On May 10, 9:13 am, James Burns <burns...@osu.edu> wrote: > Transfer Principle wrote: > > But by that count, _no_ real has an infinite representation, > > since any representation of that real, in practice, has only > > finitely many characters. > From context, it looks like you mean " _every_ real has a finite > representation". I call ".99[bar]" a finite representation, > but this does not prevent the same number from having an infinite > representation. > The rest of the real numbers -- that is, /almost all/ of them -- > have /only/ infinite representations, whether you take that either > to mean the listing of the digits in the decimal representation > or to mean any other specifiable scheme (AKA algorithm) that > will tell you what these digits are.
OK, I agree with what you're saying here re: computable reals and finite representations.
> I gathered as much from your previous post. However, it looks > like a very arbitrary distinction you make -- totally lacking in > motivation. What I mean to say is: WHY do you make THAT distinction > in THAT way? > byron seems to be throughly absorbed in wrestling with the Nature > of Truth. How are you advancing /byron's/ project by your distinction > between numbers representable as a/10^b and numbers representable > as a/11^b
The undecimal (base 11) reals?
It's evident that byron intends the decimal reals, and not the undecimal reals? We gather this from context: he writes .99[bar]=1, and had this referred to the undecimal reals, .99[bar] would equal the fraction nine-tenths (.9 decimal). And I bet even byron knows that nine-tenths doesn't equal unity.
So by context, we conclude that byron intends the decimal reals -- the set of reals with only finitely many nonzero decimal digits (also known as the ring Z[.1]).
> > Now we know that in classical analysis, having a finite decimal > > representation doesn't preclude a real number from having an > > infinite such representation as well, but byron isn't bound by > > what classical analysis proves. > I don't see why your comment about byron is so. Could you expand > on the reasons behind your assertion here?
Classical analysis proves that a real number can have both a finite and an infinite decimal representation. Now byron tells us that therefore, math ends in "meaninglessness."
So this leaves me wondering -- can there be a "meaning_ful_" math, one that avoids the undesirable classical result? If such a theory exists, then it's not bound by classical results -- indeed its aim would be to avoid the result .99[bar]=1 that leads to math ending in "meaninglessness."
> > If I knew how to discourage use of a word without insulting its > > users, I'd do so in a heartbeat. > You have your own way of scoring the moral righteousness of > others' posts (and even your own posts, occasionally). I don't > really expect that to change. However, let me put on my > salesman's hat for just a moment while I point out the > advantages of my own way of scoring posts for moral > righteousness. > My own method of scoring (described upthread) is directed > toward making everbody truthful.
But what is "truth," if you will? Indeed, byron himself asks this in another thread:
byron, 7th of May, approx. 8AM Greenwich "Aatu Koskensilta makes the claim that it is the mathematical community that decides what is true or a proof who is this mathematical community is it the grads or masters or doctorates or proffs who is this mathematical community that tells us what truth/proof is"
(This last line is also the title of the relevant thread.)
This is a _very_ good question. There does appear to be this "mathematical community" that decides what is true -- and nowhere is this more evident than on sci.math. And anyone who doesn't agree with the consensus on what is considered true ends up receiving a string of those overused words.