Re: query on number theory

From: Richard Henry (rphenry_at_home.com)
Date: 07/17/04


Date: Sat, 17 Jul 2004 13:28:09 -0700


"Rob Johnson" <rob@trash.whim.org> 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.

A curious aside that I noticed while fiddling with the numbers preparing my
earlier response.

In base 30, 0.99999.... = 1/3.



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