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A horse has infinite number of legs? huh? help!

  1. Nov 27, 2004 #1
    I don't get this at all :confused: If you were going to respond to this, what would be a logical answer? I don't even know what the question is asking me to do!
    Thanks,
    Jennifer :blushing:
    ----------------------------------------------

    In this paper we develop machinery sufficient to prove the following

    THEOREM: All horses have an infinite number of legs.

    The theorem may seem intuitively obvious to some, but in the interest of rigor we will give a complete proof. We begin with two Lemmas:

    Lemma 1: All horses are the same color.
    Proof: We use the Principle of Mathematical Induction on the number, n, of horses.
    Clearly, one horse is the same color, so the Lemma is true for n=1.
    Now assume k horses are the same color, and consider k+1 horses. If we remove any one horse, we are left with k horses, which, by hypothesis, are all the same color. Since we removed an arbitrary horse, all k+1 horses are the same color.

    Lemma 2: If a number is both even and odd, then it is infinite.
    Proof: Let n be a number which is both even and odd, and assume n is finite.
    As an even number, n = 2a for some integer a, and as an odd number, n = 2a+1. Thus 2a = 2a+1, whence 0 = 1. This contradiction establishes Lemma 2.

    Proof of Theorem: All horses have forelegs in front and two in back, so that all horses have six legs. Now six is an even number, but six is clearly an odd number of legs for a horse to have. Thus the number of legs on a horse is both even and odd, and so by Lemma 2 it must be infinite.
    You say, "But my horse has four legs." That, however, is a horse of a different color, which by Lemma 1 does exist.
     
  2. jcsd
  3. Nov 27, 2004 #2
    What question?
     
  4. Nov 27, 2004 #3
    That's what I said! I guess the point is to disprove the Theorem... but how?
     
  5. Nov 27, 2004 #4
    Just point out some of the logical fallacies it employs.
     
  6. Nov 27, 2004 #5

    Gokul43201

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    There's a mathematical fallacy in lemma 1 (second inductive step) and the proof of the theorem is really nothing more than a play on words, utilizing the homonyms fore (four) and odd (meaning 'strange' as well as 'not even').
     
  7. Nov 27, 2004 #6

    shmoe

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    lemma 2 isn't without it's problems either.
     
  8. Nov 28, 2004 #7

    Gokul43201

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    Ooops ! Didn't really see that "2a = 2a +1" , did I ? I guess I skipped past it because it was a reasonable statement that no finite integer is both even and odd.

    This reminds me of another riddle...try this :

    Three friends go to a restaurant and order 10 cups of tea. Each of them drinks an odd number of cups, but as many cups as have been ordered have been drunk. How many cups did each drink ?

    <cue : you're supposed to say "it's not possible" or "I give up...how many ?">
     
    Last edited: Nov 28, 2004
  9. Nov 28, 2004 #8

    shmoe

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    The conclusion that your "number" was infinite is a little troublesome too.

    For the tea drinkers-one of them drinks ten cups, which is certainly an odd number off cups of tea to drink. Or something along those lines? Similar wordplay can prove that 2 is the oddest prime.
     
  10. Nov 28, 2004 #9

    Gokul43201

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    True.

    Yup...exactly along those lines.
    Haven't heard this one. But it would sound nicer to conclude that 2 is an odd number...or is that what you meant ?
     
  11. Nov 28, 2004 #10

    shmoe

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    Duh to me, that would have been more relevant to the discussion:). I do prefer the phrase "2 is the oddest prime number" though, since it adds the absurdity that one number can be "more odd" than another.
     
  12. Nov 28, 2004 #11

    matt grime

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    2 is an odd prime because -1 =1 mod 2 and this creates all kinds of special cases of things in mathematical literature. For instance in any characteristic other than two a norm is the same as an inner product (polarization identity). A symplectic form of a field of odd characteristic can only be non-degenerate only if the underlying vector space is even dimensional, this isn't true in char 2. Lie algebras in char 2 are odd. (actually they are odd in char 3, too for other important numerical reasons.) You don't want to do elliptic curves in deg 2 (or 3 as well, I seem to recall, though I'm not sure). If we want to get fancy, and who doesn't, then the homology of elementary p-groups of the same rank is the same for all p except p=2.
     
  13. Nov 28, 2004 #12
    The "theorem" was worth a good laugh, though. Here's a proof that anything exists:

    1. X exists.
    2. Both these statements are false.

    where X can be anything from unicorns to pink elephants.
     
    Last edited by a moderator: Nov 28, 2004
  14. Dec 10, 2004 #13
    The fallacies in the proof, in order, are:
    --fallacy of composition
    --false dichotomy
    --equivocation (twice)
    --amphiboly
     
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