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Troll Math - June 30, 2011
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32 Comments

If you treat the room as a FIFO queue, the input rate is 2 and the output rate is 1 the time elapsed between input and output grows with each throw at a rate n, as n->infinity the time between input and output goes to infinity and the ball is unable to make it through in a finite time.

Which brings us to the childhood factor....infinity + 1!

cool, that means i've had sex with an infinite number of women

haha two in one day NICe

I laughed more at this comment than I did at the comic itself.

'asdf's comment, I mean.

yeah of course you would laugh at mine -_- nobody does

Ah, humour derived from a misunderstanding of sequence limits. My favourite form of humour...

Very clever, actually: the number of balls inside and outside the room both grow arbitrarily large, i.e., in the limit go to \Aleph_0. The fallacy is thus basically a form of considering the undefined cardinal \Aleph_0-\Aleph_0, with a hint of confusion about ordinal and cardinal numbers thrown in.
But what I find more intriguing is the following question: in what kind of completion (and under which topology/metric) of the set of subsets of the integers does this sequence of sets (no.s of balls in the room after the n-th step) actually converge? Or isn't there any? Or is this trivial??

I think that the balls left in the room include √2, e, π. . . .

Yarr There be no balls left in the room me matees, since each ball excited the room at some point.

It reminds me of the light wave crossing an expanding (but finite) universe. However fast the universe expands, the light can reach the "other" end, although it may take an inordinately long time to reach it. Similarly, if a person has an eternity to read through a library that keeps expanding--the same situation as this paradox--eventually the person will read every book in the library, even though more books always remain that have yet to be read.

The problem comes with going too quickly to the infinite. Odd things happen when you go from any finite number to infinity, even with induction.

This was discussed at length on the xkcd forums. Here is an interesting analogy.

Mike are you on Google+?

What if I say the 'n+1'th ball is remaining in the room?

Some metric defining "distance" between sets is needed to discuss convergence of this (these) sequence. Every metric I've constructed, leaves the sequence non-Cauchy. The first time I encountered this puzzle, I attempted to prove that any valid metric would behave so; while I was unsuccessful in this, tghe nature of the process strongly suggests this to be so.

Comment funnier than comic itself. ^^

This problem is a good example of what happens when MAthmaticians forget they are modelling a real thing.
You can't let it go to infinity because you are modelling a REAL system, and infinity just makes no damn sense. You have to pick and actual number of trials, in which case you will find there are balls left in the room.

If you start with a parameter that is meaningless and nonsensical, you get and answer to math.

a bit more interesting is what happens if instead of removing the smallest ball you always remove one at random. How many balls do you expect to have "at the end"?

lol you made my day xD

uhm okay i really don't get it .. can someone explain this one to me ? .. my mother language is french so maybe it's due to a translation problem ..

A chaque etape n, tu lances deux balles (numerotees 2n et 2n-1) dans une piece et tu en reprends une (numerotee n). Ainsi, a la fin d'une etape, il y a une balle de plus dans la piece qu'au debut de l'etape. Donc, a une etape N donnee, la piece contient N balles. A la limite, la piece contient une infinite de balles.
Pourtant, pour toute balle, je peux te dire a quelle etape elle a quitte la piece (ex. la balle numero 543 a quitte la piece a l'etape 543). Donc a l'infini, toutes les balles ont quitte la piece. Ainsi il n'y a plus de balles dans la piece. D'ou la conclusion...

d'accord j'ai compris grâce à tes explications xD c'est con .. normal que je n'avais pas pigé au départ en fait :p merci ^^

isn't this just a restatement of one of Zeno's paradoxes? The one with the race where one runner travels twice as fast as the other one, but because the distance between them is constantly being reduced by the slower runner Zeno concludes that the distance between them should reach zero, while in reality it approaches infinity?

There's an interesting perspective on it involving renumbering the balls here: http://www.faqs.org/faqs/puzzles/archive/logic/part5/

My initial reaction, which probably isn't particularly relevant, is basically that 0 and infinity are both solutions to the problem n=2n, but that doesn't mean they're equal to each other. There are of course infinity-infinity=42 balls remaining in the room, and you can buy them all with the compound interest if you deposited one cent in your bank account at the beginning of the experiment.

Consider the natural bijection from stage number to number of balls in stage. Define A_n as set of balls at stage n. Then in the limit, this set is equivalent to the natural numbers. Now consider the bijection from ball removed at stage n to 2n. This shows the equivalence of balls removed to the even numbers. Now if I took all the even natural numbers from the natural numbers, we're left with the odd numbers, which is also countably infinite. Hence removing a countable set from a countable set doesn't guarantee the removal of all of the original. Blam. And no metric needed.

And that's why the order of infinite series matters when negative terms are allowed.

there are 5 balls in total...

Actually, after a certain point there would be so many balls that they collapse into a black hole, and then it would be impossible to throw them out of the room.

I assume the balls are numbered so that you can tell which ones are which. Now, at stage one, instead of throwing out ball #1, you throw out ball #2. Then, at stage two, throw out ball #4. Similarily, at stage three throw out ball #6, at stage four, ball #8, and so on. At stage N, ALL of the odd numbered balls will remain in the room. At the limit, you will have an infinite number of odd numbered balls remaining. The only numbered balls you can identify as having been thrown out are the even numbered ones, each thrown out at (ball #)/(stage #).

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