Re: Mathematical Rabbits
From: mechmech (bbb_at_ccc.com)
Date: 11/08/04
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Date: Sun, 7 Nov 2004 20:42:48 -0500
but the quantum rabbit can do just that. It can pop its head out of the two
holes simultaneously. Quatum rabbits usually live in physics departments and
are known not to eat grass. they obey the uncertainty principle: if you know
how many you have you don't know where they are and if you know where they
are, you don't know how many there are.
the little worm
"Martin Penderis" <tafeltennis@mwpenderis.mailshell.com> wrote in message
news:e4fa4b6f.0411071107.209aeede@posting.google.com...
> Another rabbit problem:
>
> A rabbit is in a box with two holes in the lid. After 1 second after
> being put in, it pops its head out of one hole. After another 1/2
> second, it pops its head out of the other hole. After another 1/4
> second, it again pops its head out of the first hole. This pattern
> continues. After what time will the rabbit be popping its head out of
> both holes at the same time?
>
> Theoretically, the answer is 2 seconds, but in practice it would be a
> matter of splitting hares.
>
> (Not original, but I cannot remember where the credit has to go)
>
> Kira Yamato <no@mail.com> wrote in message
news:<QiTid.4789$Gm6.655@newsread3.news.atl.earthlink.net>...
> > This week I've seen solutions to some problems that just seem more like
> > magic tricks than like science. As the professor works out the
> > solutions on the board, he seems like a magician pulling rabbits out of
> > a hat. Here are the mathematical rabbits I've seen this week:
> >
> > Rabbit #1: (Analysis) Show that e is irrational.
> > Solution. He shows that for any positive integer p,
> > (1) 0 < p![e-(1+1/1!+1/2!+...+1/p!)] < 1.
> > Then if e=m/n for some integers m,n, there is a contradiction.
> >
> > Ok, showing (1) (by a geometric series comparison on its tail) and the
> > contradiction (let p=n and the middle term in (1) becomes an integer) is
> > easy. But how the heck does he ever thought of to consider (1) to begin
> > with???
> >
> > Rabbit #2: (Complex) Show that there is no analytic bijective map from
> > 1<=|z|<=4 to 1<=|z|<=2.
> > Solution. Suppose f is such function. Now define (the rabbit) g(z) =
> > (f(z)^2)/z. Then on both |z|=1 and |z|=4, |g(z)|=1. Hence by maximum
> > principle, g is constant, say c. Thus, f = sqrt(cz). But sqrt is not
> > analytic around the origin if c != 0. (c cannot be 0 for f to be
> > injective.)
> >
> > Rabbit #3: (Algebra) Show Aut Q_8 = S_4. That is, the automorphism
> > group of the quaternion group of order 8 is isomorphic to the symmetric
> > group of 4 elements.
> > Solution. This solution actually has two rabbits.
> > First rabbit: Let Q_8 = {+-1, +-i, +-j, +-k}. Now take cube and put
> > i,j,k on the faces so that i,j,k are on counterclockwise faces sharing a
> > vertex, and that (i, -i), (j, -j) and (k, -k) are on opposite faces for
> > each pair. Then this cube is the geometric representation of Q_8.
> > Thus, symmetry on the cube = symmetry on Q_8. Thus, Aut Q_8 = rotation
> > of the cube. Second rabbit: Now the rotation of the cube is just a
> > permutation of the 4 long diagonals of the cube. The professor shows
> > moreover that this association is actually an homomorphism. So,
> > rotation of the cube = S_4. Hence, Aut Q_8 = S_4.
> >
> > Obviously, I took the loooong approach to list out the conjugacy classes
> > and all the possible automorphisms. Then I look for the conjugacy
> > classes of the automorphism group to show how it must be S_4. That took
> > hours and pages while the professor's solution took only 5 minutes and
> > just a quarter of the board.
> >
> > Rabbit #4: (Combinatoric) Show that the # of combination of choosing k
> > items out of n items *with* replacement is the # of combination of
> > choosing k items out of n+k-1 items *without* replacement.
> > Solution. Label each of the n items by integers 1 to n. Suppose u've
> > chosen k items with replacement from n items. Now sort them in
> > ascending order according to their assigned integers. Add 1 to the
> > second item, 2 to the third, and so on. In this fashion, each set of k
> > non-unique items chosen out of n items corresponds to a set of k unique
> > items chosen out of n+k-1 items. Similarly, starting with choosing k
> > items without replacement from n+k-1 items, then ordering them in
> > ascending order, then subtracting 1 from the second item, 2 from the
> > third, and so on; thus each set of k unique items chosen out of n+k-1
> > items corresponds to a set of k non-unique items chosen out of n items.
> > Hence, the # of combination of choosing k items out of n items *with*
> > replacement is the # of combination of choosing k items out of n+k-1
> > items *without* replacement.
> >
> > Lesson learned this week: If you know your subject well enough, *all*
> > proofs should be at most half a page long. Perhaps Fermat did know his
> > subject well, and did have a short proof for the FLT.
> >
> > Kira
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