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
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 a: (..0)b=a a: (..1)b=c ... a: (..g)b=1 Process: 625: (..0)5=625 (..d) - counter divisions 625: (..1)5=125 625: (..2)5=25 626: (..3)5=5 625: (..4)5=1 View attachment 4537 [S42]-root [S43]-counter divide CM-[42] - know [43]-dot no know _____________________ PDF [S43] - http://www.mediafire.com/?435vx8lu7111n5f
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.