Re: infinitely many nn's = infinite nn's?
- From: David Marcus <DavidMarcus@xxxxxxxxxxxxxx>
- Date: Tue, 13 Mar 2007 04:09:26 -0400
Phil wrote:
G. Frege wrote:
Quantifier dyslexia strikes again. Seems to be epidemic among cranks.
Phil seems to be confused about those two (different) things:
Well I doubt that I am confused about the fact that there are two
proofs, one which shows that there is a FINITE distance (or difference,
if you must) between any two reals on the line segment (which is why I
can use the word "distance," duh) [0,1], and another which shows that
there are infinitely many reals -- in fact, infinitely MORE than there
are natural numbers (aleph-C versus aleph-0) -- on that same segment.
You mean "c", not "aleph-C". aleph_c is awfully big. c is probably more
like aleph_2.
The question that most of you seem unable or unwilling to address, even
though you know that it is not a GIVEN that all proofs use compatible
premises,
All the proofs we have been giving do use the same axioms. I know this
because I've checked them myself and because I have books that say the
same thing.
is whether the an infinite sum of FINITE distances (or
differences) is can result in a FINITE distance (or sum) of one,
equivalent to the line segment [0,1].
What "infinite sum of finite distances"? Do you mean something like
this: Let {a_n} be a sequence of numbers in [0,1] such that
0 = a_0 < a_1 < a_2 < a_3 < ...
Assume that lim n->oo a_n = 1. Then
1 = sum_{n=1}^oo (a_n - a_{n-1}).
Tell me G., since you at least seem to be more honest, and less of a
***, than David,
Appearances can be deceiving.
do you think there is some FINITE distance from the
door such that we can double that distance an INFINITE number of times
and wind up in the middle of the room?
Not if the room is of finite size.
We could simply ADD that distance
to itself an infinite number of times and get the same result. Are you
going to ACTUALLY CLAIM here, for all to see from now on, that you
believe that the "existence" of infinitely many reals, all a finite
distance/difference apart, does not even HINT at a contradiction?
We are still waiting for you to present the contradiction.
Sure,
this apparent contradiction may be okay, meaning that the two proofs use
completely compatible premises, but don't you HONESTLY think that it at
least hints that there be an incompatible premise or two, that we should
at least THINK about it?
What makes you think that people haven't thought about it?
I am not asking you to say that these "incompatible premises" actually
exist, mind you, I am just asking whether you want everyone to know,
from now on, that when you combine the results from two proofs, and that
combination states that (1) there is a FINITE distance between ANY two
points (real numbers) on the segment [0,1], and (2) there are INFINITELY
many points (real numbers) on that segment, that you see NO HINT OF A
CONTRADICTION AT ALL???
We are still waiting for you to present the contradiction.
Are you going to say that you see NO REASON AT
ALL to investigate a bit further, even though you think it is probably
just an apparent contradiction, and not a real contradiction?
What makes you think that people haven't investigated further? Do you
really think you know as much mathematics as all the mathematicians in
the world (not to mentioin all the deceased mathematicians)?
Please
remember, the fact that these two proofs exist does NOT prove that they
can both be applied to the same object without producing a
contradiction, since we have NOT yet proved that the premises, hidden as
well as stated, are the same in both proofs.
I have checked the proofs and the axioms used are the same. And, you
really aren't in a good position to judge, since you haven't
demonstrated much ability to construct proofs. Most of the arguments
that you've posted use all sorts of invalid assumptions (and they aren't
even hidden).
for all x,y e [0,1] with x =/= y: there is a number d e IR, d > 0,
such that |x - y| >= d.
and
there is a number d e IR, d > 0, such that for all x,y e [0,1]
with x =/= y: |x - y| >= d.
I'm not sure what you mean here. This first is obvious, there exist a
real non-zero number d which is smaller than the difference between any
two real non-equal numbers,
Are you stating #1 or #2?
but I don't understand the significance of
stating it "in reverse." Yes, I understand that we often need to do
that, but I don't see what that accomplishes here.
Do you think that #1 is true and #2 is false? Or, do you think the two
say the same thing? Or, do you think they are different, but both true?
--
David Marcus
.
- References:
- Re: infinitely many nn's = infinite nn's?
- From: George Dance
- Re: infinitely many nn's = infinite nn's?
- From: David Marcus
- Re: infinitely many nn's = infinite nn's?
- From: G . Frege
- Re: infinitely many nn's = infinite nn's?
- From: herbzet
- Re: infinitely many nn's = infinite nn's?
- From: Virgil
- Re: infinitely many nn's = infinite nn's?
- From: Phil
- Re: infinitely many nn's = infinite nn's?
- From: Virgil
- Re: infinitely many nn's = infinite nn's?
- From: Phil
- Re: infinitely many nn's = infinite nn's?
- From: David Marcus
- Re: infinitely many nn's = infinite nn's?
- From: G . Frege
- Re: infinitely many nn's = infinite nn's?
- From: Phil
- Re: infinitely many nn's = infinite nn's?
- Prev by Date: Re: help with Godel's
- Next by Date: Re: infinitely many nn's = infinite nn's?
- Previous by thread: Re: infinitely many nn's = infinite nn's?
- Next by thread: Re: infinitely many nn's = infinite nn's?
- Index(es):
Loading