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Posers and Puzzles

Posers and Puzzles

  1. 05 Feb '13 20:29
    I'm sure many of you have seen this proof, back in school, but I just recently had it shown to me and rather enjoyed it. Props to the person who can disprove the proof!

    If a = x , and a = 1 the x = 1. So...

    a + a = a + x
    simplified
    2a = a + x
    2a - 2x = a + x - 2x
    simplified
    2(a-x) = a + x - 2x
    2(a-x) = a-x
    2(a-x)/(a-x) = (a-x)/(a-x)
    2=1
  2. 05 Feb '13 20:40
    Originally posted by King Tiger
    I'm sure many of you have seen this proof, back in school, but I just recently had it shown to me and rather enjoyed it. Props to the person who can disprove the proof!

    If a = x , and a = 1 the x = 1. So...

    a + a = a + x
    simplified
    2a = a + x
    2a - 2x = a + x - 2x
    simplified
    2(a-x) = a + x - 2x
    2(a-x) = a-x
    2(a-x)/(a-x) = (a-x)/(a-x)
    [b]2=1
    [/b]
    Oh no, you broke the universe.

    I remember having this shown to me when I was in primary school. I was stumped. Neat little trick if you're not very good with division.
  3. Subscriber joe shmo
    Strange Egg
    06 Feb '13 00:40
    Originally posted by AnomalousCowturd
    Oh no, you broke the universe.

    I remember having this shown to me when I was in primary school. I was stumped. Neat little trick if you're not very good with division.
    Ah yes,... division by zero! I must shamefully say I missed that little tid bit of information.
  4. 07 Feb '13 08:15
    Yes, the last step, divides by zero-undefined. Or really, the entire expression as last shown is an indeterminate form. Can't be done.