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Are all numbers the same?

  1. Mar 12, 2012 #1
    That's false!, since this logic a + b = t
    (a + b)(a - b) = t(a - b)
    a^2 - b^2 = ta - tb
    a^2 - ta = b^2 - tb
    a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
    (a - t/2)^2 = (b - t/2)^2
    a - t/2 = b - t/2
    a = b

    gives false results there must be an error in it because we know that two things will never equal 3 things thats impossible. 2 is not 3, you know this. A person who did this proof should have doubted his logic because it produce the obviously false results.


    What do you say?
     
  2. jcsd
  3. Mar 12, 2012 #2
    The mistake is near the end:

    (a - t/2)^2 = (b - t/2)^2
    is correct, but this does not necessarily mean that
    a - t/2 = b - t/2

    Since a square root can be positive or negative, (a - t/2)^2 = (b - t/2)^2 implies that:
    either
    (1) a - t/2 = b - t/2
    or
    (2) a - t/2 = - (b - t/2)

    From (2):
    a - t/2 = t/2 - b
    a + b = t/2 + t/2
    a + b = t
    We're back where we started!
     
  4. Mar 12, 2012 #3

    mathman

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    Gold Member

    Since t/2 = (a+b)/2, |a - t/2| = |b - t/2|. Assume a < b, then a - t/2 < 0 and b - t/2 > 0.
     
  5. Mar 15, 2012 #4
    If a=b, multiplying (a-b) both sides doesn't make any sense because a-b=0. Because that way any equation could be proven true. Just multiply 0 both sides and say 0=0.
     
  6. Mar 15, 2012 #5
    No... If you say that x=Sqrt(b)+c-d, then yes this equation implies that 0*x=0*(Sqrt(b)+c-d) which means that 0=0. The opposite, going from 0=0 to x=Sqrt(b)+c-d involves division by zero, which doesn't make sense.

    I think Michael C hit the nail on the head.
     
  7. Mar 15, 2012 #6

    Well we both basically mean the same thing.
     
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