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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 ]
 Jim Burns Posts: 831 Registered: 9/26/15
Re: It is a very bad idea and nothing less than stupid to define 1/3
= 0.333...

Posted: Oct 7, 2017 10:29 AM

On 10/7/2017 4:06 AM, netzweltler wrote:
> Am Samstag, 7. Oktober 2017 01:42:15 UTC+2
> schrieb Jim Burns:

>> On 10/6/2017 6:03 AM, netzweltler wrote:
>>> Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
>>> schrieb Jim Burns:

>>>> On 10/5/2017 3:12 PM, netzweltler wrote:
>>>>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
>>>>> schrieb Jim Burns:

>>>>>> On 10/5/2017 10:00 AM, netzweltler wrote:
>>>>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
>>>>>>> schrieb Jim Burns:

>>>>>> [...]
>>>>>>>> _We don't do what you're describing_
>>>>>>>
>>>>>>> Nevertheless,

>>>>>>
>>>>>> "Nevertheless"?
>>>>>> Do you agree that what you're describing
>>>>>> is not what we're doing?

>>>>
>>>> *NETZWELTLER*
>>>> DO YOU AGREE THAT WHAT YOU'RE DOING
>>>> IS NOT WHAT WE'RE DOING?

>>>
>>> Let's say I agree. Doesn't mean that it is obvious to me
>>> what *you* are doing.

>>
>> Great. Let's say you agree. Will you stop saying that
>> "0.999... means infinitely many commands"?

>
> No.

When I showed you what 0.999... means, your reaction
(paraphrased) was "But there are no 'infinitely many
commands' there!"

Am I wrong about that paraphrase? Will you show me the
infinitely many additions which I do? Which _I_ do,
not which _you_ do.

If you can't, why do you keep saying that they're there?

> Because it is not obvious to me why the equation
> 0.999... = 0.9 + 0.09 + 0.009 + ...
> should be wrong.

By definition,
0.999... = LUB{ 0.9, 0.9 + 0.09, 0.9 + 0.09 + 0.009, ... }

If, by
0.9 + 0.09 + 0.009 + ...
one means
LUB{ 0.9, 0.9 + 0.09, 0.9 + 0.09 + 0.009, ... }
then it's not wrong. But you don't mean that.

The answer to _why_ you're wrong is: it's not defined your way.

Now that you know why, will you stop saying that
"0.999... means infinitely many commands"?

[...]
>
> You are not addressing what I mean what has to be proven.
> How do you prove that 0.999... is pointing to a location
> on the number line? To me you are just assuming that it does.

By the real number axiom that any bounded non-empty set of
reals has a least upper bound, LUB{ 0.09, ... } represents
a unique real number.

Suppose you have the set N of natural numbers. How would you
prove that there is no predecessor of 0 in it? Wouldn't using
the axiom (Ax)~(Sx = 0) be "just assuming" that there isn't?

> Then you are comparing its "location" to the locations of
> the terminating decimals in (1-1/10^n)n?N and conclude that
> it is a location to the right of all those numbers.

Basically, yes. We prove 1 is an upper bound. But we also prove
that there is no number to the left of 1 which is also an
upper bound. Thus 1 is the unique _least_ upper bound.

> I could rather show that 0.999... is definitely not pointing
> to a location coincident with point 1. Therefore we had to
> agree that 0.999... can be represented by the line segments
> {[0, 0.9], [0.9, 0.99], [0.99, 0.999], ...}.

So, if we changed the definition of 0.999... , you could show
that 0.999... does not represent 1? I'm sure you could.

Why should anyone care about 0.999... with a different
definition?

And why do you keep speaking as though the definition has
been changed to what you suggest, when it hasn't been?

> But I guess - since you don't agree that 0.999... is the
> result of infinitely many additions - you don't agree with
> that notion either.

It's not actually a question of _my_ agreement. I could
agree that there is an integer between 2 and 3, but there
still would not be an integer there.

Now, we could _change the definitions_ so that I would be
correct. Maybe we'd start counting 0, 1, 2, 4, 3, 5, 6, ...
So, it's not inconceivable. But it's a useless point to
make. When we talk about 2 and 3, we're talking about 2 and
3 _with the usual definitions_ . And when we talk about
0.999..., we talk about 0.999... _with the usual definitions_ .
The usual definitions do not have infinitely many additions.

> You don't show how 0.999... is pointing to a location on
> the number line.

The set { 0.9, 0.99, 0.999, ... } is bounded and non-empty.
Proof:

For all x e { 0.9, 0.99, 0.999, ... }, x < 2.

0.9 e { 0.9, 0.99, 0.999, ... }

Done.

The infinite decimal 0.999... (a map from negative places
to decimal digits, all 9s) points to a bounded non-empty
set of finite partial sums. That set points to its least
upper bound -- a unique location on the number line.

It's just not true that I haven't showed you this,
over and over and over.

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