Algebra trick

This actually bugged me for 10 minutes or so before I figured it out. Anyone not falling into the same trap as me?

x^2-x^2 = x^2-x^2
x(x-x) = (x+x)(x-x)
x = (x+x)
x = 2x
1 = 2

Comments

  • Gotta remember what "canceling out" actually entails.

    As I recall, this catches a good number of people.
  • I dint take algebra.
    Please explain this as its bugging the shit out of me.
    I would read it as 1=1 but.... Its not?
  • Think it doesn't work because there's division by 0.
  • Gotta remember what "canceling out" actually entails.

    As I recall, this catches a good number of people.

    I love unintentional humour. tyvm
  • Think it doesn't work because there's division by 0.

    This is the reason.

    The original equation is 0=0, which then gets factored to 0x=0x^2. You could put any number on either side and the equation would remain true.
    ie,
    1000000*(0)=1*(0)

    But then in the last step you divide both sides by 0, which isn't allowed. 1000000 != 1.
  • The division by zero happens going from:
    this step x(x-x) = (x+x)(x-x)
    to this step x = (x+x)

    You can't just divide both sides by (x-x).....you would have to take limits.
  • yup, you guys had no problem with this. 0 = 0 isn't really a great start for a mathematical proof anyway ^^'
  • Richard~ wrote: »
    yup, you guys had no problem with this. 0 = 0 isn't really a great start for a mathematical proof anyway ^^'
    neither is 1 = 1....:confused:
  • darbday wrote: »
    neither is 1 = 1....:confused:

    that's the way most proofs start <.<
  • Richard~ wrote: »
    x^2-x^2 = x^2-x^2
    x(x-x) = (x+x)(x-x)
    x = (x+x)
    x = 2x
    1 = 2

    doesn't x(x-x) = x(0) = 0, not 'x'?
    doesn't '-x^2' = -(x+x), not *(x-x)?
    therefore 0=0??
  • Richard~ wrote: »
    that's the way most proofs start <.<
    i know its fucking insane huh? 1 = 1? Wheres the fucking proof for that ??? If we can prove that, then ill agree with everything else everyone says....
  • Pretty sure the answer is "2".
  • pokerJAH wrote: »
    doesn't x(x-x) = x(0) = 0, not 'x'?
    doesn't '-x^2' = -(x+x), not *(x-x)?
    therefore 0=0??

    yes, unless you divide by zero.
    no, that's not even close
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