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Proving 2=1

  1. Dec 14, 2004 #1

    T@P

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    Hey can everyone please post any/all ways they know of "proving" 2 =1? obviously they are all wrong, but some are more "convinving" than others.

    For example, heres an old one to start:

    e^(2*pi *i) = 1, so ln (e^(2*pi *i)) = ln (1) or 2 * pi * i = 0

    clearly pi != 0 and i != 0 so 2 = 0. Although this dosent *prove* 2 = 1, from 2= 0 you can really show that almost anything is true.

    any other ideas?
     
  2. jcsd
  3. Dec 14, 2004 #2
    I haven't seen that version. The only "proof" I've seen is this one.

    If you can't spot what's wrong with it, it's the dividing by a^2 - ab as you're dividing by a(a - b), which is 0 as a = b.
     
    Last edited: Dec 14, 2004
  4. Dec 14, 2004 #3

    Zurtex

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    But of course in actuallity:

    e^(2*pi*k*i) = 1

    And when you take the log of both sides it's the k that is equal to 0.

    It's no diffrent from saying sin(2*pi) = 0, so 2=0
     
  5. Dec 14, 2004 #4

    jcsd

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    ln z is a multivalued function in C

    ln z = |ln z| + iarg(z)

    therefore ln e2πi = 2πi + n2πi for all n in Z.
     
  6. Dec 16, 2004 #5

    T@P

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    anyother ways? there are more such as:

    1/(-1) = (-1) /1
    taking the square root of both sides yeilds:
    1/i = i/1 or -1 = 1, 2 = 0
     
  7. Dec 16, 2004 #6
    You seem to enjoy these kind of "proofs" T@P. I think most of the ones here were already posted here , with a bit of changing around. I love the last one in that thread, it actually stumped my Calculus I teacher, but not my TA though, not even for a second (he's working on his doctorate, my instructor has been teaching for about 25 years). These problems are fun for me now, I finally understand enough math to get what they are saying! :rofl:
     
  8. Dec 17, 2004 #7
    Question: not a Answer....consider the honey bee cell, regular hexagon, with side length is 'R'. we can find the distance between the adjacent cells, regular hexagons, centre's are sqroot3*R. how we can find the distance between the centre's of a cell's which are not adjacent
     
  9. Dec 17, 2004 #8
    Give any group of order infinite, every element of that group is of finite order?
     
  10. Dec 17, 2004 #9

    NateTG

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    For any particular infinite cardinality [itex]C[/itex], take the subgroup of [itex]S_C[/itex] which only contains permutations that permute a finite number of elements.
     
  11. Dec 21, 2004 #10
    (1) X = Y Given
    (2) X^2 = XY Multiply both sides by X
    (3) X^2 - Y^2 = XY - Y^2 Subtract Y^2 from both sides
    (4) (X+Y)(X-Y) = Y(X-Y) Factor both sides
    (5) (X+Y) = Y Cancel out common factors
    (6) Y+Y = Y Substitute in from line (1)
    (7) 2Y = Y Collect the Y's
    (8) 2 = 1 Divide both sides by Y
     
  12. Dec 23, 2004 #11

    T@P

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    thats fancily dividing by 0, since x = y you cannot divide by (x-y)

    i was looking more for proofs that rely on little known rules, but thanks for your input anyway :)
     
  13. Dec 23, 2004 #12
    Here's one:

    (-x)^2 = x^2
    log[(-x)^2] = log(x^2)
    2 log(-x) = 2 log(x)
    log (-x) = log (x)
    -x = x
    -1 = 1
     
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