Re: query on number theory

From: Harsha (harsha.hs.81_at_gmail.com)
Date: 07/19/04 

Date: 19 Jul 2004 04:19:54 -0700

rob@trash.whim.org (Rob Johnson) wrote in message news:<20040717.031328@whim.org>...
> >3)0.9999... <> 1? Is it because we follow decimal notation. It is
> >clear that binary representation cannot represent all the fractions
> >even if the number of bits is huge as it can only represent fractions
> >as multiple of 2^(-n).
> >Can a base > 10 represent all the fractions that can be represented in
> >decimal format along with even more other precise fractions.
>
> If by .9999... you mean that the 9s repeat indefinitely, then your
> inequality is false. 0.9999... = 1. To represent a fraction exactly in
> a finite number of digits, the denominator of the fraction, in lowest
> terms, must be only divisible by primes that divide the numerical base.
>
> Certainly any base which is divisible by ten can exactly represent any
> fraction in finitely many digits that can be so represented in base ten.
> To include more fractions, the base must be divisible by other primes.
> For example, the same fractions are exactly representable in finitely
> many digits in base ten and base twenty. However, more are exactly
> representable in finitely many digits in base 30, which represents the
> same fractions as base 60.

I observed this while revisiting my primary school math of converting
recurring decimal to fractions.
say we have 0.98989898...
let x = 0.989898...
 100x =98.989898...
subtracting 99x= 98 hence x = 98/99
This method will give for 0.9999.... as x = 1 as its limiting answer.
But as pointed out in previous mails the same can be expressed as 1/3
in base 30 without any ambigiuty.

My question is does all our decimal or fractional representation are
governed by the base we use. Can we have a number system wherein all
the numbers are represented unambiguously irrespective of base.

Relevant Pages

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