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X^3 -x -x^3=0 puzzle

  1. Feb 10, 2005 #1
    This a puzzle for me.

    x^3 -x -x^3=0 Obviously x=0
    but, divide both sides by x

    x^2 -1 -x^2=0 except for when x=0 'cause undefined
    but that means

    is this apparent conflict not in fact one because you can't rule out x=0 ?
  2. jcsd
  3. Feb 10, 2005 #2
  4. Feb 10, 2005 #3
    [tex]x^3 - x - x^3 = 0[/tex] then divide by x:

    [tex]\frac{x^3 - x - x^3}{x} = \frac{0}{x}[/tex]

    [tex]x^2 - x^2 = 1[/tex]

    Even saying that [tex]\frac{0}{x}[/tex] is 0 you should see that it is not possible to get two different numbers to make 1 when the two x values will give the same to take away.

    The Bob (2004 ©)
  5. Feb 10, 2005 #4
    I thought that -1=0 is not a statement of equality, but more of a qualifier for the original equation. We want to know when say, x^2 - x^2 == 1, and that will only happen when 0 = 1, which is never. So that has no solution. I thought this meaning was always implied in mathematics. I mean I can say
    1 + 1 =5, but how can that be?! I mean I just wrote it, but its not supposed to be possible! Just becasue yo ucan write down an equation doesnt mean its true.
  6. Feb 11, 2005 #5
    You are simply breaking the rule of "not dividing by zero".
    You could state your "problem" simpler:
    x = 0
    divide both sides by x
    x/x = 0/x <-> 1 = 0
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