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 March 31st, 2017, 12:17 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 464 Thanks: 29 Math Focus: Number theory Volume paradox -- Banach–Tarski Can one construct two identical balls from a third, without topological manipulation or deformation, using only decompositions into finite sets and reassemblage? "The reassembly process involves only moving the pieces around and rotating them, without changing their shape. However, the pieces themselves are not "solids" in the usual sense, but infinite scatterings of points. The reconstruction can work with as few as five pieces." https://en.wikipedia.org/wiki/Banach...Tarski_paradox
 March 31st, 2017, 01:52 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,410 Thanks: 754 Yes, it's a standard result. Did you have a specific question about it? The reason it works is that at least one of the pieces must be non-measurable, meaning that there's no way to sensibly assign it a volume. So our intuition that rigid motions preserve volume fails for non-measurable sets. The proof isn't actually that difficult. The only tricky part is showing that you can find two independent rotations in 3-space. Once you do that, the group of rigid motions of 3-space contains a copy of the free group on two generators and you're done. I can drill down the jargon if you're interested. If I recall, the Wiki page you linked actually has the proof. Thanks from Loren Last edited by Maschke; March 31st, 2017 at 01:54 PM.
 March 31st, 2017, 10:56 PM #3 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 464 Thanks: 29 Math Focus: Number theory Do you know where I can find a visualization of the (symmetry?) process, or is there none by nature of the proof? Do the infinite scatterings have volume, or is "3-volume" not fundamental here? Is the non-measurability like uncertainty in physics, or due to the infinitesimal scatterings? Are the two independent rotations either antiparallel or orthogonal? What is meant by "the free group on two generators" in 3-space?
 April 1st, 2017, 04:01 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 I came across this by accident, and have no familiarity with the jargon. However, given that all the decompositions have an infinite number of points, a "paradox" isn't at all surprising. $\displaystyle \infty+\infty=\infty$ EDIT: Mathematicians should postulate the existence of an absolute increment of length. It would save them a lot of confusion. How many points are there in one increment of length, two increments of length? Unless you answer twice as many, you are going to collide with the real world if you use real world semantics. Last edited by skipjack; April 1st, 2017 at 10:17 AM.
April 1st, 2017, 05:18 AM   #5
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 Originally Posted by zylo I came across this by accident, and have no familiarity with the jargon. However, given that all the decompositions have an infinite number of points, a "paradox" isn't at all surprising. $\displaystyle \infty+\infty=\infty$ EDIT: Mathematicians should postulate the existence of an absolute increment of length. It would save them a lot of confusion. How many points are there in one increment of length, two increments of length? Unless you answer twice as many, you are going to collide with the real world if you use real world semantics.
And how big might that be? Why wouldn't half that length increment also exist?

As far as the real world goes, we have something called the Planck length.

Last edited by skipjack; April 1st, 2017 at 10:19 AM.

April 1st, 2017, 08:50 AM   #6
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 Originally Posted by Loren Do you know where I can find a visualization of the (symmetry?) process, or is there none by nature of the proof? Do the infinite scatterings have volume, or is "3-volume" not fundamental here? Is the non-measurability like uncertainty in physics, or due to the infinitesimal scatterings? Are the two independent rotations either antiparallel or orthogonal? What is meant by "the free group on two generators" in 3-space?

After that, questions would be appropriate. Else I'd be copy/pasting what's already in the article. The free group on two generators is described there as is the bit with the rotations. I could not explain it better than they did.

I can talk about nonmeasurability later if you are interested but you do need to read the Wiki article first because it answers some of the questions you asked.

Last edited by Maschke; April 1st, 2017 at 08:53 AM.

 April 1st, 2017, 09:26 AM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 You can resolve the paradox by referring all measurements to the original object. If at some step you find that 1+1=1, you have made a mistake. On the other hand, any paradox can be addressed by referring to the above wiki article and "Planck Length." You don't see how? You obviously didn't understand the article, a masterpiece of the genre. I particularly enjoyed "outline," because it gave me the satisfaction of filling in the missing steps myself, the key to which was "Planck Length." Thanks Romsek.
April 1st, 2017, 12:40 PM   #9
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 Originally Posted by zylo ....the key to which was "Planck Length."
interesting thing about the Planck Length....

one might think oh well it's just the resolution that we ended up with, it had to be something...

but the reason why it is the value that it is is that in order to try and resolve length scales smaller than the Planck Length you have to use high enough frequency electromagnetic radiation such that it contains so much energy that introducing it to a patch of spacetime would immediately collapse that patch of spacetime into a black hole with an event horizon beyond which you can get zero information.

You simply cannot resolve smaller than the Planck Length.

this opens up a sort of chicken and egg problem regarding the Planck Length and the gravitational constant but I think we'll find that the various constants are all related eventually.

 April 1st, 2017, 04:48 PM #10 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 464 Thanks: 29 Math Focus: Number theory I enjoyed the "Library of Babel," A version of which I believe appears in G, E and B (and also an almost infinite number of books). I'm ready for more, simplistic reasoning -- about my level.

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