural along
    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

    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
    a question for you if a function can be final and natural ?advantage of my mathematics

    1.Z÷(10^n)=?,Z-integers
    a={0,1,2,3,4,5,6,7,8,9} , b={1,2,3,4,5,6,7,8,9}
    n=1 , Z÷10={...,(-2÷10),(-1÷10),(0÷10),(1÷10),(2÷10),...}={...,-0.2,-0.1,0,0.1,0.2,...}={Z,Z.b}
    n=2 , Z÷100={Z,Z.b,Z.ab}
    n=3 , Z÷1000={Z,Z.b,Zab,Zaab}
    n=4 , Z÷10000={Z,Z.b,Z.ab,Zaab,Zaaab}
    ...
    Z÷(10^n)=R
    This evidence shows that the same rational and real numbers, that there are irrational numbers
     
  15. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    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¤
    a12.png
    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
    _______________________________
    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.png
    20m-(.10m)5m=¤10m(5m)5m¤
    How would you solve with current knowledge of mathematics

    _________________--
    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
     
  16. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Do I understand correctly by saying that I think you mean

    Find an expression for Z such that Z is an integer or
    Find all integers, Z, such that

    \frac{Z}{{{{10}^n}}} \in {\rm N}

    What do you mean by a function, a final function and a natural function?
     
  17. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    \frac{Z}{{{{10}^n}}}= {\rm R}
     
  18. studiot

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    I'm sorry I don't understand your statement.

    If you are asserting that the set of all integers, whether divided by some power of 10 or not, is isomorphic to the set of all real numbers

    This is not the case.

    There are more numbers in R than there are in Z.
     
  19. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    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
    _______________
    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
     
  20. msbiljanica

    Thread Starter New Member

    Feb 14, 2012
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    Presupposition-In the expression a-(.b)c=d , d+(.z)11¤¤e or d+(.z) number ( more ) from
    11¤¤e , e={(f),(f)(f),(f)(f)(f),...}
    Process:

    P1 4-(.3)2=¤1(2)1¤+(.z)11¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)11¤¤(2) , 4-(.3)2<31¤¤(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)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 ..., #-calculation operations
    (+,-,×,..)
    [S25]-left inequality gap number
    CM-[S25]-does no know
    _____________________________________
    Presupposition-In the expression a-(.b)c=d , d-(.z)11p¤¤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)211¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)211¤¤(2) , 4-(.3)2>011¤¤(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)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 .. ,#-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
    a12.png
    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
     
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