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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.
>
> You will probably not read this, but even so I would like to call your
> attention to the fact that 0,11111111... = 1/9. On the other hand, you
> wrote that half of an infinite number is that same number. It follows
> that (1/9)/2 = 1/9. Don't you see anything strange here?
>
> Best regards,
>
> Jose Carlos Santos
My way of thinking is non standard, to me 0.11111............ < 1/9
0.1111..... = 1/10 + 1/10^2 + 1/ 10^3 + ........... = (1/9) - ( 1/
[9 (10^Aleph-0)] )
Since ( 1/ [9 (10^Aleph-0)] ) >0
you can imagine ( 1/ [9 (10^Aleph-0)] ) as a continuually decreasing
quantity
and thus 0.1111......... as continually approaching 1/9 but never
reaching it.
so 1/9 is the limit of the sequence above and not the upper max.bound.
Then 0.1111....... <1/9
Divide that number by itself will only result in itself, since it is
infinite?!
Zuhair
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