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Srdanova Math


msbiljanica

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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)

post-27212-0-26267200-1327251107_thumb.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.

post-27212-0-87172800-1327251143_thumb.png

Marjanovic Srdan

M.Biljanica

16201 Manojlovce

Serbia

[email protected]

natural axiom

What is " nature along "?

-nature along in figure 1

What is "point"?

-start (end) natural long in figure 2

post-27212-0-91992400-1327251150_thumb.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

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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)

...

post-27212-0-09381200-1327255831_thumb.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, ...}

...

post-27212-0-72016600-1327255942_thumb.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

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I dont understand, please make a step by step repeat.

This presentation is a step by step, each new piece of evidence derives from "natural along" or previous evidence, it can be easier than this

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)}

...

post-27212-0-57119500-1327344119_thumb.png

[s6]-number points

CM-[s6]does not know

___________________________________________________________

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)}

...

post-27212-0-95170200-1327344185_thumb.png

...

 

[s7]-number opposite points

CM-[s7]does not know

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This presentation is a step by step, each new piece of evidence derives from "natural along" or previous evidence, it can be easier than this

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)}

...

post-27212-0-57119500-1327344119_thumb.png

[s6]-number points

CM-[s6]does not know

___________________________________________________________

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)}

...

post-27212-0-95170200-1327344185_thumb.png

...

 

[s7]-number opposite points

CM-[s7]does not know

 

I still dont understand it all,

but now i begin to think there might be something to understand.

 

Since you have an interest in maths, perhaps you also care for logic?

If so, I invite you to the thread "The Final Solution of the Liar".(#1)

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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

post-27212-0-96986500-1327425429_thumb.png

P3-four numbers (a, b, c, d) are comparable with each other

post-27212-0-45226300-1327425459_thumb.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.

___________________________________________________________________________________________

Presupposition-number ranges for number along

Process:

P1-image

P2-image

P3-image

post-27212-0-07837700-1327425705_thumb.png

[s9]-mobility of number

CM-[s9]-does not know

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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

post-27212-0-04315900-1327507597_thumb.png

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¤

...

post-27212-0-33571500-1327508877_thumb.png

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¤

PDF - http://www.mediafire.com/file/2c7vvarvvqaqp63/srdanova math.pdf

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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

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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

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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

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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¤

post-27212-0-34247800-1327836685_thumb.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

_______________________________________________________________________

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

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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

___________________________

PDF - http://www.mediafire.com/?jzheej0qu93jjkm

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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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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

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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:

post-27212-0-64694400-1328185580_thumb.png

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)

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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))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)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

_______________________________-

PDF - http://www.mediafire.com/?qle11bvr2yuey9o

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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

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