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What is the negation of the statement

  1. Mar 16, 2004 #1
    What is the negation of the statement "For each s in R, there exists an r in R such that if f(r) >0, then g(s) >0."
     
  2. jcsd
  3. Mar 16, 2004 #2
    I'm no expert, but here's a try:

    [tex]\neg [\forall s \in R[\exists r \in R(f(r) > 0 \supset g(s > 0)]][/tex]
     
  4. Mar 16, 2004 #3

    matt grime

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    there is an s, such that for all r, f(r)>0 and g(s) is negative (or zero).
     
    Last edited: Mar 16, 2004
  5. Mar 16, 2004 #4

    matt grime

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    remember (a)implies(b) is the same as (not a)or(b), so the negation is (a)and(not b)
     
  6. Mar 16, 2004 #5
    matt grime's answer is correct.
    thanks :)

    DeadWolfe, all you did was restate the question.
     
  7. Mar 16, 2004 #6

    matt grime

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    I didn't realize this was a test... Well, at least my students will be happy I can get their level of questions right.
     
  8. Mar 22, 2004 #7
    Original Statement:

    [tex] [\forall s \in R[\exists r \in R(f(r) > 0 \supset g(s) > 0)]][/tex]

    Prefix Of Original statement by the logical operator NOT:
    [tex]\neg [\forall s \in R[\exists r \in R(f(r) > 0 \supset g(s) > 0)]][/tex]

    DISCUSSION
    As you can see, all quantifiers used in the original statement refer to the same domain of discourse, namely the set of real numbers. In the orginal formulation of the statement, the domain of discourse is made explicit (we can see the script R symbol denoting the set of real numbers). But, when all the quantifiers used in a statement refer to the same domain of discourse, there is a shorter way to formulate the statement, but it must be made clear at the outset, that all the quantifiers used in the statement refer to the set of real numbers. The only time we must specify what set a quantifier refers to, is when we are talking about multiple domains of discourse. Since there is only one domain of discourse here, the orginal statement is more succinctly formulate as follows:

    Let the domain of discourse be the set of real numbers.

    Original statement(domain of discourse implicit, instead of explicit)

    [tex]\neg [\forall s [\exists r \( f(r) > 0 \supset g(s) > 0]][/tex]

    Now, we can pass the negation through the quantifiers (reversing them), and it operates on the conditional:

    [tex]\exists s \forall r \[ \neg [f(r) > 0 \supset g(s) > 0 ][/tex]

    if X then Y = not (X and not Y), hence
    not (if X then Y) = not (not(X and not Y)) = X and not Y

    So, we can formulate the negation of the original statement as follows:

    [tex]\exists s \forall r \ [ f(r) > 0 and \neg g(s) > 0 ][/tex]

    All the formulations are semantically equivalent, in other words, they all mean the same thing.

    Translation: Let the domain of discourse be the set of real numbers. There is at least one s such that for any r, f(r) is greater than zero and not ( g(s)>0 ).

    (by saying the domain of discourse is the set of real numbers, it is known by he who formulated the statement, that s,r are elements of the set of real numbers.

    First order logic examples:

    There is at least one dog, who is older than any cat.

    In the previous example, there are TWO domains of discourse, instead of one.

    Let D denote the set of dogs, and let C denote the set of cats.

    We can formulate the statement as follows:


    [tex] \exists d \in D\forall c \in C [d-is older than-c] [/tex]

    Let us suppose that the previous statement is true.
    Now, consider the meaning of the previous statement. There is at least one dog, who is older than any cat. So, suppose that wonderdog is such a dog, then it follows that:

    [tex] \forall c \in C [Wonderdog-is older than-c] [/tex]

    Suppose that Hairball is an element of the set of cats. If the previous statement is true, then the following statement is true:

    Wonderdog is older than Hairball.

    My point is this: That when we are talking about multiple domains of discourse, we need to write out which set a quantifier refers to explicitely, but when all the quantifiers refer to the same single set, we can use the implicit notation, which is shorter.
     
    Last edited: Mar 22, 2004
  9. Mar 22, 2004 #8

    ANSWER1:
    There is at least one s in R, such that for any r in R: f(r)>0 and not ( g(s)>0 ).

    And using trichotomy we also have:

    ANSWER2:
    There is at least one s in R, such that for any r in R: f(r)>0 and ( g(s)<0 OR g(s)= 0 ).
     
    Last edited: Mar 22, 2004
  10. Sep 15, 2007 #9
    [tex] [\neg s][/tex]
     
  11. Sep 2, 2011 #10
    How can we negate using similar steps? $\forall a,b \in R$ with $a<b$ there is an $r \in Q$ with $a<r<b$


    I do know the negation, but I need explain to smb who is learning this. Thanks
     
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