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Negation for proposition

  1. Jan 17, 2016 #1

    my attempt.

    Let P = At least one a and at least one b
    Let Q = r=a/b

    Hence the proposition is simplified to,

    For all r where P Then Q

    Not all
    r where P Then Q
    = Atleast one R When Not(P Then Q)

    Not(P Then Q) = P And Not Q

    Atleast one R When Not(P Then Q)
    = Atleast one R When P And Not Q

    After this step i substitute back P and Q.
    Is this way correct?
  2. jcsd
  3. Jan 17, 2016 #2


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    There is a general method for negating that goes from the outermost (left-most) statement : For all is negated into there is
    There is a general method for negating compound statements : the negation is done from the leftmost to rightmost statement: first you negate the "for all" into there exists
    and then you negate the A and B statement into NotA or NotB , where A is There exists an integer Z , negated into " For all a in Z "and so on. Informally, the negation says that there exists a Rational that is not the ratio of two integers.
  4. Jan 17, 2016 #3
    Hi so the ans is.

    There exists a rational number r, Where all integer a and all integer b such that r != a/b

    May i know what is "such that' converted to? For my case i tot it means "then"
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