Paradox!

[ryan]

Beero1000, would the answer to the OP be x, where x is an element of [-infinity, infinity]?

 

[Angry Floof]

Just to mess with you even more

Start with an empty vase at 11:00am, at 11:30am add a (countable) infinity of balls to the vase and remove 1 ball. At 11:45am, add another countable infinity of balls to the vase and remove 1 ball. Continue the process, halving the time step each iteration. How many balls are in the vase at noon?

My answer - still 42!

I tried to solve this with my solar calculator. I couldn't get an answer, though, on account of the sun being finite.

 

[Bomb#20]

Start with an empty vase at 11:00am. At 11:30am, place 10 balls in the vase, and remove one ball. At 11:45am, place 10 more balls in the vase and remove one ball. Continue repeating the procedure of adding 10 balls and removing one, but at each step reduce the time between steps by half. Question: At noon, how many balls are in the vase?

My answer: At noon, there are exactly 42 balls in the vase.

Of course, there are many other possible answers; this is a paradox, after all. Thoughts?

I make it either zero or one, depending on our metaphysical prejudices. Consider...

(1) The 10-to-1 ratio is immaterial. All that matters to set up the paradox is that in each phase there are more added than subtracted.

(2) The alternation of waiting periods and simultaneous instantaneous insertion/deletion is immaterial. It would be the same if we added 10 and subtracted 1 during the period from 11:00 to 11:30, then again between 11:30 and 11:45, and so on.

(3) The direction of time is immaterial. The reasoning is the same in a time-reversed version of your scenario.

Therefore,

End with a positron and an electron annihilating each other at T = 10 s, leaving a nearly empty region of space containing just one electron. Between T = 4 s and T = 8 s, nuclear reactions delete at least 4 fermions and create two. (A fermion is a particle such as an electron, positron, quark, etc. with a spin of 1/2, 1 1/2, 2 1/2, etc. Particles with spin 0, 1, 2, etc. are called bosons. Particle creation and destruction conserves total spin.) Between T = 2 s and T = 4 s, more fermions are deleted than added. Between T = 1 s and T = 2 s, more fermions are deleted than added. Between T = 0.5 s and T = 1 s, more fermions are deleted than added. The procedure of deleting and adding repeats indefinitely, but each step takes twice as long as the previous. Question: At T = 0 s, how many fermions were in the corresponding region of space?

In nearly every argument about this process it is customarily referred to by both sides as "something coming from nothing". This would seem to imply at T = 0 there were zero fermions. But some people insist that something can't come from nothing and therefore at T = 0 there was God. If God is a boson then that's still zero; but if God has half-integral spin then at T = 0 s there was one fermion.

It follows that at noon there are zero balls in your vase, unless we assume a spherical God, in which case, one. QED.

 

[Tom Sawyer]

It would depend on how long it takes you to add and remove the balls each time.

If you go up to the van, open the door, count the balls out from the bag, put them in the van, take one out, put that into your bag, close the van and walk away, it takes you longer than the interval after about four or five times at most to just do the work and noon arrives quite quickly.

Assume you have an infinite supply of balls, and that you can do the operations faster than any finite speed. Zeno's paradox is not that hard of a paradox to deal with, after all 1/2 + 1/4 + 1/8 + 1/16 + ... = 1.

Well, that doesn't make sense. There are three possible options.

The first is the pure math one where noon never arrives.

The second is mine where it takes time to switch the balls. In that case, the added time makes noon arrive after just a few rounds of switching.

The third is yours, which makes no sense. You start out by having a premise where there's infinite balls and infinite speeds, but then completely ignore those infinities when coming to your conclusion and somehow the infinitesimally small intervals which get arrived at have time spans which are somehow more relevant in comparison to infinite speeds than the 15 minute interval was, so all the intervals end up adding up to one as opposed to continuing on forever without reaching it.

 

[bigfield]

For example, if we imagine placing balls 1 - 10, and then removing ball 1 at 11:30am, then placing balls 11-20 and removing ball 2 at 11:45am, etc. It should be clear that every ball placed in the vase will have some number label, and also that every number label is removed at some point before noon. Therefore, all balls have been removed by noon, and so the vase is empty.

But you are always adding more balls than you are removing. How can you end up with no balls?

The only answer that makes sense to me is that as time approaches noon, the number of balls in the vase approaches infinity. The number doesn't also approach zero.

And damned if I understand how you got 42, other than the Douglas Adams reference.

Sure, that's the paradox. The number of balls in the vase keeps increasing, but so does the number of balls removed. The intuition says 'but balls are being placed faster than they are being removed', but why does that matter as long as every placed ball is removed before noon?

How is every ball removed before noon, if at any point in time before noon, more have been added than have been removed? I just don't understand how you are getting that conclusion.

 

[Wizardry]

For example, if we imagine placing balls 1 - 10, and then removing ball 1 at 11:30am, then placing balls 11-20 and removing ball 2 at 11:45am, etc. It should be clear that every ball placed in the vase will have some number label, and also that every number label is removed at some point before noon. Therefore, all balls have been removed by noon, and so the vase is empty.

But you are always adding more balls than you are removing. How can you end up with no balls?

The only answer that makes sense to me is that as time approaches noon, the number of balls in the vase approaches infinity. The number doesn't also approach zero.

And damned if I understand how you got 42, other than the Douglas Adams reference.

Sure, that's the paradox. The number of balls in the vase keeps increasing, but so does the number of balls removed. The intuition says 'but balls are being placed faster than they are being removed', but why does that matter as long as every placed ball is removed before noon?

How is every ball removed before noon, if at any point in time before noon, more have been added than have been removed? I just don't understand how you are getting that conclusion.

If you don't think that every ball is removed, then label the balls 1, 2, 3, .... What is the smallest numbered ball remaining at noon?

 

[Juma]

If don't think that every ball is removed, then label the balls 1, 2, 3, .... What is the smallest numbered ball remaining at noon?

Since we seem to be are able to label the balls freely we just assure that the removed balls has number 1, 11, 21 etc. Thus 2, 3, 4, 5, 6, 7, 8, 9 , 10, 12 etc will remain..

 

[Wizardry]

If don't think that every ball is removed, then label the balls 1, 2, 3, .... What is the smallest numbered ball remaining at noon?

Since we seem to be are able to label the balls freely we just assure that the removed balls has number 1, 11, 21 etc. Thus 2, 3, 4, 5, 6, 7, 8, 9 , 10, 12 etc will remain..

Well, sure, if you do it that way, you'll end up with an infinite number of balls at the end. But if you're trying to understand how you can add 10 balls at every step and remove 1 ball at every step and end up with 0 balls at the end, then you want to add the next 10 numbers at every step and remove the smallest remaining number at every step.

That way the steps go:
1: +10 -> 1..10; -1 -> 2..10; t = 11 1/2
2: +10 -> 2..20; -1 -> 3..20; t = 11 3/4
3: +10 -> 3..30; -1 -> 4..30; t = 11 7/8
...
100: +10 -> 100..1000; -1 -> 101..1000; t = 12 - 1/2^100
...

So under that procedure, at 12, what is the smallest number left on that list?

 

[steve_bnk]

[FONT=Arial Black, sans-serif]Beero[/FONT]

[FONT=Arial Black, sans-serif]'...Assume you have an infinite supply of balls, and that you can do the operations faster than any finite speed. Zeno's paradox is not that hard of a paradox to deal with, after all 1/2 + 1/4 + 1/8 + 1/16 + …'[/FONT]

[FONT=Arial Black, sans-serif]The ratio of remaining distance to total distance at each increment the sequence is ½, 1/4 ,1/8. In the limit as x → inf 1/2^n → 1, but for any finite value the sum is always < 1.[/FONT]

[FONT=Arial Black, sans-serif]The function of distance covered at any step is an exponential equation of the form A-A*(e^-x/k).[/FONT]

[FONT=Arial Black, sans-serif]Given A = 100 feet, distance(x) = 100 -100*e^-x/k where x = 1,2,3... are the increments.[/FONT]

[FONT=Arial Black, sans-serif]Given the distance is 100 feet, the distance at x = 1 is 50 feet. Solving for k, 0.442695. The exponential equation and the numerical simulation correlate. The exponential term will always be > 0 for any finite x so there is always be a remaining gap to cross.. In the limit as x → inf y → 100 feet.[/FONT]


[FONT=Arial Black, sans-serif]//scilab[/FONT]

[FONT=Arial Black, sans-serif]clear;[/FONT]

[FONT=Arial Black, sans-serif]total_distance = 100;[/FONT]
[FONT=Arial Black, sans-serif]current_distance = 0;[/FONT]
[FONT=Arial Black, sans-serif]remaining_distance = 100;[/FONT]

[FONT=Arial Black, sans-serif]for i = 1:20; [/FONT]

[FONT=Arial Black, sans-serif]current_distance = current_distance + (remaining_distance/2);[/FONT]
[FONT=Arial Black, sans-serif]remaining_distance = total_distance - current_distance; [/FONT]
[FONT=Arial Black, sans-serif]t(i,1) = i;[/FONT]
[FONT=Arial Black, sans-serif]t(i,2) = current_distance;[/FONT]
[FONT=Arial Black, sans-serif]d(i) = current_distance;[/FONT]

[FONT=Arial Black, sans-serif]z = -(i)/1.442695;[/FONT]
[FONT=Arial Black, sans-serif]d_exp(i) = 100 - (100*%e^z);[/FONT]
[FONT=Arial Black, sans-serif]t(i,3) = d_exp(i);[/FONT]
[FONT=Arial Black, sans-serif]t(i,4) = remaining_distance/total_distance;[/FONT]

[FONT=Arial Black, sans-serif]end;[/FONT]

[FONT=Arial Black, sans-serif]plot2d([d_exp d]);[/FONT]

[FONT=Arial Black, sans-serif]t[/FONT]
[FONT=Arial Black, sans-serif]Step - Numerical - Equation - Ratio Remaining to Total[/FONT]
[FONT=Arial Black, sans-serif]1. 50. 50.000001 0.5 [/FONT]
[FONT=Arial Black, sans-serif]2. 75. 75.000001 0.25 [/FONT]
[FONT=Arial Black, sans-serif]3. 87.5 87.500001 0.125 [/FONT]
[FONT=Arial Black, sans-serif]4. 93.75 93.75 0.0625 [/FONT]
[FONT=Arial Black, sans-serif]5. 96.875 96.875 0.03125 [/FONT]
[FONT=Arial Black, sans-serif]6. 98.4375 98.4375 0.015625 [/FONT]
[FONT=Arial Black, sans-serif]7. 99.21875 99.21875 0.0078125 [/FONT]
[FONT=Arial Black, sans-serif]8. 99.609375 99.609375 0.0039063 [/FONT]
[FONT=Arial Black, sans-serif]9. 99.804688 99.804688 0.0019531 [/FONT]
[FONT=Arial Black, sans-serif]10. 99.902344 99.902344 0.0009766 [/FONT]
[FONT=Arial Black, sans-serif]11. 99.951172 99.951172 0.0004883 [/FONT]
[FONT=Arial Black, sans-serif]12. 99.975586 99.975586 0.0002441 [/FONT]
[FONT=Arial Black, sans-serif]13. 99.987793 99.987793 0.0001221 [/FONT]
[FONT=Arial Black, sans-serif]14. 99.993896 99.993896 0.0000610 [/FONT]
[FONT=Arial Black, sans-serif]15. 99.996948 99.996948 0.0000305 [/FONT]
[FONT=Arial Black, sans-serif]16. 99.998474 99.998474 0.0000153 [/FONT]
[FONT=Arial Black, sans-serif]17. 99.999237 99.999237 0.0000076 [/FONT]
[FONT=Arial Black, sans-serif]18. 99.999619 99.999619 0.0000038 [/FONT]
[FONT=Arial Black, sans-serif]19. 99.999809 99.999809 0.0000019 [/FONT]
[FONT=Arial Black, sans-serif]20. 99.999905 99.999905 0.0000010 [/FONT]

[FONT=Arial Black, sans-serif]A limit at infinity is not reachable.[/FONT]

 

[Juma]

Well, sure, if you do it that way, you'll end up with an infinite number of balls at the end. But if you're trying to understand how you can add 10 balls at every step and remove 1 ball at every step and end up with 0 balls at the end,

That your labeling scheme results in that the lower index of the remaining balls increases beyond all values does not show that there are 0 balls left.

As I showed there are other labeling schemes wich doesnt have behave in that way.

The labeling scheme has nothing to do with the number of remaning balls.

The singularity catastrophe at 12 makes such arguments as yours invalid.

You are just revisiting "Hilberts hotel".

 

[steve_bnk]

Beero

Thanks for sparking a pleasant diversion.

 

[Wizardry]

Well, sure, if you do it that way, you'll end up with an infinite number of balls at the end. But if you're trying to understand how you can add 10 balls at every step and remove 1 ball at every step and end up with 0 balls at the end,

That your labeling scheme results in that the lower index of the remaining balls increases beyond all values does not show that there are 0 balls left.

That is exactly what it shows. If you think there are some balls left, give me the number of any ball that survives this procedure.

As I showed there are other labeling schemes wich doesnt have behave in that way.

I'm not claiming that every scheme for adding and removing balls results in 0 balls remaining. I'm claiming that there exists at least one scheme that results in 0 balls remaining.

The labeling scheme has nothing to do with the number of remaning balls.

You can't just assume that. You have to prove it with a logical argument or disprove it with a counterexample. A counterexample like you exhibiting a scheme that leaves an infinite number of balls and me exhibiting a scheme that results in 0 balls remaining.

The singularity catastrophe at 12 makes such arguments as yours invalid.

Or rather it would, if "singularity catastrophe" was a real thing and not something you made up just now.

You are just revisiting "Hilberts hotel".

Yes, I am, I'm glad you noticed. This problem and Hilbert's Hotel are just fanciful popularizations of basic set theory, which is about as well established a bit of human knowledge as has ever existed in the history of the world.

 

[Juma]

I'm not claiming that every scheme for adding and removing balls results in 0 balls remaining. I'm claiming that there exists at least one scheme that results in 0 balls remaining.

And I'm still waiting for a proof of that.
The only thing you shown is that when the clock strike twelve your labeling scheme is fucked.
That has nothing to do with number of remaining balls.

The labeling scheme has nothing to do with the number of remaning balls.

You can't just assume that. You have to prove it with a logical argument or disprove it with a counterexample. A counterexample like you exhibiting a scheme that leaves an infinite number of balls and me exhibiting a scheme that results in 0 balls remaining.

I already did that. Here is another: number the balls with even numbers. You will then have all odd balls left.

The singularity catastrophe at 12 makes such arguments as yours invalid.

Or rather it would, if "singularity catastrophe" was a real thing and not something you made up just now.
[/QUOTE]
It is a very real thing:
You get a singularity when the time between each action goes to zero in a bounded interval.(that is: at noon).
It is catastrophic because it would require infinite number of balls.

 

[bigfield]

For example, if we imagine placing balls 1 - 10, and then removing ball 1 at 11:30am, then placing balls 11-20 and removing ball 2 at 11:45am, etc. It should be clear that every ball placed in the vase will have some number label, and also that every number label is removed at some point before noon. Therefore, all balls have been removed by noon, and so the vase is empty.

But you are always adding more balls than you are removing. How can you end up with no balls?

The only answer that makes sense to me is that as time approaches noon, the number of balls in the vase approaches infinity. The number doesn't also approach zero.

And damned if I understand how you got 42, other than the Douglas Adams reference.

Sure, that's the paradox. The number of balls in the vase keeps increasing, but so does the number of balls removed. The intuition says 'but balls are being placed faster than they are being removed', but why does that matter as long as every placed ball is removed before noon?

How is every ball removed before noon, if at any point in time before noon, more have been added than have been removed? I just don't understand how you are getting that conclusion.

If you don't think that every ball is removed, then label the balls 1, 2, 3, .... What is the smallest numbered ball remaining at noon?

The answer cannot be defined, because as t approaches noon, the number of balls remaining in the vase approaches infinity.

It doesn't matter how you name the balls.

 

[Deepak]

At noon there are no balls since they've all been transmogrified into cigars.

 

[beero1000]

The posts here are what the paradox was designed to cause. It is supposed to expose the flawed nature of human intuition when dealing with infinity.

If one process starts listing the numbers 1,2,3,4,... and another process lists the same numbers 1,2,3,4,... except it operates 10 times faster, is there ever a number that the fast process reaches that the slow process does not? What does it even mean to go to infinity 'faster' than something else? Do these infinite processes even make sense as thought experiments?

Are there more even numbers than odd numbers? Are there more numbers divisible by three or numbers not divisible by three? Is the operation 'add 10 and remove 1' underspecified for an infinity of operations? Do we need to specify which ball to remove at each step? Why?

For those interested, this is called the  Ross–Littlewood paradox.

 

[Angry Floof]

I don't understand this paradox, but it reminded me of this, which I also don't understand, but both prove that math is a motherfucker.

 

[zorq]

The total number of balls in any given container is unaffected by the different operations of adding 10 balls and removing one vs. just adding nine balls.

So regarding the total number of balls at any given step after the first, even the infinite step, it will always be a positive number and will never be zero, or negative.

This is not a paradox.

Since the question assumes that we have infinite balls, and , I presume, a vase capable of holding any number of balls, We are clearly not dealing with a problem based in the real world, as such, I will treat the problem as a pure math problem and the answer is that at 12:00 there will have passed an infinite number of transactions each adding nine balls to the vase yielding a total of infinity balls in the vase. The number of balls in the vase remains undefined.

 

[Kharakov]

108.

 

[bigfield]

The posts here are what the paradox was designed to cause. It is supposed to expose the flawed nature of human intuition when dealing with infinity.

If one process starts listing the numbers 1,2,3,4,... and another process lists the same numbers 1,2,3,4,... except it operates 10 times faster, is there ever a number that the fast process reaches that the slow process does not? What does it even mean to go to infinity 'faster' than something else? Do these infinite processes even make sense as thought experiments?

Are there more even numbers than odd numbers? Are there more numbers divisible by three or numbers not divisible by three? Is the operation 'add 10 and remove 1' underspecified for an infinity of operations? Do we need to specify which ball to remove at each step? Why?

For those interested, this is called the  Ross–Littlewood paradox.

I read the solutions proposed using the numbering systems on the wiki page, and they all seem to be flawed in that they are trying to subtract infinity from infinity to get a non-infinite result.