neilist
Posts:
454
Registered:
5/11/07
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Re: Proof 0.999... is not equal to one.
Posted:
May 31, 2007 11:48 AM
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On May 31, 9:37 am, bassam king karzeddin <bas...@ahu.edu.jo> wrote:
> > > "bassam king karzeddin" <bas...@ahu.edu.jo> wrote
> > in
> > > message
>
> >news:22019165.1180601903492.JavaMail.jakarta@nitrogen.
>
> > > mathforum.org...
> > > > Re: What is wrong between decimal and fraction?
> > > > Posted: May 28, 2007 6:01 PM Plain Text
> > > Reply
>
> > > > Dear All
>
> > > Mr King.
>
> > > > [...]
> > > > Any positive real number (except one) is a
> > unique
> > > production of prime
> > > > numbers with each prime raised to a non-zero
> > > integer and therefore of
> > > > unique decimal representation
>
> > > Factorisation is good.
>
> > > > Hence, the irrational numbers are all those
> > numbers
> > > that have endless
> > > > decimal digital expansion in any number system,
> > > provided that their
> > > > terminating digits are not all zero
>
> > > Why?
>
> > From the early definition of the rational numbers, we
> > can simply extend their concept, but with infinite
> > integers, so the real number definition becomes as a
> > ratio of two finite or infinite coprime integers
>
> > And this definition doesn't count (zero, one,
> > infinity) as real numbers except by CONVENTION
>
> > > > From this you can see now why (0.999...) is an
> > > irrational number even we
> > > > don't know its prime factorization and therefor
> > > can't be equal to one
>
> > > > [...]
>
> > > I'm not sure what this is, but it's not a sound
> > > nd proof.
>
> > In my opinion, the proof is straight foreword from
> > the definition only
>
> > > --
> > > Glen
>
> > Regards
>
> > B.Karzeddin
>
> The proof
>
> First-Consider a number (N) with finite number of digits say (M), of string (999...), this of course can be factored into prime numbers only
>
> Second-Now add one to the previous number (N), (999..+1), you will get another number (N+1) with COMPLETELY deferent prime factorization
>
> Third repeat the process for (M+1) digits, and this is the principle of Induction method of the proof, where you would find that always applicable, then
>
> You will always get two sets of prime factorization for (N), and (N+1), where can never be considered exactly equal, there fore their division (N/(N+1)) can not be equal to one except by consideration or limit or convention
>
> Is not this is a rigorous proof for such a SILLY problem?
>
> B.Karzeddin
>
> Al Hussein bin Talal University
> JORDAN- Hide quoted text -
>
> - Show quoted text -
Hey Harris, I see you dropped your cockamamie fake bad English in your
other "Bassam" posts.
Hahahahahahahahahahahahahahahahahahahahahaha
... 'round the moons of Nibia, and 'round the Antares maelstrom, and
'round Perdition's flames ...
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