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Thread: negation of statement

  1. #1
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    negation of statement

    Here is the statement about set of real numbers:
    $\displaystyle (\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$

    I need to find the negation of it.

    This is what I have so far. Am I correct?

    $\displaystyle (\exists x \in R) (x = 5 \wedge (\forall y \exists R) (y > x))$
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  2. #2
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    Quote Originally Posted by relyt View Post
    Here is the statement about set of real numbers:
    $\displaystyle (\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$

    I need to find the negation of it.

    This is what I have so far. Am I correct?

    $\displaystyle (\exists x \in R) (x = 5 \wedge (\forall y \exists R) (y > x))$
    i think the $\displaystyle x \ne 5$ should stay

    the negation of $\displaystyle P \implies Q$ is $\displaystyle P \wedge \sim Q$

    so the antecedent does not change
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    $\displaystyle (\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$

    So, would you say this like:
    "For the set of all real numbers x, if x does not equal 5 then their exists (at least) one y in the set of real numbers such that y is less than x."

    A is "For the set of all real numbers x, x not equal to 5."
    B is "There exists one y in the set of real numbers such that y is less than x"

    A $\displaystyle \Rightarrow$ B $\displaystyle \equiv$ B $\displaystyle \vee \neg$ A.
    The negation is A $\displaystyle \wedge \neg$ B

    $\displaystyle \neg $ B $\displaystyle \equiv (\forall y \in R) (y > x)$
    so,

    $\displaystyle \equiv (\forall x \in R) (x \neq 5) \wedge (\forall y \in R) (y > x)$

    And you would say, "For the set of all real numbers x, x not equal to 5 and the set of all real numbers y, y is greater than x."

    So, can you simplify this to, "for the set of all real numbers, y is greater than x, with the exception of x equals 5, where x is not defined"?
    Last edited by bmp05; Mar 29th 2009 at 04:25 AM.
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  4. #4
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    Quote Originally Posted by relyt View Post
    Here is the statement about set of real numbers:
    $\displaystyle (\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$
    I need to find the negation of it. This is what I have so far. Am I correct?
    $\displaystyle (\exists x \in R) (x = 5 \wedge (\forall y {\color{red}\exists R}) (y {\color{red}>} x))$
    Almost, two corrections.
    $\displaystyle (\exists x \in R) (x \not= 5 \wedge (\forall y {\color{blue}\in R}) (y {\color{blue}\ge} x))$
    Last edited by Plato; Mar 29th 2009 at 04:36 AM.
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  5. #5
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    $\displaystyle A \Rightarrow B \equiv B \vee \neg A$

    Are thes two statements equivalent?
    $\displaystyle (\forall x \in \mathbb{R})(x \Rightarrow (\exists y \in \mathbb{R})(y)) \equiv (\exists y \in \mathbb{R})(y \vee (\exists x \in \mathbb{R})(x))$
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  6. #6
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    Quote Originally Posted by bmp05 View Post
    $\displaystyle A \Rightarrow B \equiv B \vee \neg A$
    Are thes two statements equivalent?
    $\displaystyle (\forall x \in \mathbb{R})(x \Rightarrow (\exists y \in \mathbb{R})(y)) \equiv (\exists y \in \mathbb{R})(y \vee (\exists x \in \mathbb{R})(x))$
    I have no idea what those statements could possibly mean.
    But I do know that they are not equivalent.
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  7. #7
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    Quote Originally Posted by Plato View Post
    I have no idea what those statements could possibly mean.
    But I do know that they are not equivalent.

    Yes, I just found this link, which might be helpful: An Elementary Introduction to Logic and Set Theory: Methods of Proof

    And there's a section Rules of Replacement for Quantified Predicates with an example, which shows that what I wrote in my last post is definately wrong, but now, I wonder what the negation of (x) is?!
    $\displaystyle \exists x \in \mathbb{R}: \neg (x)$ does that even make sense?
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  8. #8
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    Quote Originally Posted by bmp05 View Post

    I wonder what the negation of (x) is?!
    $\displaystyle \exists x \in \mathbb{R}: \neg (x)$ does that even make sense?
    $\displaystyle \neg (x)$ is just not proper notation. What does not x mean?
    Usually we have a propositional function.
    Example: Let P(x) mean that x is a rational number.
    Now $\displaystyle \neg P(x)$ reads “x is not a rational number.”
    $\displaystyle \neg \left( {\forall x} \right)\left[ {P(x)} \right]$ reads: “It is false that every x is rational.
    $\displaystyle \left( {\exists x} \right)\left[\neg {P(x)} \right]$ reads: “Some x is not rational.
    Clearly those two statements are equivalent.
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  9. #9
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    Yes, this is really quite interesting, so you can negate a predicate, but is there so much different between P(x):= (x$\displaystyle \neq$5) and P(y):=(x)..?

    So, the nagation of P $\displaystyle \forall x: x \neq 5 \equiv \exists x: x=5 \equiv x=5$? But then the negation of Q $\displaystyle \forall x: x \equiv \neg $Q $\displaystyle \equiv $ Because, you can negate all x, that would be for any x, False.



    Quote Originally Posted by Plato View Post
    $\displaystyle \neg (x)$ is just not proper notation. What does not x mean?
    Usually we have a propositional function.
    Example: Let P(x) mean that x is a rational number.
    Now $\displaystyle \neg P(x)$ reads “x is not a rational number.”
    $\displaystyle \neg \left( {\forall x} \right)\left[ {P(x)} \right]$ reads: “It is false that every x is rational.
    $\displaystyle \left( {\exists x} \right)\left[\neg {P(x)} \right]$ reads: “Some x is not rational.
    Clearly those two statements are equivalent.
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