Re: Multiple infinities - one more look

Dave Seaman <dseaman@xxxxxxxxxxxx> writes:
On Sun, 9 Dec 2007 08:51:48 -0800 (PST), Venkat Reddy wrote:
On Dec 9, 8:17 pm, Venkat Reddy <vred...@xxxxxxxxx> wrote:
On Dec 9, 7:48 pm, Dave Seaman <dsea...@xxxxxxxxxxxx> wrote:





On Sun, 9 Dec 2007 05:02:53 -0800 (PST), Venkat Reddy wrote:
On Dec 9, 1:34 am, mike3 <mike4...@xxxxxxxxx> wrote:

<snip>

Removing points of the form k/(2^n), where k and n are natural
numbers,
removes only a countably infinite subset, FAILS to remove even certain
rationals like 1/3, and utterly fails to remove ANY of the irrationals
like you need. You've blasted the continuum full of holes, but have
failed to empty it of it's points. So every time you cut at k/(2^n),
removing it from the set, there are still points that you miss. It
only removes such points, and "continuing it to infinity" simply
removes ALL such points, but 1/3 is not such a point and hence is not
removed by the cutting process.
Why don't you believe that there could be two values n and 1/2^m
whoose product is 1/3 where n and m are not equal and aproach infinity
at different rates of growth, while I'm supposed to believe that same
number of zero width points can add up to 1 or 2 when they are
"unionized" in uncountably infinite number?

The fundamental theorem of arithmetic. Every number has a unique
factorization into primes. In the case of 2^m, all of the prime factors
are 2's. Therefore, there are no 3's in the factorization.

How about the infinite sum:
sum_(n=1->oo) 1/2^(2n) = 1/3

Here, if we add up all the fractions, the denominator will always be
in 2^k form, but still the ratio equals 1/3.

One more ratio with 2^k in denominator but equals sqrt(2):

Sqrt(2) is not a ratio of integers.

Consider the infinite series

x_1 = 1; x_2 = x_1/2 + 1/x_1; .....

The denominator here is always in 2^k form where k is a positive
integer, but the value converges to sqrt(2).

It's possible for a sequence to converge to sqrt(2) even though each term
is a rational of the form m/2^n. The limit is not a member of the
sequence and is not of that form.

Neither are the x_i above beyond i=2.

I'm not sure if the infinte sums and series obey the fundamental
theorem of arithmetic.

The fundamental theorem of arithmetic applies to integers. Each integer
has a unique factorization into primes. A number of the form 2^k does
not have 3 as a factor.

I suspect you've told him that already.

Phil
--
Dear aunt, let's set so double the killer delete select all.
-- Microsoft voice recognition live demonstration
.

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