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~(p->q) of part of A Probabilistic Proof of Wallis's Formula for pi

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Write in symbols the negation of the theorem stated in part a.

    part a:We immediately see that if g(x) is a nonnegative continuous function whose integral is finite, then there exists an a>0 such that a*g(x) is a continuous probability distribution (take a=1/[tex]\int g(x)dx[/tex] from -[tex]\infty[/tex] to [tex]\infty[/tex]).

    The negation of P implies Q is equivalent to not P or Q.

    The issue is with the P part.

    Is ~P= g(x) isn't a nonnegative continuous fuction whose integral isn't finite?

    I am not sure if both is parts need to be negated or just one.

    Then for the Q part it is just itself starting with the existential quantifier.
     
  2. jcsd
  3. Feb 13, 2010 #2
    ~(P => Q) is equivalent to "P and not Q."

    Since you don't need the negation of P, does that solve your problem?
     
    Last edited: Feb 13, 2010
  4. Feb 13, 2010 #3
    Just for completeness sake:

    I suppose P is the statement "g(x) is a non-negative continuous function whose integral is finite." I would re-write this as "g(x) is a function which is non-negative and continuous and whose integral is finite" to emphasize the internal logical structure. Since ~(a and b and c) = ~a or ~b or ~c, the negation of P is:

    "g(x) is a function which is not continuous or assumes negative values or whose integral diverges."
     
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