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Math Help - Writing down propositions symbolically & state truth values.

  1. #1
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    Writing down propositions symbolically & state truth values.

    Hey guys, I need help with a study question. "Write down the following propositions symbolically and state their values. Then write down the negation of each propositions."

    a) There is a real number x such that x^2 - 3x + 2 = 0.

    My Answer: \exists x \epsilon  \mathbb{R} (x^2 - 3x + 2 = 0) and this is TRUE for when x=1. The negation to this is \forall x \epsilon  \mathbb{R}  (x^2 - 3x + 2 \neq 0)

    b) For every real number x there is a real number y such that x=y^2.

    My Answer: \forall x \epsilon  \mathbb{R} \exists y \epsilon  \mathbb{R} (x=y^2) and this is FALSE because not a single y real value for y^2 can equal all real values of x. The negation to this is \exists x \epsilon  \mathbb{R} \forall y \epsilon   \mathbb{R} (x \neq y^2)

    Are my answers OK? thanks
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by jvignacio View Post
    Hey guys, I need help with a study question. "Write down the following propositions symbolically and state their values. Then write down the negation of each propositions."

    a) There is a real number x such that x^2 - 3x + 2 = 0.

    My Answer: \exists x \epsilon \mathbb{R} (x^2 - 3x + 2 = 0) and this is TRUE for when x=1. The negation to this is \forall x \epsilon \mathbb{R}  (x^2 - 3x + 2 \neq 0)
    Right, on both counts.

    b) For every real number x there is a real number y such that x=y^2.

    My Answer: \forall x \epsilon \mathbb{R} \exists y \epsilon \mathbb{R} (x=y^2) and this is FALSE
    Right.
    because not a single y real value for y^2 can equal all real values of x.
    Wrong! The point is not that there must exist a single y that does it for all x. The idea is that if the statement were true, then one could provide an arbitrary, but fixed, x\in\mathbb{R}, and you, the defender of that statement, would have to come up with a y\in\mathbb{R} such that x=y^2 holds. If someone comes up with a different x, you are allowed to choose a different y. This is because the existential quantifier for y occurs in the scope of the quantifier of x. So at the point where the y must be show to exist, the value of x has already been decided upon and cannot be changed anymore, so to speak.

    This is impossible only if x<0, otherwise it is quite easily possible to come up with such a y

    The negation to this is \exists x \epsilon \mathbb{R} \forall y \epsilon \mathbb{R} (x \neq y^2)
    Correct, and this statement is, of course, true.
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  3. #3
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    Quote Originally Posted by Failure View Post
    Right, on both counts.


    Right.

    Wrong! The point is not that there must exist a single y that does it for all x. The idea is that if the statement were true, then one could provide an arbitrary, but fixed, x\in\mathbb{R}, and you, the defender of that statement, would have to come up with a y\in\mathbb{R} such that x=y^2 holds. If someone comes up with a different x, you are allowed to choose a different y. This is because the existential quantifier for y occurs in the scope of the quantifier of x. So at the point where the y must be show to exist, the value of x has already been decided upon and cannot be changed anymore, so to speak.

    This is impossible only if x<0, otherwise it is quite easily possible to come up with such a y


    Correct, and this statement is, of course, true.
    thank you so much! Understood
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