Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Re: It is a very bad idea and nothing less than stupid to define 1/3
= 0.333...

Replies: 42   Last Post: Oct 9, 2017 11:53 AM

 Messages: [ Previous | Next ]
 netzweltler Posts: 473 From: Germany Registered: 8/6/10
Re: It is a very bad idea and nothing less than stupid to define 1/3
= 0.333...

Posted: Oct 8, 2017 5:40 PM

Am Sonntag, 8. Oktober 2017 22:12:22 UTC+2 schrieb Jim Burns:
> On 10/8/2017 8:20 AM, netzweltler wrote:
> > Am Sonntag, 8. Oktober 2017 13:23:05 UTC+2
> > schrieb FromTheRafters:

> >> It happens that netzweltler formulated :
>
> >>> I'd have to look it up: Did you say that 0.999... IS the
> >>> result of infinitely many addition operations or IS NOT
> >>> the result of infinitely many addition operations?

> >>
> >> If you had an oracle with enough time which could do the
> >> arithmetic and hand you an answer, then yes. Without such
> >> an oracle, then I'd have to say no. That's why I said "after"
> >> doing infinitely many steps you would have that number
> >> exactly. John Conway used language similar to 'after infinitely
> >> many of these steps, there is an explosion of sets...' to
> >> describe a similar notion in describing his construction of
> >> the surreals, so I'm not exactly breaking any new ground here.

>
> > I don't see the "infinite time" problem. What is the time a
> > single addition operation takes? 0? More? Even if you don't
> > allow 0 time for a single addition operation, try this:
> > t = 0: Add 0 + 0.9
> > t = 0.9: Add 0.9 + 0.09
> > t = 0.99: Add 0.99 + 0.009
> > ...
> > Every operation take some time greater than 0. Nonetheless,
> > we have done infinitely many additions by t = 1.
> > No "infinite time" involved.

>
> Not infinite time, but "enough time" for the oracle to operate,
> as Mr Rafters said. It just happens that your oracle takes
> 1 [unit of time] to operate.
>

> > No oracle needed.
>
> I don't see how you get this answer without an oracle. Maybe
> you have an oracle and you don't realize it, it's invisible
> or something.
>
> However, in some way you are satisfied that you can get an answer
> infinitely many additions, wasn't it? That there was no answer to
> the _infinite_ sum? But now you _do_ have an answer. Somehow.
>
> (Sadly, I do not have an oracle, so I will continue to
> define 0.999... the same way.)
>
> ----
> have) compare to the standard answer?
>
> One thing we all know, including you, is that the sum of _all_
> of 0.9, 0.09, 0.009, ... must be strictly larger than any
> _partial_ sum. So it can't be less than or equal to any of
> 0.9, 0.99, 0.999, ...

This statement makes sense only if you can show that the sum of _all_ of 0.9, 0.09, 0.009, ... is a _point_on_the_number_line_.

> So, we know the oracle's answer is a bound of the set
> { 0.9, 0.99, 0.999, ... }

I strongly disagree here. It cannot be a bound, because the numbers of that set represented as line segments in this list

[0, 0.9]
[0, 0.9]?[0.9, 0.99]
[0, 0.9]?[0.9, 0.99]?[0.99, 0.999]
...

already contain all the segments that you can find in 0.999...
There is nothing to the right of all of those segments in 0.999...

> Suppose that we have two _distinct_ bounds b and c of all the
> finite partial sums, with b < c. Whether or not b is the
> oracle's answer to the infinite sum, we know that c _is not_
>
> Suppose c _was_ the infinite sum 0.9 + 0.09 + ...
> Then we know what the "error term" in the infinite sum
> is _exactly_ .
>
> For example, the sum of all the infinitely many terms after
> the (10^100)th one
> 9/10^(10^100+1) + 9/10^(10^100+2) + ...
> is simply
> (1/10^(100)*( 0.9 + 0.09 + 0.009 + ... ) = c/10^100
>
> This gives us (by oracular means) a second way to calculate
> the first 10^100 terms of the sequence:
> 0.9 + 0.09 + ... + 9/10^100
> = (1 - 1/10^100)*c
>
> But this is a finite sum!
> And b is also a bound of finite sums!
> It had better be the case that
> (1 - 1/10^100)*c < b < c
>
> But Wait! There's More!
> Whatever non-zero distance b is below c, there is some
> finite stage in the infinite sum 0.9 + 0.09 + ...
> (if not 10^100, then 10^200, or 10^(10^100) or more)
> where we can repeat this argument and contradict the
> assumption that c is the _infinite_ sum.
>
> Therefore, if b and c are bounds and b < c, then c
> _is not_ the oracle's answer.
>
> In fact, if the oracle answers with anything _other than_
> the least upper bound of finite partial sums, it will be
> provably wrong.
>
> ----
> So, netzweltler, what answer did you come up with? Is it
> different from 1?

Date Subject Author
10/2/17 Guest
10/2/17 netzweltler
10/2/17 Jim Burns
10/3/17 netzweltler
10/3/17 FromTheRafters
10/3/17 Jim Burns
10/3/17 FromTheRafters
10/3/17 Jim Burns
10/3/17 FromTheRafters
10/3/17 netzweltler
10/3/17 bursejan@gmail.com
10/4/17 netzweltler
10/3/17 FromTheRafters
10/3/17 Jim Burns
10/3/17 FromTheRafters
10/3/17 netzweltler
10/3/17 Jim Burns
10/4/17 netzweltler
10/4/17 Jim Burns
10/4/17 netzweltler
10/5/17 Jim Burns
10/5/17 netzweltler
10/5/17 Jim Burns
10/5/17 netzweltler
10/5/17 Jim Burns
10/5/17 netzweltler
10/5/17 Jim Burns
10/5/17 FromTheRafters
10/6/17 netzweltler
10/6/17 Jim Burns
10/7/17 FromTheRafters
10/8/17 FromTheRafters
10/8/17 netzweltler
10/8/17 Jim Burns
10/8/17 netzweltler
10/8/17 Jim Burns
10/9/17 netzweltler
10/9/17 Jim Burns
10/9/17 netzweltler
10/9/17 Jim Burns
10/7/17 Jim Burns