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 ]
 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 4, 2017 12:27 PM

On 10/4/2017 4:19 AM, netzweltler wrote:
> Am Mittwoch, 4. Oktober 2017 01:29:50 UTC+2
> schrieb Jim Burns:

>> <Burns<netzweltler>>
>>

>> > Do you agree that 0.999... means infinitely many
>> > commands
>> > Add 0.9 + 0.09
>> > Add 0.99 + 0.009
>> > Add 0.999 + 0.0009
>> > ...?

>>
>> 0.999... does not mean infinitely many commands.
>>
>> </Burns<netzweltler>>

>
> To me it looks like that we don't even agree, that there
> are infinitely many 9s following.

Maybe we agree, maybe we don't. We might be using the same
words and meaning different things by them.

I say there are infinitely many nines following the '.'
What I mean by "infinitely many" here is that there is a
map, one-to-one but not onto, from those after-dot decimal
places to those after-dot decimal places. And '9' is in every
place.

I could say more, and I should, in order to say what 0.999...
means, but that is what "infinitely many 9s following" means.

What do you mean? That there is a '...' at the end?
How would someone reason from that to any conclusion?

> If there are infinitely many 9s following then there is a
> bijection between N and the decimal places in 0.999...

True enough for our purposes here. "Infinitely many" is broader
than we really want here. There are uncountable ordinals,
for example. If "9s following" is a map from an uncountable
ordinal to {0,1,2,3,4,5,6,7,8,9}, then there would not be
such a bijection, N to "decimal places".

But there is a bijection between N and decimal places.
Even better, there is an order-preserving bijection.
I think we mean the same thing here.

> If there are infinitely many 9s following then we are
> dealing with an infinite stepwise process as described
> above and I can't see why this shouldn't mean
> "infinitely many commands".

Even though you don't see why, it doesn't mean that.
I'm trying to explain, but if you still don't get it, that
will not mean that 0.999... means infinitely many commands.

Your argument is basically "This thing that you guys mean
does not make sense". You're right (I think) that "this thing"
does not make sense, but you're wrong that "this thing" is
what we mean.

----
When I write
0.999...
I mean that there is a map (very uninteresting) from N to digits
0 |-> 9
1 |-> 9
2 |-> 9
...
and this map represents the infinite decimal 0.999...

(Notice that I used '...' here to continue my definition
of the map. I did that in order to improve its readability.
I don't need '...' and I shouldn't use '...' to reason about
that map. I may have described that map or one similar without

For every infinite decimal (which means "map from N to digits"),
there is a map from N to the finite initial decimal expansions
of 0.999...
0 |-> 0.9
1 |-> 0.99
2 |-> 0.999
...

(I don't need to use '...' to describe this map, either.)

None of those involve infinitely many addition operations.
This is provable by induction. Induction is how we can
validly reason about these infinitely many finite sums on a
finite page.

The value of an infinite decimal ("map from N to digits")
is _defined_ as the least upper bound of the set of all its
finite initial decimal expansions. This doesn't involve

When you say
"we are dealing with an infinite stepwise process"
you are wrong. I am being careful here. I am not just saying
that you and I disagree, though we do. Saying that we
are dealing with an infinite stepwise process is as wrong
as saying that there is an integer between 2 and 3.

We are dealing with infinitely many finite stepwise processes.
Each process generates a number, and the least upper bound
of those infinitely many numbers is the value of 0.999...

----
Compare this to the set N of natural numbers.
Every natural number is result of finitely many applications
of the successor operation to 0. We can prove this by induction.

There are infinitely many natural numbers, and we could say,
reasonably, that each non-zero number is the result of a
different application of the successor. So, someone _could_
say that the set of natural numbers is the result _in some_
_sense_ of infinitely many successor operations. But it would
be wrong to say that _therefore_ there are naturals which are
the result of infinitely many successor operations.

It is correct to say that, with the natural numbers, we are
dealing with infinitely many finite applications of the successor.
finite applications of the successor. (Infinitely many
finite sums are not a problem either, because we already

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