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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.
>
> I don't know what an "infinite number" is. Do you have a reference?
>
> By the way, is 0,111111111111... an infinite number too?
According to my way of thinking , Yes. But these are only my dam
nonsensical misconceptions as standard mathematicians say. The whole
idea is not so serious.
>
> > 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
>
> I have no idea what most of this means.
>
> Best regards,
>
> Jose Carlos Santos
Dear Santos. All of these ideas are only mine, they are not accepted by
standard mathematicians like Arturo , they consider them nonsense, Even
I don't have a solid believe in them, I am just figuring them out. So
don't take them seriouselly.
Yours
Zuhair
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