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Topic: 1/3 partitioned expansion in different bases
Replies: 28   Last Post: Feb 3, 2013 11:57 PM

 Messages: [ Previous | Next ]
 JT Posts: 1,434 Registered: 4/7/12
Re: 1/3 partitioned expansion in different bases
Posted: Feb 1, 2013 11:40 PM

On 2 Feb, 03:54, forbisga...@gmail.com wrote:
> On Friday, February 1, 2013 1:10:12 PM UTC-8, JT wrote:
> > You are confused the zeros in 1/2 1/4 etc etc is imaginary not real.
> > And for the endless expansion of digits in 1/3, 1/7 i offered a
> > solution using bases without zero.

>
> And what of x in
>   x = 1 - 1

Well what i try to say with all this is that the empty set do not need
a placholder, so =y-y as you can see there is no reminants of y it is
completly depleted.
> Now construct 1/9th in base 3.

Well i have not thought it thru for empty positions in the digit
expansion but i guess it will end up be.
1/3=,1
1/9=,(2)1
1/27=,(3)1

2/3=,2
2/9=(2)2
2/27=(3)2

And so on, well you get the hang of it although i am not finished
thinking it thru for fractional digit places, i think i can do all
this for all bases with just 3 or 4 lines of codes,
and it will of course translate back to standard decimal base within
that code.

>
> 0.[9] in base 10(given 9 is followed by 10)
> 0.[2] in base 3 is the same.
>
>  x = 0.[2]

I am not sure i understand your notation does it say x=0,2?
>  10x = 2.[2]
How can then 10x be 2,2?
>  10x - x = 2
This i follow
>  2x = 2
This follow
>  x = 1
But what do you want to say or ask with this?

> I can disagree with Frederik Williams about
>   .3 base 3 = 1 base 3 in your numbering system
> Now what was the benefit again?  Why do I want a
> 3 in base 3?

Date Subject Author
2/1/13 JT
2/1/13 JT
2/1/13 JT
2/1/13 JT
2/1/13 JT
2/1/13 JT
2/1/13 JT
2/1/13 forbisgaryg@gmail.com
2/1/13 JT
2/1/13 forbisgaryg@gmail.com
2/1/13 JT
2/2/13 JT
2/1/13 Frederick Williams
2/1/13 JT
2/1/13 JT
2/1/13 JT
2/1/13 JT
2/2/13 JT
2/2/13 JT
2/2/13 JT
2/2/13 JT
2/2/13 JT
2/2/13 JT
2/2/13 JT
2/2/13 JT
2/2/13 JT
2/3/13 JT
2/3/13 Virgil
2/3/13 JT