Re: Proof 0.999... is not equal to one.

"bassam king karzeddin" <bassam@xxxxxxxxxx> wrote
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Re: What is wrong between decimal and fraction?
Posted: May 28, 2007 6:01 PM Plain Text
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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
.

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