Paradox!

[beero1000]

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?

 

[ryan]

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?

If we keep putting 9 balls in the vase an infinite number of times, wouldn't we get an infinite number of balls?

 

[Juma]

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?

There should be -9/12 balls...

 

[beero1000]

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?

If we keep putting 9 balls in the vase an infinite number of times, wouldn't we get an infinite number of balls?

I never said that you have to remove one of the ten just placed. Adding ten and removing one isn't the same thing as adding nine.

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.

 

[rosy tetra]

At noon, I have run out of balls. And my hands are tired.

 

[rosy tetra]

Every number label is removed before noon? Depends. Suppose we color-code the balls, and suppose the ball we remove is from the set we just put in. The first set of 10 balls is green, and we remove one green ball. The second set of 10 balls is yellow, and we remove 1 yellow ball, and so on. The other 9 green balls will never be removed, the other 9 yellow balls will never be removed, etc.

 

[ryan]

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?

If we keep putting 9 balls in the vase an infinite number of times, wouldn't we get an infinite number of balls?

I never said that you have to remove one of the ten just placed. Adding ten and removing one isn't the same thing as adding nine.

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.

Doesn't the sum of 10n - 1 go to infinity "faster" than the just sum of n?

 

[beero1000]

Every number label is removed before noon? Depends. Suppose we color-code the balls, and suppose the ball we remove is from the set we just put in. The first set of 10 balls is green, and we remove one green ball. The second set of 10 balls is yellow, and we remove 1 yellow ball, and so on. The other 9 green balls will never be removed, the other 9 yellow balls will never be removed, etc.

Sure. Given the thread title, it shouldn't be surprising that two seemingly sound analyses yield conflicting answers. You can trivially modify the 'coding' to argue that there are n balls at noon, for pretty much any number n you'd like.

The question remains - how many are there at noon? Which number is the 'right' answer?

 

[beero1000]

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?

If we keep putting 9 balls in the vase an infinite number of times, wouldn't we get an infinite number of balls?

I never said that you have to remove one of the ten just placed. Adding ten and removing one isn't the same thing as adding nine.

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.

Doesn't the sum of 10n - 1 go to infinity "faster" than the just sum of n?

In what sense?

Of course, it is perfectly reasonable to argue that the answer should be infinity. But if the answer is infinity, why is the 0 analysis wrong? Where is the logical error?

 

[ryan]

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?

If we keep putting 9 balls in the vase an infinite number of times, wouldn't we get an infinite number of balls?

I never said that you have to remove one of the ten just placed. Adding ten and removing one isn't the same thing as adding nine.

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.

Doesn't the sum of 10n - 1 go to infinity "faster" than the just sum of n?

In what sense?

Of course, it is perfectly reasonable to argue that the answer should be infinity. But if the answer is infinity, why is the 0 analysis wrong? Where is the logical error?

Yeah, I'm going leave this one alone.

 

[rosy tetra]

My answer still remains: I have run out of balls, and my hands are tired. Also, I can't move my hands fast enough to keep up with my clock. This will happen no matter how many balls I have, and no matter how strong and fast my arms are.

 

[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.

 

[ryan]

Beero1000, isn't the answer 0 because aleph-naught - aleph-naught = 0?

 

[steve_bnk]

That is Zeno's Paradox is it not?


If you evaluate on a computer with afinite floating point length eventually it will round to 3600 seconds
or one hour.


Mathematically the sequence that takes a proportion ofwhat remains and adds it to the current value defines an asymptoticexponential function. Another example is the voltage vs time ofresistor-capacitor circuit.

If you plot the half time steps you get an asymptotic exponential function.

I think the logical contradiction is trying to perform an infinite number of operations in a finite time span of 1 hour.



// scilab

clear;

//One hour = 3600 seconds so thestarting hour is irrelevant.


_time = 0; // time counter
time_remaining = 3600; // timeremaining
vase = 0; // ball counter
t_stop = 3600; // end point


for i = 1:30;
vase = (vase + 10) -1;
_time = _time + (time_remaining/2);
time_remaining = t_stop - _time;
tt(i,3) = _time;
tt(i,1) = i;
tt(i,2) = vase;
t(i) = _time;
end;

plot2d(t);


tt




Iteration – vase count - time
1. 9. 1800.
2. 18. 2700.
3. 27. 3150.
4. 36. 3375.
5. 45. 3487.5
6. 54. 3543.75
7. 63. 3571.875
8. 72. 3585.9375
9. 81. 3592.9688
10. 90. 3596.4844
11. 99. 3598.2422
12. 108. 3599.1211
13. 117. 3599.5605
14. 126. 3599.7803
15. 135. 3599.8901
16. 144. 3599.9451
17. 153. 3599.9725
18. 162. 3599.9863
19. 171. 3599.9931
20. 180. 3599.9966
21. 189. 3599.9983
22. 198. 3599.9991
23. 207. 3599.9996
24. 216. 3599.9998
25. 225. 3599.9999
26. 234. 3599.9999
27. 243. 3600.
28. 252. 3600.
29. 261. 3600.
30. 270. 3600.

 

[rizdek]

Is time continuous in your experiment? If so how can noon ever arrive? You always have another "halfway" point in time with which to contend.

 

[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.

 

[beero1000]

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?

As for how to get 42, place balls 1 - 10 at the first step and remove ball 1, place balls 11-20 at the second step and remove ball 2, place balls 21-30 and remove ball 3, place balls 31-40 and remove ball 4, place balls 41-50 and remove ball 5, then place balls 51-60 and remove ball 47, place 61-70 and remove ball 48, etc. Every ball except 5, 6, ..., 45, 46 is removed, so those are exactly the balls remaining. Therefore, there are exactly 42 balls in the vase at noon.

Beero1000, isn't the answer 0 because aleph-naught - aleph-naught = 0?

The key to the different solutions is that you cannot subtract cardinals like that. In particular, aleph-naught - alepha-naught can be any cardinal aleph-naught or smaller, not just zero (or infinity).

My answer still remains: I have run out of balls, and my hands are tired. Also, I can't move my hands fast enough to keep up with my clock. This will happen no matter how many balls I have, and no matter how strong and fast my arms are.

That is Zeno's Paradox is it not?

Is time continuous in your experiment? If so how can noon ever arrive? You always have another "halfway" point in time with which to contend.

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.

 

[beero1000]

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!

 

[4321lynx]

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 guess we are to assume that the vase is big enough hold all these balls, no matter what the size of the balls is. You did not state that, but if we don't the whole thing is just a lot of balls.

 

[Loren Pechtel]

None.

As we get sufficiently close to noon the balls are going to be moving *VERY* fast. Eventually the energy liberated this way is going to explode the vase. Thus, come noon it won't hold balls at all.