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06 Oct 10
4 edits
Consider the functions:
Y = X
and
Z = 2X
Z - Y is obviously equal to X and increases continually to infinity with increasing x.
Assume these function start at X = 1
Now assume A = 1/X
A now starts at 1 and descends to 0, much like the time in this question
the two functions at the start become
Y = 1/A
Z = 2/A
The functions now curve away from the A axis but since the two lines have merely been squashed horizontally, at all points:
Z-Y=1/A
Z and Y are infinitely apart at A=0 (which was X=Infinity)
Taken at suitable points where A = 1/2^2 these functions are our lower and upper ball limits.
07 Oct 10
Originally posted by iamatigerThe question about "how many elements are in a set" is a question of cardinality, not addition. The two are not the same.
Consider the functions:
Y = X
and
Z = 2X
Z - Y is obviously equal to X and increases continually to infinity with increasing x.
Assume these function start at X = 1
Now assume A = 1/X
A now starts at 1 and descends to 0, much like the time in this question
the two functions at the start become
Y = 1/A
Z = 2/A
The functions now curv ...[text shortened]...
Taken at suitable points where A = 1/2^2 these functions are our lower and upper ball limits.
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07 Oct 10
Originally posted by wolfgang59I don't know much about cardinalities and number theory...but
[b]How can "What is left?" not be the question?
It was not the original question.
"What is left?" has no answer.[/b]
lets let the time interval be t
ball#1 will be removed at t/2
ball#2 will be removed at t/4
ball#3 will be removed at t/8
.
.
.
ball#N will be removed at t/(2^n) (which is a convergent geometric series)
so try look in the bag, pick up a ball...
what is the # on it
its not ball 1, it was removed at t/2
its not ball 100, it was removed at t/1.26765E30
its not ball N, as it was removed at t/(2^n)
its not ball # infinity, because infinity is not a number
so all balls have been removed from the bag.
07 Oct 10
Originally posted by joe shmoWhen ball #N is removed there are still balls in the bag N+1, N+2, et cetera
I don't know much about cardinalities and number theory...but
lets let the time interval be t
ball#1 will be removed at t/2
ball#2 will be removed at t/4
ball#3 will be removed at t/8
.
.
.
ball#N will be removed at t/(2^n) (which is a convergent geometric series)
so try look in the bag, pick up a ball...
what is the # on it
its not ball 1 ...[text shortened]... ll # infinity, because infinity is not a number
so all balls have been removed from the bag.
as you point out infinity is notb a number so we cannot talk about that ball being removed.
So at any moment we can prove there are balls still in the bag.
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07 Oct 10
1 edit
Originally posted by joe shmoHowever, how many balls are in the bag just after ball N is removed?
I don't know much about cardinalities and number theory...but
lets let the time interval be t
ball#1 will be removed at t/2
ball#2 will be removed at t/4
ball#3 will be removed at t/8
.
.
.
ball#N will be removed at t/(2^n) (which is a convergent geometric series)
so try look in the bag, pick up a ball...
what is the # on it
its not ball 1 ...[text shortened]... ll # infinity, because infinity is not a number
so all balls have been removed from the bag.
We can easily prove that N more balls are in the bag at that point. (I can't be bothered proving it *again* (sigh).
So, unless the ball being removed is 0, there are more balls to remove in the bag, hence, the last ball can *never* be removed.
The problem is that you suppose there is a last ball, this is wrong (because the balls are infinite last does not apply)
Then you say that you can prove that any ball N is removed, this is right, but you don't consider at what step ball N is removed. It is easy to show that it is removed at step 2N so it is severely lagging the addition of balls.
This whole argument is really about people arguing that a non-convergent series converges, which is wrong and leads to many logical inconsistencies, for instance if this series converges to 0 we have diverging lines that meet, and triangular sections that change size when rotated.
07 Oct 10
1 edit
Originally posted by iamatigerSet of ALL balls inserted : {1,2,3,...}
(sigh)
Set of ALL balls removed: {1,2,3,...}
Same set. You're also welcome to *sigh* disprove the argument that it's impossible that there are infinite balls in the bag and there isn't a lowest numbered ball.
Again *sigh* your mistake is extrapolating cardinality follows from infinite sequences, but anyone would agree that the sequences
{1,2} ; {2}
{1,2,3,4} ; {2,4}
...
{1,2,3,4,...,n,...,2n} ; {2,4,6,...,2n}
...
seem to be diverging in terms of cardinality yet by creating a one-to-one correspondence we can show that the sets
{1,2,3,...} ; {2,4,6,...}
have exactly the same cardinality.
*sigh*
07 Oct 10
Originally posted by PalynkaIf you are right then you can tell me flaw in this argument.
Set of ALL balls inserted : {1,2,3,...}
Set of ALL balls removed: {1,2,3,...}
Same set. You're also welcome to *sigh* disprove the argument that it's impossible that there are infinite balls in the bag and there isn't a lowest numbered ball.
Again *sigh* your mistake is extrapolating cardinality follows from infinite sequences, but anyone would agree ...[text shortened]... n show that the sets
{1,2,3,...} ; {2,4,6,...}
have exactly the same cardinality.
*sigh*
1. After step n the number of balls in bag is (2n - n).
2. As n-->inf then (2n-n) -->inf
We are NOT concerned with naming what is left in the bag!!
07 Oct 10
1 edit
Originally posted by wolfgang59For finite sets one can represent such operations on cardinality easily with standard operations, but for infinite sets many standard things break down. You can add and remove elements (including sets of countably infinite sets) and the cardinality will remain unchanged and other particularities.
If you are right then you can tell me flaw in this argument.
1. After step n the number of balls in bag is (2n - n).
2. As n-->inf then (2n-n) -->inf
We are NOT concerned with naming what is left in the bag!!
I find it surprising that so many of you readily accept that the natural numbers and even numbers are equinumerous, but find it so hard to accept this.
Here's a challenge: try replicating that known fact of equinumerosity by going from finite to infinite sets. For every increase in n you'd see cardinality diverging (with finite sets) and yet when you consider both infinite sets then cardinality becomes the same. S Why? Because only then can you generate a one-to-one correspondence. This is exactly the same.
With which of these do you disagree:
1. The set of balls inserted is: {1,2,3,4,...}
2. The set of balls removed is: {1,2,3,4,...}
3. The set {1,2,3,4,...} is the same set as {1,2,3,4,...}
08 Oct 10
Originally posted by PBE6You may be able to boil an egg, but it is impossible to put balls in and take them out that quickly (i.e. infinitely fast). You can only see what the limit is as you approach infinity (t=0 in this case). That limit is"If you've done six impossible things this morning, why not round if off with breakfast at Millyways?" - Douglas Adams
Suppose you and your friend come into possession of a bag of infinite capacity, along with an infinite supply of numbered balls and a stopwatch with an infinite number of gradations. After watching the universe explode ...[text shortened]...
[b]Question: How many balls will be left in the bag when the stopwatch reads 0 seconds?[/b]
B = 2S - S = S
The number of balls equals the number of steps. As time approaches zero, the number of balls put in approaches infinity twice as quickly as the number taken out. For example, the value of
(2N +1)/(N-1) as N approaches infinity is 2, not 1. That's because the numerator approaches infinity twice as quickly as the denominator.
08 Oct 10
Originally posted by Palynka4. The balls are inserted at twice the rate they are removed.
For finite sets one can represent such operations on cardinality easily with standard operations, but for infinite sets many standard things break down. You can add and remove elements (including sets of countably infinite sets) and the cardinality will remain unchanged and other particularities.
I find it surprising that so many of you readily accept tha ...[text shortened]... of balls removed is: {1,2,3,4,...}
3. The set {1,2,3,4,...} is the same set as {1,2,3,4,...}
Thus at step S the set of balls inserted is {1,2,3,4,...,2S} and the set of balls removed is {1,2,3,4,...,S}.
08 Oct 10
Originally posted by AThousandYoungI like it how you guys always completely ignore my post and decide to repeat the same arguments.
4. The balls are inserted at twice the rate they are removed.
Thus at step S the set of balls inserted is {1,2,3,4,...,2S} and the set of balls removed is {1,2,3,4,...,S}.
I've addressed that. Your statement is correct, but that doesn't preclude the fact that at t=0 there are no balls in the bag. If you expand each of those sets to infinity you'd see that they are exactly the same. Do you disagree?