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Topic: JS coding
Replies: 27   Last Post: Feb 24, 2013 6:17 AM

 Messages: [ Previous | Next ]
 JT Posts: 1,448 Registered: 4/7/12
Re: JS coding
Posted: Feb 24, 2013 6:17 AM

On 23 Feb, 14:43, JT <jonas.thornv...@gmail.com> wrote:
> On 23 Feb, 08:46, Virgil <vir...@ligriv.com> wrote:
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> > In article

>
> >  JT <jonas.thornv...@gmail.com> wrote:
> > > On 11 Feb, 00:17, Virgil <vir...@ligriv.com> wrote:
> > > > In article

>
> > > > JT<jonas.thornv...@gmail.com> wrote:
> > > > > Below ternary fractions in NyaN
> > > > > Paragraphs for empty preceding
> > > > > entrys to keep track of multiple * bas (Ternary=1,3,9,27,81....)

>
> > > > > Fractions = NyaN ternary fraction
> > > > > 1/3 = .1
> > > > > 2/3 = .2
> > > > > 1/9 = .(1)1
> > > > > 2/9 = .(1)2
> > > > > 1/27 = .(2)1
> > > > > 2/27 = .(2)2
> > > > > 1/81 = .(3)1
> > > > > 2/81 = .(3)2 ...

>
> > > > Since 0's in standard notation are merely placeholders, you still
> > > > apparently need placeholders.

>
> > > > HOW IS 1/81 = .(3)3 BETTER THAN 1/81 =.0001 in base 3?
> > > > --

>
> > > No there is no placeholders needed for the naturals because they are
> > > discrete incremental entities

>
> > In decimal notation, 10 and 100 and 1000, and so forth all require
> > placeholders.

It is not the partitioning/grouping into bases itself that are in
question, what is in question if we really should allow write empty
positions in bases.

Example 21 decimal equals 210 in ternary (2*9)+(1*3)+(0*1) while in
NyaN numbers represented in bases do not have void positions =(0), so
using NyaN writing 21 in ternary equals 133 (1*9)+(3*3)+(1*3)
this way to write numbers is of course working for anybase so ternary
1000000000000000000100000000000000000001/201 would be much easier
perform using NyaN

> No they do not need ask the romans they used X, but for us maybe A
> would be more suitable.
> Now you have 123456789A or 123456789X instead of 0123456789 you may
> notice the zero is gone.
>
> zero is just a distraction performing math, especially when dividing
> writing without zeros have great computational advantages when it come
> to factoring, it may also come with some insight into prime number
> distribution so finally someone can crack the mystery of the Riemann
> zeta function.
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> > > \ the fractions though are parts of the
> > > continuum and expressed as part of one whole. This does not mean there
> > > is really any numerical value to the preceding zeros as you as well
> > > could write it as 10^-3 or A^-3.

>
> > And how is writing 5*10^-3 any less placeholder dependent that 0.005?
>
> > In one way or another, the existenceof places holding 0's must be noted.
> > --

Date Subject Author
2/3/13 JT
2/3/13 JT
2/3/13 JT
2/3/13 Virgil
2/3/13 JT
2/3/13 JT
2/3/13 JT
2/3/13 JT
2/3/13 JT
2/4/13 JT
2/6/13 JT
2/6/13 Virgil
2/6/13 JT
2/7/13 JT
2/10/13 JT
2/10/13 Virgil
2/22/13 JT
2/23/13 Virgil
2/23/13 JT
2/23/13 JT
2/23/13 JT
2/24/13 JT
2/23/13 JT
2/6/13 JT
2/6/13 JT
2/6/13 JT
2/3/13 JT
2/3/13 JT