
Re: five reasons why mathematics ends in meaninglessness
Posted:
May 10, 2011 3:13 PM


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 ninetenths (.9 decimal). And I bet even byron knows that ninetenths 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.

