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"Math is delicious!"

 According to a theorem called the BanachTarski paradox, it's possible to divide a sphere into five "pieces" (subsets), then reassemble them into a sphere with twice the volume using only rotations and translations. (The catch is that the "pieces" are "nonmeasurable sets"  they're constructed in such a way that they literally don't have a volume.)
 This paradox was the subject of an April Fool's joke in Scientific American magazine.
 Also seen expressed thuslywise: "What's an anagram of 'BanachTarski'? 'BanachTarski BanachTarski'."
 Another catch is that this stems from an axiom that's commonly omitted from set theory, being totally consistent with and independent of the others, but frequently yielding paradoxical results like these. You can choose to use it or not use it, similar to the parallel postulate.
 However, that is underselling its use. While the axiom of choice (which is the axiom that is vital in the construction) is used all over the place in analysis, and used in a couple mindnumbingly crucial places in algebra. Almost all ordinary mathematicians (read: nonFOMers) accept choice as true, and see the paradoxical results as unfortunate things that are true, and see the benefit from the more normal use of the axiom.
 Or you can accept the fact that mathematics is simply a lot less intuitive than we'd guess at first glance, and that there's nothing unfortunate or wrong about BanachTarski's paradox... in fact, in hindsight I would be much more surprised if such things were not possible, since there's no reason why splitting a sphere in nonmeasurable sets would need to preserve any intuitive notion of volume. Poor Axiom of Choice, always wronged for no real reason.
 You'd think they could do it in just four pieces though.
 And then there's Whitehead and Russell's Principia Mathematica, a multivolume opus intended to construct all of mathematics from the most basic of axioms (such as "every statement is either true or false"). After 379 pages of incomprehensibly dense notation, it succeeds in proving that 1+1=2. And then Gödel's Incompleteness Theorem undermined the whole thing using a reallife Logic Bomb.
 "The above proposition is occasionally useful" was their comment when the 1+1=2 proof was finally complete (presumably after arithmetical addition had been defined) in the second volume.
 It bears pointing out that Gödel's Incompleteness Theorem essentially says that no axiomatic system can be both complete and consistent, i.e., your system can be complete or consistent but not both. Given that, most mathematicians choose the latter, accepting an incomplete axiom system whose incompleteness is tempered by consistency (so that, e.g., 1+1 does indeed equal 2).
 A googol is equal to 10^100, or a 1 followed by 100 zeroes. A googolplex is equal to 10^(10^100), or a 1 followed by a googol zeroes; this is bigger than the number of elementary particles in the known universe (10^80) and the number of Planck times since the big bang (8*10^60)... times each other.
 And then there is the googolplexian (10^(10^(10^100))). It has a googolplex of zeroes after the first digit. But even that is nothing compared to...
 Graham's number. A number so large that it could not even be written with conventional scientific notation, and a new form of mathematical notation had to be developed just so it could be expressed. It is often regarded as the largest finite number that pure mathematics actually takes seriously. What's worse? The answer to the problem it was created to solve might actually be as low as 13.
 Graham's number is so ridiculously, mindbogglingly huge that there aren't enough elementary particles in the whole universe to write out the number of digits of the size of the power tower (3^3^3^3^3^3...3^3^3^3^3 E.T.C.)" in the first element of the generating series, of which each number is obtained by performing the operation on the previous number that got that first number from four, and Graham's number is the sixtyfourth.
 TREE(3). Graham's number, an upper bound is dwarfed by a lower bound!
 Möbius strips. It's one thing to have a ridiculous, mindbending shape like that on paper, but you can make them in real life. It's the weirdest feeling in the world to hold one.
 If you cut a Möbius strip in half lengthwise it stays in one piece. If you give it not one but three halftwists and cut it in half lengthwise, not only does it stay in one piece, it twists itself into a knot. If you cut a Möbius strip in thirds lengthwise (possibly by making the guide lines on the strip of paper beforehand), you'll finish cutting it in one go, and end up with a regular loop with another Möbius strip attached.
 Möbius strips can be made using legos  if you use the thin treads that have the chain link, give them a twist and connect them. the fun part, is if you run them with a gear, they still have full mobility.
 Incidentally, that's how some machines actually work they wind the leather motor belts into Möbius strips between wheels to ensure the belt wears out evenly.
 Not just Legos either, you can also make them out of an odd amount of Bucky balls, you just need to stagger them next to each other in a line of 2, then when you have a complete line, twist one end and connect them.
 Klein bottles take it a step further by doing away with those pesky edges. And you can cut them in half into a pair of Möbius strips and it's possible to cut a Klein bottle into a single Möbius strip.
 Here's someone making them in glass.
 And from the same site, Klein Hats.
 In a similar vein, how about some crochet models of hyperbolic space?
 New use for the Mobius strip: the world's smallestdenomination polyhedral die. (OK, monohedral die...is that even a word?)
 0.999..., which is a decimal with an infinite number of nines behind it, is exactly equal to 1. Not almost, not "so close there's no practical difference," exactly equal. Here are three proofs of this:
 First, from the accepted decimal expansions of some fractions, such as:
1/3=0.333... 
 Second, by simple algebraic manipulation:
Let 0.999...= x 
 Third: implicitly, 0.999... represents an infinite geometric sum of its digits, 9/10 + 9/100 + 9/1000 + ..., for which we can use the standard formula for the sum of a geometric series^{[1]}:
x = 0.999... 
 There are the same number of odd numbers (1, 3, 5, ...) as there are natural numbers (1, 2, 3, ...). What's even more bizarre is that there are the same number of fractions as there are natural numbers. What's more bizarre even than that is that this is not a trivial fact. There are multiple types of infinity (an infinite number, actually). The set of real numbers (everything that can be plotted on a continuous number line) is a larger type of infinity. (The set of all possible subsets of the real numbers is larger still.) However, most numbers that we care about"Computable numbers"fit into a countable set.
 As do any numbers recognized by a countable ordinal Hypercomputation degree.
 Which brings up that for almost all numbers, there is literally no process of computation that doesn't have some finite limit to its precision, and never can be.
 All infinities are infinite, but some are more infinite than others. The study of infinite sets is so fantastically weird that when the mathematician Georg Cantor first came up with it, he was accused of challenging God.
 This can be extraordinarily confusing, and tends to prompt discussion when it shows up, likely because it involves things like "all odd numbers" which are widely understood, while the more difficult formal ideas (like cardinality) aren't expressly stated. In short, two sets are defined as having the same cardinality ("size") if you can match up their members in some onetoone fashion. This is perfectly intuitive for finite sets, but a bit confusing for infinite ones. It is consistent, though, and more sensible than any other definition. Possibly the best example is Hilbert's paradox of the Grand Hotel: in a hotel with an infinite number of rooms, the propositions "every room is occupied" and "no more people can fit" are not equivalent.
 Is there a set with cardinality (size) that is strictly larger than the set of natural numbers but smaller than the set of real numbers? There is no answer (at least in ZFC), this is possible because of Gödel's incompleteness.
 If you buy the Continuum Hypothesis, there is no such set.
 Even trippier? All ndimentional real spaces have the same cardinality. That is, the number line (x), the Cartesian plane (x,y), threedimensional space (x,y,z), etc.
 What's more, the closed interval [0,1] over the reals, and the corresponding open interval (0,1), both have the same cardinality as as the entire real line itself. In other words, there are the same number of real numbers in toto as there are real numbers between zero and one, inclusively OR noninclusively.
 How about a geometric figure that has an infinite surface area, but a finite volume? Gabriel's Horn is such a figure. And then you get the Koch Snowflake, which has an infinite perimeter but a finite area. Similarly, the Menger Sponge has infinite surface area and finite volume, but unlike Gabriel's Horn, it can fit within a finite bounding box. Sierpinski's Carpet has an infinite perimeter and but zero area.
 Create your own Menger Sponges with Magic: The Gathering cards
 By using the concept of "inner product spaces," you can do absolutely absurd things with anything that can be considered a vector space. Like calculating the angle between two matrices. Or finding the distance between two polynomials.
 That is basically what is going on in modern mathematics, and has been for over a century. Mathematicians have been trying to find the correct generalizations, and see where it gives useful ideas. There is a germ of usefulness in the latter one that you gave for sure.
 The reason you can do this sort of thing is that any finitedimensional vector space V over a given field F is isomorphic to F × F × … × F for dim(V) factors F. In other words, no matter what your vector space is, any vector in it can be represented as a tuple (i.e., a fieldvalued vector in the classic sense) as long as it is finitedimensional.
 Topology has some fun ones, but there is no way to explain why they're fun unless you know a little topology yourself. The simplest one is that there is a notion of "closeness" on the natural numbers so that the sequence 1, 2, 3, ... is close to every natural number. Moreover, this notion of closeness doesn't seem all that unintuitive before you start digging.
 e^(iπ) = 1.
 And the graph of that [[media:180pxEuler Identity 2.svg.pnglooks like this]]. Makes a nice Tshirt.
 Slightly cooler (it's the generalization) is that e^(i* x)=cos(x)+i* sin(x).
 What's even cooler is that (i^i)^i=i
 The principal value  in fact, every value  of i^i is a real number.
 10^2+11^2+12^2=13^2+14^2=365. Looks like someone had fun screwing with our neighborhood.
 If you want proof that Evil Is Cool, play around with the Number of the Beast. 666 is not only the sum of the squares of the first seven primes, but it's also a palindrome. 1^3+2^3+3^3+4^3+5^3+6^3+5^3+4^3+3^3+2^3+1^3 is another palindrome, which equals 666. And that middle number, 6^3? That's three sixes, of course. Cool, right?
 Want another fun game with 666? You'll discover that it is the sum of all cardinal numbers from 1 to 36. Since 36 = 6^2, that means you can fit it into a Magic Square. Let he who hath understanding reckon the number of the beast.
 Some sources say the number is 616, though...
 Probability can be a very strange thing to behold at times. For example: You can have a situation where the probability of any given event is 0, yet events still happen (thanks to our good old friend "continuum").
 That's defined more precisely as probability density. Instead of probability of specific values, you calculate the probability of landing within selected boundaries by finding a slice of area under the density function (the total area under it being 1 = 100%).
 If you have a 50% chance of winning one cent, a 25% chance if winning two, etc., the expected winnings is infinite, despite the fact that it's impossible to win an infinite amount of money.
 However, if the winnings grow linearly rather than exponentially (1/8 chance to get 3 cents, 1/16 to win 4), then that expected winnings is both finite and readily calculable: two cents.
 You can justify it by saying that if we pick a real number at random, we can never be sure what exactly we picked. Even if we could somehow express arbitrary irrational numbers in the real world, calculating them requires an infinite number of steps  in practice, we stop somewhere and declare margins of error. And we can't even express them in the physical world  the circumference of a real physical circle with a diameter of 1 cm would never be exactly π cm (and on that matter, we cannot expect the diameter to be exactly 1 cm either). If we choose numbers by, for example, marking a point on a ruler, then not only does the ruler only have a finite number of atoms, but the mark would have finite thickness, specifying a whole range of numbers (presumably much wider than the size of an atom), and ranges have nonzero probability.
 Another weird probability distribution is the Cauchy Distribution, which does not have a welldefined mean because the integral does not converge.
 Here's a strange one, the minimum number of people you need such that any two of them have a 50% probability of having the same birthday (including leap years) is 23.
 There's also the Pidgeonhole Principal, complete with proof, on explaining that if you have x+1 elements and x slots, one slot will have at least two elements.
 "A point is that which has no part"Euclid, The Elements.
 For the entirely useless, but interesting spectrum of math we have the weird number bases. If you know decimal (the normal base) and get your head around binary or hexadecimal, the jump to the the other natural bases is not so great. However, consider a number base of 10. Or 2i. Or even base φ.
 Quiz: What is 1+1 in base φ? 10.01
 There are infinitely many primes, which the abovementioned Gödel's theorem implicitly relies on, along with integers having unique factorizations.
 And the proof is ridiculously easy: Suppose there were finitely many primes. If you multiply them all together you get a number that can be divided by any prime. If you add one you then get a number that cannot be divided by any prime. Therefore this big number is either prime and not a member of the set of all primes, or has prime factors that aren't members of the set of all primes. That's absurd. Hence there aren't finitely many primes.
 New Math. Yes folks, the brilliant folks in charge of our education systems tried to throw matrix ops, set theory, number bases, and abstract algebra at elementary schoolers. Many of the elementary teachers ordered to do the throwing didn't know the material themselves and wouldn't have been able to teach it to high school and college students of appropriate level and background.
 I believe that our youth can handle this!
"You can't take 3 from 2, 
 And a truly trivial piece of trivia: you know prime numbers, the numbers which can't be divided by any integer but 1 and themselves? One such number is ... eight six seven five three oh niiine!
 It is also the sum of two squares, (422^2+2915^2) forms a Pythagorean triple with 2460260 and 8319141, also twin prime with 8675311. (Don't look at me like that, I just looked it up on Wolfram Alpha)
 Though strictly speaking, this isn't that unusual  precisely half the primes can be expressed as sums of two squares (an interesting theorem in itself).
 It is also the sum of two squares, (422^2+2915^2) forms a Pythagorean triple with 2460260 and 8319141, also twin prime with 8675311. (Don't look at me like that, I just looked it up on Wolfram Alpha)
 e to the power of π is a strange number. When you take π from it, it's ridiculously close to 20.
 The fraction 355/113 is a far better approximation to π than any of us will ever need (sevenfigure accuracy). Two fives, two threes, two ones, in a really simple fraction.
 But a far more fun approximation is 42/1337 is approximately π/100, in other words π% of 1337 almost exactly 42 (actually 42.003...).
 A fun, notapproximationbutactuallyanexactanswer is the fourth root of nine squared plus 19 squared over 22 here.
 But a far more fun approximation is 42/1337 is approximately π/100, in other words π% of 1337 almost exactly 42 (actually 42.003...).
 Multiply nine by any "natural" number. Any at all. Now add up the digits. If it leaves you with multiple digits, keep adding up digits (from that and so on) until you're left with only one. That remaining digit will be nine, every time, no matter what number you picked for the multiplication. This works because 9=101. If we used a different base, say, trinary, then all even numbers would have a digital sum of two. The sideeffect of all multiples of three having a digital sum that is also a multiple of three (the only three of which are less than ten, thus being possible values for digital sums^{[2]}, are 3, 6, and 9) is because nine is a multiple of three.
 9x4=36. 3+6=9.
 9x26=234. 2+3+4=9.
 9x117=1,053. 1+0+5+3=9.
 9x42,361=381,249. 3+8+1+2+4+9=27. 2+7=9.
 9x8,675,309=78,077,781. 7+8+0+7+7+7+8+1=45. 4+5=9.
 Plug in anything. Try it yourself.
 The best part is, no other numbers have this property. if a number's digits don't add up to 9, you can know it isn't a multiple of nine.
 Take any natural number. Divide it by 9 and write down the remainder. Then do the trick of adding the digits until you get a single number. You'll get the remainder unless the original number was a multiple of 9 (in which case you'll get a sum of 9 and a remainder of 0).
 204/9 = 22 with remainder 6. 2+0+4=6.
 For those who can do Modular arithmetic: 10000a*+1000b+100c+10d+e mod 9 = 9999a+999b+99c+9d+(a+b+c+d+e) mod 9 = a+b+c+d+e mod 9. And it works with any base too, like with F in hexadecimal or 3 in quaternary.
 Along the same vein, if you sum the digits of any multiple of three, you get a multiple of three.
 This is probably really obvious but write 0123456789 down a paper and write 9876543210 next to it (also down). Hey look, it's the 9 times table!
 The sum of any two consecutive, positive, whole numbers is equal to the difference of their squares. E.g.
 3+4=7, 3x3=9, 4x4=16, 169=7.
 7+8=15, 7x7=49, 8x8=64, 6449=15.
 123456+123457=246913, 123456^2=15241383936, 123457^2=15241630849, 1524163084915241383936=246913.
 1,000,000+1,000,001=2,000,001. 1,000,000^2=1,000,000,000,000. 1,000,001^2=1,000,002,000,001. 1,000,002,000,0011,000,000,000,000=2,000,001.
 (n+1) + n = 2*n + 1 = (n^2 + 2*n + 1)  n^2 = (n+1)^2  (n)^2
 y^2  x^2 = (y+x)(yx), so if yx = 1, then ...
 More simply, imagine a square. To get to the next square, you add a row on the side, making it one wider. Then add a row on the top, and its length is the side of the original square plus one for the row just added on the side. So you add a side (the square root), and then you add another side +1.
 Like the above trick, the product of two integers two away from each other is one less than the square of the integer between them: for example, 2^2=1*3+1. 5^2=4x6+1. 126^2=125*127+1.
 Why? By factorizing, x^21 = (x+1)(x1).
 Using a trick with factorizing, you can make it seem like 2=1.
 A=B
 A^2=AB
 A^2B^2=ABB^2
 (AB)(A+B)=B(AB)
 A+B=B
 B+B=B
 2B=B
 2=1
 The trick is that since A = B, A  B is equal to zero; dividing by zero is not allowed.
 You can play a similar trick using integral calculus  in this case there's no division by zero and the error is more subtle. It works like so:
 INTEGRAL 1/x dx = INTEGRAL u dv, where u = 1/x, v = x.
 So, integrating by parts, INTEGRAL 1/x dx = uv  INTEGRAL v du = (1/x) x  INTEGRAL x*(1/x^2) dx.
 Simplifying this gives INTEGRAL 1/x dx = 1 + INTEGRAL 1/x dx, so 1=0. ^{[3]}
 2+2=5 is actually very simple by abusing definitions of rounding (that is, 0.4 and below is rounded low and 0.5 and above is rounded up). For 2.4=2, 2.4+2.4=4.8, which can be rounded off to 5, so 2+2=5 for rounded off numbers.
 2+2=5 for sufficiently large values of 2.
 I've always liked this for 1=1
 1=1
 1/1=1/1
 1/i=i/1
 1=1
 In this case, the trick is that the third line attempts to use the property that sqrt(a/b)=sqrt(a)/sqrt(b)  which is only true when a and b are both positive real numbers. If you allow a or b to take on negative or complex values (like, say, 1), this is no longer a valid manipulation.
 Want to find the square of a number ending in 5? Multiply the numbers 5 greater and 5 fewer than it, and then add 25, e.g. 35^2 = (40 * 30) + 25 = 1225. From here, you can move on to squares of other numbers, or just numbers that are relatively close in value.
 The trick? (x5)(x+5) = x^2  25.
 More generally: (xy)(x+y) = x^2  y^2.
 The trick? (x5)(x+5) = x^2  25.
 Contrary to what most overzealous math teachers might suggest, it is possible to divide by zero, provided you're in a wheel. The consequences though? x  x is not always zero, 0 * x is not always zero, and x / x is not always 1.
 So if xx is not zero, and you add x to both sides, x=x+ (some nonzero number)...?
 Wait, the reflexive property has a point?
 It does. So does welldefinition, for that matter.
 The overzealous math teachers to whom you refer can fall back on the fact that essentially all math that's taught from kindergarten through high school is done in rings and fields, wherein division by zero is not defined (assuming that division is even possible).
 Here's a fun one: 1+2+3+4+5+6... etc. can in some sense be shown to be equal to 1/12. In two different ways.
 Behold, Polish hand magic. A visual method for multiplying any two integers without a calculator. And it even comes with an explanation.
 This troper tried the 7*8 example and got 53... primarily because the troper's right hand is not like the other. (Google image Brachysyndactyly, I dare you.) Then again, "imagine negative fingers and extra fingers!" Writers Cannot Do Math anyways, so...
 There's a joke proof that states that there are no uninteresting numbers. Here it is: Assume that there is some set of uninteresting numbers. Of this set, one of these numbers must be the smallest one. Thus, it would be the smallest uninteresting number, and therefore interesting, which would remove it from the set. Then the next smallest uninteresting number would become the smallest, which would be interesting, and so on, until the set of uninteresting numbers is empty. Therefore, all numbers must be in some way interesting!
 Somehow, I don't believe that the set of "interesting numbers" is welldefined.
 Also, you can only pick the smallest if the set of all numbers you're considering is well ordered, which is true if it's the set of integers. As for real or complex numbers well....
 There exists a wellordering of the reals in ZFC. However, the usual ordering—i.e., <—is not a wellordering on the reals. Furthermore, the complex numbers do not form an ordered field, so wellordering them is even harder.
 Most people know about 5 number systems: Natural Numbers, Integers, Rational Numbers, Real Numbers, and Complex Numbers. However some other interesting ones are:
 the Algebraic Numbers, which consist of numbers that are solutions of polynomials with rational coefficients — e.g., all rational numbers, φ (the socalled Golden ratio, √2, etc.
 the Quaternions, Octonions, and Sedenions: Complex numbers extended to quadruples, octuples, 16tuples etc.
 the Hyperreal numbers: includes infinite numbers and infinitesimal numbers
 the Surreal numbers: which is the largest possible totally ordered set allowed by set theory
 Here's a fun one to amuse your friends: 111111111 X 111111111 = 12345678987654321
 So you need to find the square root of a number, but you have one of those cheap calculators that don't have a square root function, or maybe you don't have a calculator. No worries, there's a quick, simple and Babylonian way to solve this problem.
1) Take a wild guess at what the square root of the number might be. Let's call the guess G and the original number N. 
 Example: What is √65536^{[4]}? (Uh... I don't know, 376?)
Let G_{0} = 376, N = 65536 
 Fun fact: If your initial guess is a negative number, you'll get the negative root as the answer. That is, if the initial guess for √65536 was 376, the final answer would have been 256.
 Out of all the undefined values (commonly occurring a mix of 0 or ∞), the reason why 1^∞ is undefined, as opposed to 1 (as 1 raised to anything equals 1, right?), is because 1^∞ can equal Euler's constant. Or to be more precise, the naive solution to how to get e, lim x > ∞ ( 1 + 1/x ) ^ x is 1 ^ ∞. But there's some rules that rearrange the equation, such that you can approach e.
...my brain hurts. I think I'll stick to biology.
 Back to Mathematics
 ↑ which, incidentally, is proven using a construction similar to the second proof
 ↑ digital sums can only result in a onedigit number. If it has more than one digit, it's not the digital sum
 ↑ Note, however, that these are indefinite integrals  that is, they're only defined up to a constant of integration anyway. Essentially, you just have to pick the constant on one side of the equation to be 1 greater than the constant on the other side, and then everything works out nicely.
 ↑ Programmers, don't blurt out the answer