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Discrete fallacious proof

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Alright here it is:

    Theorem: if there exists an x belonging to reals such that (x^2)-x-2=(x^2)-4 then 1=2.

    Remark: note that there is such an x belonging to reals.

    Proof:

    1) by hypothesis assume there exists an X belonging to reals such that (x^2)-x-2=(x^2)-4

    2)factor each side,

    3)resulting in (x-2)(x+1)=(x-2)(x+2)

    4)divide each side by (x-2),

    5)resulting in x+1=x+2

    6)subtract x from each side, resulting in 1=2

    1) What terminology (quantifiers, predicates) can be used to express the entire statements 1,3,5

    2)Why is this proof fallacious, refer to statements by their numbers. Hint: What are the domains for each statement?



    2. Relevant equations



    3. The attempt at a solution

    Ok, my attempt at part one (is
    Theorem: ([tex]\exists[/tex]X [tex]\in[/tex][tex]\Re[/tex]) [tex]\right arrow[/tex] ((x^2)-x-2=(x^2)-4))

    Statement 1: [tex]\exists[/tex]X [tex]\in[/tex][tex]\Re[/tex] ((x^2)-x-2=(x^2)-4))

    Statement 3:[tex]\exists[/tex]X [tex]\in[/tex][tex]\Re[/tex](x-2)(x+1)=(x-2)(x+2)

    Statement 5:[tex]\exists[/tex]X [tex]\in[/tex][tex]\Re[/tex]x+1=x+2

    For part 2, i am quite lost, the problem is the use of existential quantifier. My guess is a division by zero somewhere, but otherwise, I need a bigger hint.
    1. The problem statement, all variables and given/known data
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2008 #2

    HallsofIvy

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    The theorem starts "there exists an X belonging to reals such that (x^2)-x-2=(x^2)-4". Okay, is that true? If not then the conclusion is false because the hypothesis is false! If it is true, what is that x? In other words, solve x^2- x- 2= x^2- 4. I will tell you right now that the hypothesis is true but once you have determined what that x is, you will see why dividing both sides of the equation by x- 2 is an error.
     
  4. Sep 29, 2008 #3
    Aha! got it. Thank you so much for that. In retrospect I really should have noticed this x value.
     
  5. Sep 29, 2008 #4

    HallsofIvy

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    In retrospect, I should have been a genius!
     
  6. Sep 30, 2008 #5

    statdad

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    Actually, in logic, when there are several hypotheses, if any single hypothesis is false the implication (hypotheses imply conclusion) is always true, simply by the nature of the formal definition of a valid argument.

    the problem with this "proof" is simply that dividing by [tex] x-2 [/tex] is division by zero, since 2 is a solution to [tex] x^2 - x -2 = x^4 - 4 [/tex]
     
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