srdan math

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
View attachment 4506

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¤
View attachment 4507
General form
a×(s.0)b=c
a×(s.1)b=c
...
a×(c.d)b=c

[S32b]-multiplication
CM-[S32b]-know a×(s.0)b other do not know , forms withuut any gaps numbers not known
 
PDF - 5 days
___________________________________________
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
View attachment 4510
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
 
rpenner said:
Still waiting for you to explain the purpose of this thread.
Click to expand...
different mathematics, different bases, all the geometry ...
__________________________________________----
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
View attachment 4511
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
View attachment 4512
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
View attachment 4516
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-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¤
View attachment 4517
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
Note - this is a two-way calculation, and therefore has two equals signs left(srki) right(gap srki)
PDF - http://www.mediafire.com/?ffy4ravu1gw4xkd
 
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)c-...-(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
________________________
PDF - srdanova math [S36].pdf
 
Presupposition-Two ( more) multiplying the same numbers can be for short write
Process:
P1 a×(s.c)a=a[sup](s.c)2[/sup], a[sup](s.a)b[/sup]
P2 a×(s.c)a×(s.c)a=a[sup](s.a)3[/sup] , a[sup](s.a)b[/sup]
P3 a×(s.c)a×(s.c)a×(s.c)a=a[sup](s.a)4[/sup] ,a[sup](s.a)b[/sup]
...
General form
a[sup](s.a)b[/sup]
[S37]-graduation
CM-[S37]-know,axiom
__________________________________
Presupposition-In terms a[sup](s.a)b [/sup] , b can be number 0 (1)
Process:
P1 a[sup](s.a)0[/sup]
P2 a[sup](s.a)1[/sup]
[S37a]-graduation -amendment
 
Presupposition-Graduation
Process:
P1- 4[sup](s.0)5[/sup]=1024
P2- 4[sup](s.1)5[/sup]=769
P3- 4([sup]s.2)5[/sup]=514
P4- 4([sup]s.3)5[/sup]=259
P5- 4[sup](s.4)5[/sup]=4
General form
a[sup](s.0)b[/sup]=c
a[sup](s.1)b[/sup]=d
...
a[sup](s.e)b[/sup]=a
[S37b]-graduation
CM-[S37b]-know a[sup](s.0)b[/sup]=c , other do not know
 
Presupposition-Two ( more) same number are abbreviated
Process:
P1- {4,4}=4[sub]f2[/sub]
P2-{4,4,4}=4[sub]f3[/sub]
P3-{4,4,4,4}=4[sub]f4[/sub]
....
General form
{a,a}=a[sub]f2[/sub]
{a,a,a}=a[sub]f3[/sub]
...
{a,...,a}?=a[sub]fn[/sub]
[S38]-the same numbers
CM-[S38]- do not know
 
Presupposition- Part gap number of repetitive two ( more) times
Process:
P1-¤5(1)1(1)1(1)1¤ - ¤5((1)1)[sub]f3[/sub]¤
P2-¤6(2)3(4)3(4)3(4)3(4)5¤ -¤(6(2)(3(4))[sub]f4[/sub]5¤
...
General form
¤b((c )d)[sub]fn[/sub]¤
¤b(c )(a(d))[sub]fn[/sub]g¤
...
[S39]-the same part gap number
CM-[S39]- do not know
 
Presupposition-Graduation, contact is deleted
Process:
P1- 4[sup](s.0)5[/sup]=¤4((0)4)[sub]f255[/sub]¤
P2- 4[sup](s.1)5[/sup]=¤3((1)2)[sub]f254[/sub]3¤
P3- 4[sup](s.2)5[/sup]=¤2(508)2¤
P4- 4[sup](s.3)5[/sup]=¤1(254)1¤
P5- 4[sup](s.4)5[/sup]=0
General form
a[sup](s.0)b[/sup]=c
a[sup](s.1)b[/sup]=d
...
a[sup](s.e)b[/sup]=0


[S40]-graduation subtraction
CM-[S40]-dot no know
 
Presupposition-Graduation, contact remanins , others will be deleted
Process:
P1- 4[sup]-(s.0)5[/sup]=¤(0(1))[sub]f254[/sub]1¤
P2- 4[sup]-(s.0)5[/sup]=¤1((2)1)[sub]f254[/sub]¤
P3- 4[sup]-(s.0)5[/sup]=510
P4- 4[sup]-(s.0)5[/sup]=255
P5- 4[sup]-(s.0)5[/sup]=4
General form
a[sup]-(s.0)b[/sup]=e
a[sup]-(s.1)b[/sup]=d
...
a[sup]-(s.f)b[/sup]=c
[S41]-graduation opposite subtraction
CM-[S41]-dot no know
 
Presupposition-Term written in abbreviated form , number b is replaced with other numbers
,alternating
a: (..0)b=a
a: (..1)b=c
...
a: (..g)b=1
Process:
400: (..1)4=100
400: (..2)5=20
400: (..3)4=5
400: (..4)5=1
View attachment 4547
[S44]-inequality root
CM-[S44] -dot no know
 
Presupposition-Part number (the number of gaps) negates the second number(gap number)
-one of then must be gaps numbers
Process:
View attachment 4548
P1- 0=¤3(1)2(1)2¤n(.0)¤2(2)2¤=1
P2- 1=¤3(1)2(1)2¤n(.1)¤2(2)2¤=1
P3- ¤1(2)1¤=¤3(1)2(1)2¤n(.2)¤2(2)2¤=2-image
P4- 1=¤3(1)2(1)2¤n(.3)¤2(2)2¤=1
P5- 0=¤3(1)2(1)2¤n(.4)¤2(2)2¤=1
P6- 1=¤3(1)2(1)2¤n(.5)¤2(2)2¤=2
P7- 0=¤3(1)2(1)2¤n(.6)¤2(2)2¤=1
P8- 0=¤3(1)2(1)2¤n(.7)¤2(2)2¤=0
P9- 0=¤3(1)2(1)2¤n(.8)¤2(2)2¤=0
P10- 0=¤3(1)2(1)2¤n(.9)¤2(2)2¤=0
General form
negates 1=a n (.c)b=negates 2
or
negates 1 =a n (.c)b
negates 2
[S45]-negates
CM-[S45] -dot no know
 
Presupposition-Part number (the number of gaps) negates the second number(gap number)
-one of then must be gaps numbers, other parts are added
Process:
View attachment 4549
P1- 0=¤3(1)2(1)2¤n+(.0)¤2(2)2¤=1
¤2(2)2(1)2¤=
P2- 1 =¤3(1)2(1)2¤n+(.1)¤2(2)2¤=1
¤3(2)1(1)2¤=
P3- ¤1(2)1¤ =¤3(1)2(1)2¤n+(.2)¤2(2)2¤=2
¤3(4)2¤=
P4- 1 =¤3(1)2(1)2¤n+(.3)¤2(2)2¤=1
¤3(1)1(2)2¤=
P5- 0 =¤3(1)2(1)2¤n+(.4)¤2(2)2¤=1
¤3(1)2(2)2¤=
P6- 1 =¤3(1)2(1)2¤n+(.5)¤2(2)2¤=2
¤3(1)2(3)2¤=
P7- 0 =¤3(1)2(1)2¤n+(.6)¤2(2)2¤=1
¤3(1)2(1)1(2)2¤=
P8- 0 =¤3(1)2(1)2¤n+(.7)¤2(2)2¤=0
¤3(1)2(1)2(2)2¤=
P9- 0 =¤3(1)2(1)2¤n+(.8)¤2(2)2¤=0
¤3(1)2(1)3(2)2¤=
P10- 0 =¤3(1)2(1)2¤n+(.9)¤2(2)2¤=0
¤3(1)2(1)4(2)2¤=
General form
negates 1 =a n+(.c)b=negates 2
added=
or
negates 1 =a n+ (.c)b
negates 2=
added=
[S46]-negates added
CM-[S46] -dot no know
 
If you would attempt to verbalize a bit about what you are trying to accomplish you may have some responses to your thread.
 
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