Banach-Tarski Paradox Talk Essay Example

Banach Banach-Tarski Paradox Talk Essay Banach-Tarski Paradox Talk Essay Essay Topic: Talk The catch, of course, is that each of these four pieces are so complex that they do not have any measure (i.e., their respective volumes are not well-defined), and that we do not know how to mathematically describe them other than the fact that they exist and exhibit the strange re-assembly property. In fact, it is quite possible that each of those pieces consists of isolated points spread out throughout the entire volume of the original sphere S. Now, accepting this is really pretty tough, but its possible to attempt to make an intuitive rationalization of it. Ill do it by way of an analogy with a physical sphere.  Lets forget for the moment the mathematical sphere S, which has infinite density. Lets consider a real, physical sphere B (for ball), also of radius 1. B is identical to S except that it consists of a finite number of atoms. These atoms are lain out in a crystalline structure. Usually, the crystalline structure of an object is a simple geometric relationship between neighbouring atoms.  Notice that although the geometric relationship between atoms define its crystalline structure, the precise distance between atoms in different molecules may vary. This leads to materials of different densities. Now, we perform the equivalent of a Banach-Tarski decomposition on our physical sphere B: we atomize B into four spherical clouds of atoms, lets call them C1, C2, C3, and C4. (Well ignore the central atom in B, just as in the mathematical version of this decomposition.) Lets assume that each of these clouds are sparse enough that they are gaseous, no longer solid by themselves (to try and imitate the immeasuarability of the four pieces of S ). Furthermore, lets say that the atoms in each of these clouds are laid out in a regular pattern, so that if we rotate C1 by some angle G, and put it together with C2 in the same spherical region, the atoms in both clouds line up into the same crystalline structure as B, except that now the distance between atoms is greater (to account for the missing atoms now in clouds C3 and C4). Similarly, assume we can do the same with C3 and C4: we just translate them away from the original spherical region of B so that they dont interfere with C1 and C2, and reassemble them into another sphere. Now, we have successfully built two (physical!) spheres with the same radius as B, using only material from B itself. Each of the two spheres have the same crystalline structure as B. The only difference between these spheres and B is that they each have only half the density of B.  To bring this analogy back to the mathematical sphere S: we can think of the infinitely complex pieces A1, A4 as the equivalent of atom clouds, which are non-solid (immeasurable). For those that know something of topology, which excludes myself, you can think of the crystalline structure of S as the topological structure of points in R3. These clouds lack this crystalline structure (i.e., they are unmeasurable); but by suitable rearrangement of them, we can form them into two identical spheres, with half the density of the original, so that they do have the same crystalline structure (i.e., the resultant two spheres are well-behaved, measurable sets). These two spheres are identical to S, except for having only half the density of S. However, S is infinitely dense, and so are its pieces A1, A4. This means the two resultant spheres are still infinitely dense. That is to say, they are identical to S. Et voila!There is no paradox here after all. We are merely seeing the logical consequence of mathematical sets like S being infinitely dense. In fact, if you think about it, this is not any stranger than how we managed to duplicate the set of all integers, by splitting it up into two halves, and renaming the members in each half so they each become identical to the original set again. It is only logical that we can continually extract more volume out of an infinitely dense, mathematical sphere S. Epilogue Now, having convinced you that the Banach-Tarski Paradox isnt really that strange after all, Id like to mention that the derivation of this paradox depends on the Axiom of Choice, and although most mathematicians accept the Axiom of Choice, not all agree with it. There has been much debate over the merit of adopting this axiom, as well as research into the consequences of choosing either way: it does simplify a lot of mathematical proofs, but it also introduces strange results like the Banach-Tarski paradox which we just discussed. If youre unfamiliar with the Axiom of Choice, it basically goes like this: if you have a collection of sets C (which may potentially contain an uncountably large number of sets), then there exists a set H, called the choice set, which contains precisely one element from each (non-empty) set in C. H is called the choice set because you are essentially going through each set in C and choosing one element from it. One feature of the Axiom of Choice is that H is simply assumed to exist; there is no algorithm given which might tell you how to construct an example of H. In the case of the Banach-Tarski paradox, each of the infinitely complex pieces of the sphere S is built from these choice sets. Since we do not know of any algorithm to actually construct these sets, we can only indirectly infer some of the properties of the pieces, such as their not having a (Lebesgue-) measure (i.e., they have intractible geometric complexity). Some of the debate surrounding the Axiom of Choice revolves around whether these non-constructible sets are mathematically admissible. The reader is encouraged to make good use to Google for more information about this debate; it is too vast a topic to explore in this article. It suffices to say that most mathematicians adopt the Axiom of Choice, simply because of the usefulness of results that can be derived. One might wonder, then, about what would happen if we didnt assume the Axiom of Choice. We do know that we would likely be unable to derive the Banach-Tarski paradox; however, we also know that paradoxical sets do exist even without the Axiom of Choice. These paradoxical sets exhibit the same weird behaviour of the Banach-Tarski spheres, in that you can decompose these sets into a finite number of parts, and reassemble them into multiple copies of the original.

How to Use the Perl Array Push() Function

How to Use the Perl Array Push() Function The Perl  push()  function is used to push a value or values onto the end of an array, which increases the number of elements. The new  values  then become the  last elements  in the array. It returns the new total number of elements in the array. Its easy to confuse this function with  the  unshift()  function, which adds elements to the  beginning  of an array. Heres an  example of the Perl push() function: myNames (Larry, Curly);push myNames, Moe;print myNames\n; When this code is executed, it delivers: Larry Curly Moe Picture a row of numbered boxes, going from left to right. The push() function pushes the new value or values onto the right side of the array  and increases the elements.   The array can also be thought of as a stack. Picture a stack of numbered boxes, starting with 0 at the top and increasing as it goes down. The push() function pushes the value onto the bottom of the stack  and increases the elements, like this: myNames (Larry,Curly);push myNames, Moe; You can also push multiple values onto the array directly ... myNames (Larry, Curly);push myNames, (Moe, Shemp); ... or by pushing on an array: myNames (Larry, Curly);moreNames (Moe, Shemp);push (myNames, moreNames); Note for beginning programmers:  Perl arrays begin with an symbol. Each complete line of code must end with a semicolon. If it doesnt, it wont execute. In the stacked example in this article, the lines without a semicolon are values contained in an array and enclosed in parentheses. This isnt an exception to the semicolon rule, as much as a result of the stack approach. The values in the array are not individual lines of code. It is easier to picture this in the horizontal approach to coding. Other Functions for Manipulating Arrays Other functions are also used to manipulate arrays. These make it easy and efficient to use a Perl array as a stack or as a queue. In addition to the push function, you can use: Pop function – removes and returns the last element of an arrayShift function – moves the whole array to the left. The element that is the first element of the array falls off the array and becomes the return value of the functionUnshift function – the opposite of the shift function, places a value at the beginning of an array and moves all the other element to the right.