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God Created the Integersall the rest is the work of man -Leopold Kronecker July 25 Some explanation of the lyrics of "Finite Simple Group of Order Two"It probably hardly makes any sense for us at our level of math, but I believe there're always some super geeks out there at Marianopolis interested in this kind of stuffs.
I received these stuffs by emails from anti and david. Thanks guys.
Well done. A few additional comments... > My heart was open but too dense
> But we're a mirror pair, me and you,
> And be a finite simple group, a finite simple group,
> I've proved my proposition now, as you can see,
> And by corollary, this shows you and I to be
November 09 cOmiCsMatrix Transformation
In fact, draw all your rotational matrices sideways. Your professors will love it! And then they'll go home and shrink.
I have never been totally satisfied by the explanations for why e to the ix gives a sinusoidal wave.
With reasonable assumptions about latitude and body shape, how much time might she gain them? Note: whatever the answer, sunrise always comes too soon. (Also, is it worth it if she throws up?)
The graph is of the magnitude of the function with the real value between 0 and 2 and the imaginary between about 35 and 40. I've misplaced the exact parameters I used.
It's science!
i love you
My most famous drawing, and one of the first I did for the site
You spin me right round, baby, right round, in a manner depriving me of an inertial reference frame. Baby.
cited from http://www.xkcd.com/archive/ October 26 First Math League CompetitionHello members! What did you all think of the first math league competition? Easy? Difficult? What do you think would have helped you ace it? More knowledge of geometry theorems? Number theory? Or maybe just a pencil? MATH CLUB cares what YOU THINK, so post away! October 04 Proof TechniquesProof is one the most important skills required in more advanced math learning, you will encounter them mostly in university, however, Cegep math teachers are using these techniques in their teaching. Moreover, the logic and the way of thinking that ground those techniques can be relevent to any intellectual activity.
Proof by Contradiction. Example: Theorem. There are infinitely many prime numbers. Proof. Assume to the contrary that there are only finitely many prime numbers, and all of them are listed as follows: p1, p2 ..., pn. Consider the number q = p1p2... pn + 1. This number is not divisible by any of the listed primes since if we divided pi into q, there would result a remainder of 1 for each i = 1, 2, ..., n. Well then, we must conclude that q is a prime number, not among the primes listed above, contradicting our assumption that all primes are in the list p1, p2 ..., pn.Proof by contradiction is often used when you wish to prove the impossibility of something. You assume it is possible, and then reach a contradiction. In the examples below we use this idea to prove the impossibility of certain kinds of solutions to some equations.
2. Mathematical Induction Example: Theorem. For any positive integer n, 1 + 2 + ... + n = n(n+1)/2. Proof. (Proof by Mathematical Induction) Let's let P(n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P(n) should be an assertion that for any n is verifiably either true or false.) The proof will now proceed in two steps: the initial step and the inductive step. Initial Step. We must verify that P(1) is True. P(1) asserts "1 = 1(2)/2", which is clearly true. So we are done with the initial step. Inductive Step. Here we must prove the following assertion: "If there is a k such that P(k) is true, then (for this same k) P(k+1) is true." Thus, we assume there is a k such that 1 + 2 + ... + k = k (k+1)/2. (We call this the inductive assumption.) We must prove, for this same k, the formula 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2. This is not too hard: 1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2. The first equality is a consequence of the inductive assumption.
3. Constructive Proof Example: Theorem. There is a rational number that lies strictly between the square root of 10100 and the square root of 10100+1. Proof. The square root of 10 100 is 10 50. After a little bit of trial and error, we let x = 10 50 + 10 -51, which is clearly a rational number bigger than the squre root of 10 100. To prove that x is less than the square root of 10100+1, we compute x2 = (10 50 + 10 -51)2 = 10 100 + (2) 10 -1 + 10 -102 which is clearly less than 10100+1.
4. Case by case Example: Theorem. If n is a positive integer then n7 - n is divisible by 7. Proof. First we factor n7 - n = n(n6 - 1) = n(n3 - 1)(n3 + 1) = n(n-1)(n2 + n + 1)(n+1)(n2 - n + 1). Now there are 7 cases to consider, depending on n = 7 q + r where r = 0, 1, 2, 3, 4, 5, 6, 7. Case 1: n = 7q. Then n7 - n has the factor n, which is divisible by 7. Case 2: n = 7q + 1. Then n7 - n has the factor n-1 = 7q. Case 3: n = 7q + 2. Then the factor n2 + n + 1 = (7q + 2)2 + (7q+2) + 1 = 49 q2 + 35 q + 7 is clearly divisible by 7. Case 4: n = 7q + 3. Then the factor n2 - n + 1 = (7q + 3)2 - (7q+3) + 1 = 49 q2 + 35 q + 7 is clearly divisible by 7. Case 5: n = 7q + 4. Then the factor n2 + n + 1 = (7q + 4)2 + (7q+4) + 1 = 49 q2 + 63 q + 21 is clearly divisible by 7. Case 6: n = 7q + 5. Then the factor n2 - n + 1 = (7q + 5)2 - (7q+5) + 1 = 49 q2 + 63 q + 21 is clearly divisible by 7. Case 7: n = 7q + 6. Then the factor n + 1 = 7q +7 is clearly divisible by 7.
5. The Pigeon Hole Principle
Theorem. Among any N positive integers, there exists 2 whose difference is divisible by N-1. Proof. Let a1, a2, ..., aN be the numbers. For each ai, let ri be the remainder that results from dividing ai by N - 1. (So ri = ai mod(N-1) and ri can take on only the values 0, 1, ..., N-2.) There are N-1 possible values for each ri, but there are N ri's. Thus, by the pigeon hole principle, there must be two of the ri's that are the same, rj = rk for some pair j and k But then, the corresponding ai's have the same remainder when divided by N-1, and so their difference aj - ak is evenly divisble by N-1.
Sources: September 30 Mathematical Love Song (Acapella)Finite Simple Group (of Order Two) The path of love is never smooth But mine's continuous for you You're the upper bound in the chains of my heart You're my Axiom of Choice, you know it's true But lately our relation's not so well-defined And I just can't function without you I'll prove my proposition and I'm sure you'll find We're a finite simple group of order two I'm losing my identity I'm getting tensor every day And without loss of generality I will assume that you feel the same way Since every time I see you, you just quotient out The faithful image that I map into But when we're one-to-one you'll see what I'm about 'Cause we're a finite simple group of order two Our equivalence was stable, A principal love bundle sitting deep inside But then you drove a wedge between our two-forms Now everything is so complexified When we first met, we simply connected My heart was open but too dense Our system was already directed To have a finite limit, in some sense I'm living in the kernel of a rank-one map From my domain, its image looks so blue, 'Cause all I see are zeroes, it's a cruel trap But we're a finite simple group of order two I'm not the smoothest operator in my class, But we're a mirror pair, me and you, So let's apply forgetful functors to the past And be a finite simple group, a finite simple group, Let's be a finite simple group of order two (Oughter: "Why not three?") I've proved my proposition now, as you can see, So let's both be associative and free And by corollary, this shows you and I to be Purely inseparable. Q. E. D. http://video.google.ca/videoplay?docid=2465730089162373019&q=finite+simple+group+of+order+two |
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