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

On May 31, 4:02 am, chaja...@xxxxxxxx wrote:
On May 31, 12:33 am, William Hughes <wpihug...@xxxxxxxxxxx> wrote:



I disagree with the definition given:

let a = 0.999... be a real number. We do not need

to give a full definition at this point
a<=1
and
a>(1-(1/10^n) for any natural number n

You have defined 0.999... to be a real number without jusitification.
I can make no such assumption.



You do however have to give some sort of definition for 0.999...
Whatever you define it to be it will either be equal to 1
or it will not be a real number.



<snip>



The fact that 0.999... is or is not a real number is of little concern
to my proof - it makes no assumption and has no need one way or the
other. What it demonstrates is that if x = 0.999... then 10x - x must
be less than 9 within the confines of multiplication upon real
numbers.


No. If you use the standard limit definition, 10x-x = 9.
You are only correct if you use some other definition for
0.999... in which case the "subtraction" is not in the
real numbers. You need to define a new set of "numbers".
When you do so you will not get a field.

- William Hughes


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