neilist
Posts:
454
Registered:
5/11/07


Re: Proof 0.999... is not equal to one.
Posted:
May 31, 2007 11:48 AM


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 nonzero > > > 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 > > FirstConsider a number (N) with finite number of digits say (M), of string (999...), this of course can be factored into prime numbers only > > SecondNow 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 ...

