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here's the right solution ,
- GeoGebra solution https://onedrive.live.com/?authkey=%21ANn523SIEkl70Bc&cid=48F411219265CF17&id=48F411219265CF17%21194&parId=48F411219265CF17%21105&o=OneUp
for AE greater than AB
- instruction manual, point D changes the length of the side AD, point B changes the length AB, point E changes the length of AE, turn on (design description) and check that I have followed the rules.
q1 q2 ABCD and AEFG rectangle surfaces are accuracy checks
- there is a version when AE is less AB, if you know how to set it
only ruler and compass , no hyperbola
Rectangle ABCD, along an AE that is smaller or larger than the page AB, construct an Rectangle AEFG that has the same surface area as the rectangle ABCD, only unmarked ruler and conifer are allowed, preferably in GeoGebra
theme - inverted proportionality solved by geometric process
https://www.geogebra.org/m/CukhmEVy
straightedge slip on the point B - line i
divider FIG , slides on straightedge , after slipping point F straighte line BC , point G describes lokus1
section lokus1 and line k point J , when changing the angle, the point J must be manually set to the intersection
construction of regular polygon is possible by means of the proportion of the angle
PDF - Mathematics - a new base
https://skydrive.live.com/?cid=48f41121 … CF17%21105
Hello msbiljanica
Should it be that you are from the Balkan area?
I am from Serbia - South Eastern Europe
because I do not get it, here's the work of the occurrence arranged by the sheer number of mathematical notation, mathematical journals refused to publish it, see if it's worth.
https://docs.google.com/file/d/0BzkWG0xdRpPYVTZxVThoUkJlRWs/edit
Presupposition-Two ( more) multiplying the same numbers can be for short write
Process:
P1 a×(s.c)a=
Hi;
Can you pause and explain your ideas before moving on? If you are just intending to publish your book then vixra is a much better place to do that.
what I've realized that the whole thing in math geometry, and that everything is a ratio of two (more) geometrical object in a line, plane, n-volume, I am totally rejected the historical legacy that has a current math, I started from the first geometry object (along natural) and the concept of point and a basic rule to connect the two (more) along the natural area this is the first axiom and it does not prove,each new object (concept, calculation, ...) occurs along the natural object or arising out along natural ...
If something is unclear ask
Presupposition-Srki and gap srki merger (multiplication )
Process:
P1 ¤1(7)1¤=¤1(1)1(1)1(1)1¤×(s.5¤|ß3)6=¤1(1)1(1)1(1)1¤
¤1(1)1(1)1¤=¤1(1)1(1)1(1)¤×(s.5¤|ß4)6=0
P2 4=¤1(1)1(1)1(1)1¤×(s.6¤|ß3)6=2 image
P3 ¤1(1)1(1)1(1)1¤=¤1(1)1(1)1(1)1¤×(s.7¤|ß6)6=¤1(1)1(1)1¤
[attachment=2682:a20.png]
General form
w=a#(s.1¤|ß3)b=c
w=a#(s.2¤|ßd)b=c
...
w=a#(s.e¤|ßf)b=c , #- calculation operations (×,...)
[S33]-two srki
CM-[S33]-does no know
[size="4"]Note - this is a two-way calculation, and therefore has two equals signs left(srki) right(gap srki)[/size]
__________
Presupposition-Subtraction form a-(s.0/s.0)b=0,a-(s.0/s.0)b-(s.0/s.0)b=0 , ...,a-(s.0/s.0)b-
...-(s.0/s.0)b=0 can be written differently counting the subtraction
Process:
P1 12-(..0)4=12 (..c) -counting the subtraction
12-(..1)4=8
12-(..2)4=4
12-(..3)4=0
12:4=3
General form
a-(..0)b=a
a-(..1)b=c
...
a-(..d)b=0
a:b=d
[S34]-counting the subtraction
[S35]-divide
CM-[S34]-does no know ,[S35]-know, axiom
_____________________________
Presupposition-Subtraction form a-(s.0/s.0)b=0,a-(s.0/s.0)b-(s.0/s.0)b=0 , ...,a-(s.0/s.0)b-
...-(s.0/s.0)b=0, number b can replaced b= ((s.0/s.0)c-(s./s.0)d) , b= ((s.0/s.0)c-(s.0/s.0)d-
(s./s.0)e),...,b= ((s.0/s.0)...(s.0/s.0)w) , can be written differently counting the subtraction
Process:
P1 12-(..0)1[SUB]1[/SUB]2=12
12-(..1)1[SUB]1[/SUB]2=11
12-(..2)1[SUB]1[/SUB]2=9
12-(..3)1[SUB]1[/SUB]2=8
12-(..4)1[SUB]1[/SUB]2=6
12-(..5)1[SUB]1[/SUB]2=5
12-(..6)1[SUB]1[/SUB]2=3
12-(..7)1[SUB]1[/SUB]2=2
12-(..8)1[SUB]1[/SUB]2=0
12:1[SUB]1[/SUB]2=8
General form
a-(..0)b[SUB]e[/SUB]f=a
a-(..1)b[SUB]e[/SUB]f=c
...
a-(..d)b[SUB]e[/SUB]f=0
a:b[SUB]e[/SUB]f=d ...
[S36]-inequality divide
CM-[S36]-does no know
Presupposition-The numbers are multiplication , contact remains the rest is deleted
Process:
P1 4 × (s.0)4=0
P2 4 × (s.1)4=¤1(2)1(2)1¤ image
P3 4 × (s.2)4=6
P4 4 × (s.3)4=5
P5 4 × (s.4)4=4
--image--
General form
a × (s.0)b=c
a ×(s.1)b=c
...
a × (s.d)b=c
[S30]-multiplication opposite subtraction
CM-[S30]-does no know
________________________
Presupposition-Three (more) merger (multiplication )
Process:
P1 5×(s.3ß3)5=¤1(1)1¤ - image
P2 5×(s.4ß3)5=¤1(3)1¤ - image
5×(s.4ß4)5=¤1(1)1¤ -image
5×(s.4ß5)5=1- image
P3 5×(s.5ß5)=5
--image--
General form
a#(s.1ß3)b=c
a#(s.2ßd)b=c
...
a#(s.eßf)b=c , #- calculation operations (×,...)
[S31]-srki
CM-[S31]-does no know
___________________
Presupposition-Three (more) gap merger (multiplication )
Process:
P1 ¤1(4)1¤×(s.5¤ß3)5=¤1(2)1¤ - image
¤1(4)1¤×(s.5¤ß4)5=2-image
P2 ¤1(4)1¤×(s.6¤ß4)5=4
--image--
General form
a#(s.1¤ß3)b=c
a#(s.2¤ßd)b=c
...
a#(s.e¤ßf)b=c , #- calculation operations (×,...)
[S32]-gap srki
CM-[S32]-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
...
[S28]-multiplication
CM-[S28]-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
[S28a]-multiplication-amendment
_____________
Presupposition-In terms a×(s.c)b , multiplication
Process:
P1 4×(s.0)4=16
P2 4×(s.1)4=13 image
P3 4×(s.2)4=10
P4 4×(s.3)4=7
P5 4×(s.4)4=4
--image--
P1 ¤1(2)1¤×(s.0)4=¤1(2)2(2)2(2)2(2)1¤
P2 ¤1(2)1¤×(s.1)4=¤1(2)1(2)1(2)1(2)1¤ image
P3 ¤1(2)1¤×(s.2)4=¤1(1)6(1)1¤
P4 ¤1(2)1¤×(s.3)4=7
P5 ¤1(2)1¤×(s.4)4=¤1(2)1¤
--image--
General form
a×(s.0)b=c
a×(s.1)b=c
...
a×(c.d)b=c
[S28b]-multiplication
CM-[S28b]-know a×(s.0)b other do not know , forms withuut any gaps numbers not known
___________________________________________
Presupposition-The numbers are multiplication , where a contact is deleted
Process:
P1 4 × (s.0)4=¤4(0)4(0)4(0)4¤
P2 4 × (s.1)4=¤3(1)2(1)2(1)3¤ image
P3 4 × (s.2)4=¤2(6)2¤
P4 4 × (s.3)4=¤1(5)1¤
P5 4 × (s.4)4=0
--image--
General form
a × (s.0)b=c
a × (s.1)b=c
...
a × (c.d)b=c
[S29]-multiplication subtraction
CM-[S29]-does no know
×You will see a sign in the PDF
Presupposition-In the expression a-(.b)c=d , d+(.z)1[SIZE="1"]1[/SIZE]¤¤e or d+(.z) number ( more ) from
1[SIZE="1"]1[/SIZE]¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 4-(.3)2=¤1(2)1¤+(.z)1[SIZE="1"]1[/SIZE]¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)1[SIZE="1"]1[/SIZE]¤¤(2) , 4-(.3)2<3[SIZE="1"]1[/SIZE]¤¤(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)1[SIZE="1"]1[/SIZE]¤¤e , a#(.b)c<s¤¤e+(.z)1[SIZE="1"]1[/SIZE]¤¤e , a#(s.b)c<g[SIZE="1"]1[/SIZE]¤¤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[SIZE="1"]p[/SIZE]g¤¤e , a#(.b)c<s¤¤e+(.z)k[SIZE="1"]p[/SIZE]g¤¤e , a#(.b)c<h[SIZE="1"]i[/SIZE]j¤¤e ..., #-calculation operations
(+,-,×,..)
[S25]-left inequality gap number
CM-[S25]-does no know
_____________________________________
Presupposition-In the expression a-(.b)c=d , d-(.z)1[SIZE="1"]1[/SIZE]p¤¤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)2[SIZE="1"]1[/SIZE]1¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)2[SIZE="1"]1[/SIZE]1¤¤(2) , 4-(.3)2>0[SIZE="1"]1[/SIZE]1¤¤(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)1[SIZE="1"]1[/SIZE]p¤¤e , a-(.b)c>s¤¤e-(.z)1[SIZE="1"]1[/SIZE]p¤¤e , a-(s.b)c>g[SIZE="1"]1[/SIZE]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[SIZE="1"]p[/SIZE]g¤¤e , a-(.b)c>s¤¤e-(.z)k[SIZE="1"]p[/SIZE]g¤¤e , a-(.b)c>h[SIZE="1"]i[/SIZE]j¤¤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
--image--
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
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¤ --0[sub]1[/sub]3¤¤(2)
¤2(2)0¤,¤1(2)1¤,¤0(2)2¤
¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤
...
2[sub]1[/sub]¤¤(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
Presupposition-Parts number (a) and mobile number (b ) have a contact , the contact is delete
Process:
P[sub]1[/sub] 4-(.0)2=2
P[sub]2[/sub] 4-(.1)2=¤1(2)1¤ image
P[sub]3[/sub] 4-(.2)2=2
P[sub]4[/sub] 4-(.3)2=¤3(1)1¤
P[sub]5[/sub] 4-(.4)2=¤4(0)2¤
--image--
General form
a-(.0)b=c
a-(.1)b=c
...
a-(.d)b=c
[S[sub]19[/sub]]-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))1[sub]1[/sub]f or d-(s.0/s.0))number (more) from 1[sub]1[/sub]f
Process:
P[sub]1[/sub] 3+(.s.0)5=8-(s.0/s.0))1[sub]1[/sub]8 , 3+(s.0)5>0[sub]1[/sub]7
P[sub]2[/sub] 5+(.0)5=5-(s.0/s.0))2[sub]2[/sub]4 , 5+(.0)5>3[sub]2[/sub]1
...
General form - a+(.b)c=d-(s.0/s.0))1[sub]1[/sub]f ,a+(s.b)c>0[sub]1[/sub]e
a+(.b)c=d-(s.0/s.0)e , a+(.b)c>f
a+(.b)c=d-(s.0/s.0)e[sub]f[/sub]g , a+(.b)c>h[sub]i[/sub]j ...
[S[sub]20[/sub]]-right inequality addition
CM-[S[sub]20[/sub]]-know
____________________
Presupposition-Two ( more) addition (left and right inequalities) can be short to write
Process:
P1 3+(.0[sub]1[/sub]3)4=y , 3+(.0[sub]1[/sub]3)4>y ,3+(.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 ,...
[S21]-function addition
CM-[S21]-does no know
Presupposition-Two ( more ) srcko (pendand srcko) are combined into one unit
Process:
P[sub]1[/sub] 10[sub]6 [/sub]and 11[sub]8[/sub] , 10[sub]6[/sub]11[sub]8[/sub]
P[sub]2[/sub] 10[sub]5[/sub]65 and 70[sub]3[/sub] ,10[sub]5[/sub]65_70[sub]3[/sub]
P[sub]3[/sub] 30[sub]3[/sub]60 and 45[sub]2[/sub]77_78 ,30[sub]3[/sub]60_45[sub]2[/sub]77_78
...
General form -a[sub]b[/sub]c[sub]d[/sub] , a[sub]b[/sub]c_d[sub]e[/sub] ,a[sub]b[/sub]c_d[sub]e[/sub]f_g ,...
[S[sub]16[/sub]]-two ( more) srcko
CM-[S[sub]16[/sub]]-does no know
_________________-
Presupposition-Two ( more ) srcko have the first ( last) common number
Process:
P[sub]1[/sub] 10[sub]5[/sub]30 and 3[sub]3[/sub]30 , 10[sub]5[/sub]3[sub]3[/sub](_30)
P2 4[sub]4[/sub]44 and 44[sub]10[/sub]94 and 44[sub]2[/sub]56 , 4[sub]4[/sub](_44_)[sub]10[/sub]94[sub]2[/sub]56
...
General form -a[sub]b[/sub]c[sub]d[/sub](_e) , a[sub]b[/sub](_c_)[sub]d[/sub]e[sub]f[/sub]g , ...
[S[sub]17[/sub]]-two ( more) first-last srcko
CM-[S[sub]17[/sub]]-does no know
______________________________________________________
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]
Process:
P[sub]1[/sub] 3+(.s.0)5=8+(s.0)1[sub]1[/sub] , 3+(s.0)5<9[sub]1[/sub]
P[sub]2[/sub] 5+(.0)5=5+(s.0)2[sub]2[/sub]4 , 5+(.0)5<7[sub]2[/sub]9
...
General form - a+(.b)c=d+(s.0)1[sub]1[/sub] ,a+(s.b)c<e[sub]1[/sub]
a+(.b)c=d+(s.0)e , a+(.b)c<f
a+(.b)c=d+(s.0)e[sub]f[/sub]g , a+(.b)c<h[sub]i[/sub]j ...
[S[sub]18[/sub]]-left inequality
CM-[S[sub]18[/sub]]-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
5[sub]5[/sub]50={5,10,15,20,25,30,35,40,45,50}
38[sub]3[/sub]50={38,41,44,47,50}
50[sub]10[/sub]90={50,60,70,80,90}
50[sub]7[/sub]92={50,57,64,71,78,85,92}
two(more) first-last srcko
5[sub]5[/sub]38[sub]3[/sub](_50_)[sub]10[/sub]90[sub]7[/sub]92
remains part of the function, when we come to it
Presupposition-Gap number is comparable with the gap number and number
Process:
P[sub]1[/sub] ¤a(b)c¤ , a+(s.0)c=z
P[sub]2[/sub] ¤a(b)c(d)e¤ , a+(s.0)c+(s.0)e=z
P[sub]3[/sub] ¤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
[S[sub]13[/sub]]-comparability gap numbers
CM-[S[sub]13[/sub]]-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:
P[sub]1[/sub] - 3+(s.0)0=3 , 3+(s.0)4=7 , 3[sub]4[/sub]7
3+(s.0)0=3 , 3+(s.0)4=7 , 3+(s.0)4+(s.0)4=11 , 3[sub]4[/sub]11
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 , 3[sub]4[/sub]15
...
3+(s.0)0=3 , ... , 3+(s.0)4+...+(s.0)4=d , 3[sub]4[/sub]
P[sub]2[/sub] - 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 , 15[sub]4[/sub]3
...
3+(s.0)4=7 , 3+(s.0)0=3 , 7[sub]4[/sub]3
General form -a[sub]b[/sub]c , a[sub]b[/sub]
[S[sub]14[/sub]]-srcko
CM-[S[sub]14[/sub]]-does no know
____________________________________________
Presupposition-Srcko can join a number not that can not be in the structure srcko
Process:
P[sub]1[/sub] 10[sub]10[/sub]70 and 5 , 5_10[sub]10[/sub]70
P[sub]2[/sub] 5[sub]5[/sub]20 and 22 ,5[sub]5[/sub]20_22
P[sub]3[/sub] 7[sub]5[/sub] and 25 , 7[sub]5[/sub]_25
P[sub]4[/sub] 6[sub]8[/sub] and 2 ,2_6[sub]8[/sub]
...
General form -a[sub]b[/sub]c_d , d_a[sub]b[/sub]c , a[sub]b[/sub]_d ,d_a[sub]b[/sub]...
[S[sub]15[/sub]]-pendant srcko
CM-[S[sub]15[/sub]]-does no know
Note-only one number can be pendand , number two goes into a complex srcko
Presupposition-number ranges for number along
Process:
P1-image
P2-image
P3-image
--image--
[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¤
...
--image--
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
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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
form (.a/0) and (s.a/0) well continue this write (.a) and (s.a)
--image--
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
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)}
...
--image--
...
[S7]-number opposite points
CM-[S7]does not know
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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
--image--
P3-four numbers (a, b, c, d) are comparable with each other
--image--
...
[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.
A question to you, in the last month or so you have posted this message to about 75
forums. Those forums are mostly about philosophy and a smattering of other subjects. But no maths forums
except this one. Why?Welcome to the forum.
who wants me to discuss and learn new things in mathematics is welcome who wants me to listen to his will
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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)
...
--image--
[S3]-along (natural basis)
Definition[along]-the first and last point and the distance between points
CM-[S3]-does not know
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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, ...}
...
--image--
[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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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)}
...
--image--
[S6]-number points
CM-[S6]does not know
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--
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
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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--
Marjanovic Srdan
[ contact information removed by moderator ]
natural axiom
What is " nature along "?
-nature along in figure 1
What is "point"?
-start (end) natural long in figure 2
--a2--
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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