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zuhair wrote:
Doug wrote:
"zuhair" <zaljohar@xxxxxxxxx> wrote in message
news:1165095466.854647.289380@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Doug wrote:
"zuhair" <zaljohar@xxxxxxxxx> wrote in message
news:1165082833.528329.76060@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Doug wrote:
"zuhair" <zaljohar@xxxxxxxxx> wrote in message
news:1165081409.372959.245100@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Hi All,
Suppose that we have a countably infinite number of balls put in a
large container. If we start by removing 2 balls from the container
in
the first step, then we removed 2^2 in the second step, and we
continued removing 2^n balls at the n-th step.Now if we continue
like
this infinitelly, should this empty the container?
Zuhair
no,
if the container had an odd number of balls in it, you are only
removing
even numbers, so one ball must be left.
But the container intially had Aleph-0 of balls in it, and Aleph-0 is
neither odd nor even, since division is not defined for Aleph-0, even
if we intuitivelly say that Aleph-0 divided by n were n is a finite
cardinal would result in Aleph-0, then Aleph-0 would be even and odd at
the same time, since it would be dividable by all finite cardinals ,
only to result in itself. So your argument do not apply in this case.
Zuhair
WTF is Alpha-0 ? it cannot be a number, as all numbers are either odd or
even.
Therefore your rejection of my argument is false.
who's talking about numbers here, we are talking about quantities of
balls,
which is a number.
the balls that are removed can have odd or even numbers writtin
on them, Aleph-0 is the quantity of all balls in the container,
which is a number
you
argument only hold if this quantity is odd.
You see If in the first step I remove balls 0 and 1, in the second step
I will remove balls 2,3,4,5, you see I am removing balls that have odd
numbers and balls that have even numbers, your argument is:_ since
every 2^n is an even number, and since the sum of any
2^n and 2^m were n,m are natural numbers is an even number, then the
quantity of balls that I am removing has an even number describing it,
so if the total quantity of balls in the container was intially odd,
then my removal process will leave at least one ball their,
therefor you agree with me that your rejection of my argument is false.
but if it
was even then We'll have an empty container at the end. But the problem
is that the total amount of balls in the container are Aleph-0 which is
a transfinite NUMBER, it is the cardinality of any countablly infinite
set, like the set of all the balls in this question. Aleph-0 is a
number, but it is not a natural number you mean.
Either it is a number or it is not. Make up your mind.
Still my refutation of
your argument holds.
nope.
you are only taking out even numbers of balls, if there is an odd number to
start with, their will always be one left.
refute all you want, but you can't get around that.
Zuhair
To resolve the confusion. Aleph-0 IS a NUMBER. Period.
and Aleph-0 is neither an odd nor an even number, but it is a number,it
is a transfinite NUMBER.
Therefore your argument based on the total intial number of balls in
the container either being odd or even is erronous, because the total
intial number of balls in the container is Aleph-0 which is a number
that is neither odd nor even.
Therefore your argument is irrelavent.
Zuhair
Wow, man! Zuhair, are you sure aleph_0 isn't prime? I coulda swore it
was prime, and so, odd....
;)
Tony
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