March 31st, 2017, 12:17 PM  #1 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 435 Thanks: 28 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: 2,342 Thanks: 731 
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 nonmeasurable, meaning that there's no way to sensibly assign it a volume. So our intuition that rigid motions preserve volume fails for nonmeasurable sets. The proof isn't actually that difficult. The only tricky part is showing that you can find two independent rotations in 3space. Once you do that, the group of rigid motions of 3space 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. 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: 435 Thanks: 28 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 "3volume" not fundamental here? Is the nonmeasurability 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 3space? 
April 1st, 2017, 04:01 AM  #4 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,495 Thanks: 1369  Quote:
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  
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731  Quote:
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: 125 
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, 11:52 AM  #8 
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731 
The easiest way into the proof sketch of the BanachTarski paradox is to start with the free group on two letters. But before I do that, I want to tell you a variation of a story first told by Jorge Luis Borges, called the Library of Babel. In our version of the story, there is a great university with a lot of wealthy alumni. The university decides to build a library that contains not only every book ever written in the English language, but every book that could ever be written in the English language. For simplicity let's assume a book is just a finite sequence of the letters a through z. We could do the same thing with punctuation and upper and lower case letters if we wanted to. They start printing out all the books that start with the letter 'a', then all the books that begin with 'b', and so forth. So the 'a' section of the library has all the books: a, aa, aaa, aaaa, aaaaa..., every finitelength string of 'a's. Then there's ab, abba, aeoruweorljfsd, aeiurieiere, and so forth. Every finite string of letters starting with 'a'. Then there are the b's: b, bb, bbb, bsjfsjfks, and so forth. When they are all done printing all these books, most of them are nonsense. But Moby Dick is in there, and so is the Bible, and all the Harry Potter books. Every possible variation of each book is in there too. One in which Moby Dick is the captain of the Pequod and Ahab is the whale. Another version in which Ahab kills Moby Dick and eats him for dinner. Every possible finitelength string of letters is in the library. Now the university as I said has a lot of wealthy alumni, so the university president comes up with a scheme to soak the alumni for cash that doesn't involve the football team. The president says: We will put up 26 magnificent buildings, one for the a's, one for the b's, and so forth. In this way we can soak the alumni for years. And so it was done. The alumni were soaked, and in time the campus had 26 brand new magnificent buildings, one building for the books starting with 'a', one for the books starting with 'b', and so forth. Then one day something terrible happened. 25 of the buildings burned down. All the b's, c's, d's, through z's burned down. The books were destroyed. Only the building containing the a's remained standing. The faculty were terribly upset. All those books, lost forever. But then someone noticed something amazing. If you take any book starting with k, say, like keiruwiou, that book is already in the 'a' building as akeiruwiou. To rebuild the entire library, all you have to do is take all the books starting with 'a', erase the first 'a', and what's left is a book that was in one of the 26 buildings. No books were lost at all. The 'a' books already contain every other book in the library, but with an extra 'a' in front. For example Moby Dick is in there, starting with acallmeishmael. You just delete the leading 'a' and you have all the books starting with 'c'. The president of the university was overjoyed. No books were lost. The president had only one instruction for the faculty: Just don't tell the alumni! After we have contemplated this story for a while, I'll talk about the free group on two letters. But there's already a discussion of that, including a nice picture, in the Wiki page already linked by the OP. Last edited by Maschke; April 1st, 2017 at 11:56 AM. 
April 1st, 2017, 12:40 PM  #9 
Senior Member Joined: Sep 2015 From: USA Posts: 2,495 Thanks: 1369  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: 435 Thanks: 28 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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