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