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zuhair wrote:
> 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
Dear Santos,
You asked me before is 0.111......... is an infinite number? I answered
you yes.
I think I was wrong since 0.1111.... is a combination of a real number
and an infinite number.
see that 0.1111.....= 1/9 - ( 1/ [9 (10^Aleph-0)] )
1/9 is a real number while ( 1/ [9 (10^Aleph-0)] ) is an infinite
number.
Now halfing 0.111...... would be like the following
1/2 * 0.111...... = 1/2 * 1/9 - 1/2* ( 1/ [9 (10^Aleph-0)] )
= 1/18 - ( 1/ [18 (10^Aleph-0)] )=1/18 - ( 1/
[18 (19^Aleph-0)] )
since n^Aleph-0 = m^Aleph-0 , for m , n are finite numbers.
= 1/19 + 1/19^2 + 1/19^3 + ..................
Using the 19th numeral system (number system with base =19). this
number is 0.1111...........
Zuhair
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