All,

 I have sort of followed this thread for some time now, and there
 have been some good arguments both ways concerning this issue.

 But, out of all the arguments and all the explanations, I still
 have to wonder what all the fuss is about.   Each person on this
 list is going to handle the situation as he or she sees fit, and
 according to their own rules.  And I would also venture to say that
 no amount of discussion is going to sway anyone from their current
 way of thinking to the other persons viewpoint.  We're all human,
 and if we were perfectly honest with ourselves, we each think that
 we're right.  That's just human nature and ego.   So, therefore,
 why discuss it any further?  I mean it just seems to me that this
 is a fruitless pursuit.  You can discuss it till the snow comes in
 June, and all you will have done was used up a lot of time.  The
 basic issue will still be there, more or less evenly divided between
 the two camps.

 I'm not trying to dictate what you should and shouldn't do.  Just
 pointing out the fact that I doubt that anyones mind will be changed
 drastically from what they think and believe right now about how to
 handle math problems involving this issue.

 I know my mind hasn't been changed.  I would still handle these
 problems the way I was taught, and have handled them for years.

 Anyway, that's my take on the situation.   You may continue (or not).
 I'm done ranting now.

                                               Regards,

                                                 Jim




On Fri, 08 June 2001, David Minkler wrote:

>
> 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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>
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jim@jpes.com

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