Lets try (1/11) + (10/11)

First fraction is 0.090909(09)r,
Second fraction is 0.909090(90)r
Sum of fractions is 0.999999(99)r

A little re-org on the original expression

(1 + 10)/11 yields 11/11 is 1

Did I miss part of this thread?

Regards,
Dave

Bob Ammerman wrote:
>
> All repeating numbers are rational. There is a relatively simple proof of
> this.
>
> Given a number N which ends in a repeating group of 'p' digits:
>
> Then the number (10^p) * N ends with the same repeating block.
>
> Now if we subtract:
>
>      (10^p)*N - N
>
> the 'repeating blocks' cancel out and we have a terminating decimal (the
> part of N before the repeating block). Note, of course that this terminating
> decimal is rational.
>
> Thus, N(10^p-1) == N(10^p-1) is rational. If we now divide this by (10^p-1)
> we have:
>
>             N(10^p-1)
> N == -------------------
>               10^p-1
>
> Note that both the numerator and denominator are rational, therefore the
> entire value is rational, therefore N is rational.
>
> By the way: this proof is from my 11th grade trigonometry book.
>
> Bob Ammerman
> RAm Systems
> (contract development of high performance, high function, low-level
> software)
>
> --
> http://www.piclist.com hint: The PICList is archived three different
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