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José Carlos Santos wrote:
> zuhair wrote:
>
> >>>>> The only representation of 1 is 1.0000......... and to me
> >>>>> 0.9999........... < 1.
> >>>> Then you are wrong. If that expression represented a number strictly
> >>>> smaller than 1, call it x, then (1+x)/2 would be a number strictly
> >>>> smaller than 1, and strictly larger than x.
> >>>>
> >>>> What, pray tell, is the decimal representation of this number?
> >>> How come I didn't read that post?
> >>>
> >>> Anyhow I should answer it.
> >>>
> >>> I will change your terminology.
> >>>
> >>> Let x = 1- 0.9999..............
> >> That's a rather strange way to answer to Arturo, since he told you to
> >> call _x_ to 0.9999... and you choose to call _x_ to 1 - 0.9999....
> >>
> >>> Now 1+ 0.99999... = 2 -x
> >>>
> >>> Now (2 - x)/2 = 2/2 - x/2
> >>>
> >>> now x/2= x
> >> How did you get that? But if you think (as I do) that this is a true
> >> statement, then the conclusion is, of course, that x = 0. Since you
> >> defined _x_ as 1 - 0.9999..., it follows that 1 = 0.999....
> >>
> >> Now, what about answering Arturo's question. If x = 0.9999..., what is
> >> the decimal representation of (1 + x)/2?
> >
> > I answered Arturo's question.
> >
> > Let me repeat it again but I will use Arturo's terminology.
> >
> > x= 0.9999.......
> >
> > Let y = 1-x
> >
> > it follows that x= 1-y
> >
> > Now 1 + x = 1+ 1- y = 2 - y
> >
> > Now y/2 = y
>
> Two questions here:
>
> 1) How do you know that y/2 = y?
because y is an infinite number since, 0.9999...... is infinite, and if
you divide this
infinite number by any finite number you will get the same number.
1- 0.999........ = 1/(10^Aleph-0)
[1/(10^Aleph-0)]/n = 1/(10^Aleph-0)
This comes from the equation
[x/n] + [x/(n^2) ] + [x/ (n^3)] +.....+ [x/(n^k) ]= {x/(n-1)} - {
x/[(n^k)(n-1)]}
for n=2,3,4,5,........
K=1,2,3,4,.........
if we regard k as the number of terms in the sequence above.
Then when that sequence is infinite then = Aleph-0
so for x=9, n=10 and k=Aleph-0 we will have 0.9999...........= 1-
[1/(10^Aleph-0)]
Now whatever is the division operator in 1/(10^Aleph-0) is, the result
cannot equal zero.
and 1/(10^Aleph-0) > 0
So 0.9999........ <1
Zuhair
>
> 2) How come you don't deduce from this that y = 0 and therefore that
> 1 = 0.9999...?
>
> > Then ( 1+x)/ 2 = (2-y)/2 = 1 -( y/2) = 1- y = x = 0.99999...........
> >
> > You see the same number.
>
> Best regards,
>
> Jose Carlos Santos
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