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 4, 2017 3:58 PM


Am Mittwoch, 4. Oktober 2017 18:27:18 UTC+2 schrieb Jim Burns: > 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, onetoone but not onto, from those afterdot decimal > places to those afterdot 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?
There is no end. Nothing follows after infinitely many 9s. "infinitely many 9s following" replaces '...'.
> 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 orderpreserving 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 > '...' already elsewhere, but if you ask, I'll do it again.) > > 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 > infinitely many addition operations, either. > > 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 nonzero 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. > And that is not a problem, because we already know about > finite applications of the successor. (Infinitely many > finite sums are not a problem either, because we already > know about finite sums.
While it is true that there is no line on the list
0 > write 0.9 1 > append another 9 2 > append another 9 ...
that produces other than a finite string, it is true that the complete list produces an infinite string. "We are dealing with infinitely many finite stepwise processes" is true for each line of the list. "we are dealing with an infinite stepwise process" is true for the complete list. We wouldn't get the resulting 0.999... otherwise.

