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

On May 31, 9:37 am, bassam king karzeddin
<bas...@xxxxxxxxxx> wrote:
"bassam king karzeddin" <bas...@xxxxxxxxxx>
wrote
in
message


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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 :-)

LLLOOOOOLLL

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)

etc

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 :-)

like

1/h= 0.00000000...1

1/3= 0.333333...+(1/(3h))

so 1/3 * 3 = 0.99999...+3*(1/(3h))=0.9999...+(1/h) = 1

nice hmm :-D

but of course useless.

tommy1729
.

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