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Thread: 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 $\displaystyle x$ such that $\displaystyle x^2 - 3x + 2 = 0$.

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

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

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

    Are my answers OK? thanks
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  2. #2
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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 $\displaystyle x$ such that $\displaystyle x^2 - 3x + 2 = 0$.

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

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

    My Answer: $\displaystyle \forall x \epsilon \mathbb{R}$ $\displaystyle \exists y \epsilon \mathbb{R}$ $\displaystyle (x=y^2)$ and this is FALSE
    Right.
    because not a single $\displaystyle y$ real value for $\displaystyle y^2$ can equal all real values of $\displaystyle x$.
    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, $\displaystyle x\in\mathbb{R}$, and you, the defender of that statement, would have to come up with a $\displaystyle y\in\mathbb{R}$ such that $\displaystyle x=y^2$ 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 $\displaystyle x<0$, otherwise it is quite easily possible to come up with such a $\displaystyle y$

    The negation to this is $\displaystyle \exists x \epsilon \mathbb{R}$ $\displaystyle \forall y \epsilon \mathbb{R}$ $\displaystyle (x \neq y^2)$
    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, $\displaystyle x\in\mathbb{R}$, and you, the defender of that statement, would have to come up with a $\displaystyle y\in\mathbb{R}$ such that $\displaystyle x=y^2$ 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 $\displaystyle x<0$, otherwise it is quite easily possible to come up with such a $\displaystyle y$


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