> 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 > h 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 ... > >
hey neilist, well spotted :-) also note that since i mentioned Robert Israel from the university Bassam is "suddenly" from a university too :-)
i guess JAMES HARRIS is a field medalist ?
as for the 0.999999999 :
0.999999999... = 1-(1/h) the limit is 1.
but you could say its different too.
like this 1-0.9999999999= 1/h
(1-0.9999999999)h = 1
on the other hand 0 * infinity = 1 (possible depending on limit type)
and you could also say 0.99999...=1-(2/h)
in any case 0.999999... is just a bad way of saying lim->1
decimal is an inperfect system.
just the paradox 0.99999999=1=1-(1/h)=1-(2/h)
shows the inperfection for describing numbers infinitely precise.
this solves the question.
since 1/3 = best possibly written in decimal like 0.33333...
0.99999... must be 3*1/3 EQUALS 1.
however if you consider 0.9999... as less than one e.g. 1-(1/h) in a kind of calculus way or because the first digit is 0.
and you dont want to abbandon decimal methods.
and you dont wanna give up limits and fractions either.
then you can combine them :-)
so 1/3 * 3 = 0.99999...+3*(1/(3h))=0.9999...+(1/h) = 1