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Math Help - Simple logic proof, need help interpreting

  1. #1
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    Simple logic proof, need help interpreting

    Let R be the statement "x^3 +32 = 5 for all real numbers x such that x^2 +32 = 0"
    Is R true? Why or why not?

    To me, this says that x must be real and must satisfy both equations. Since there is no real x that satisfies the second equation, the entire statement must be false. The only problem is, I have no idea if I'm interpreting the statement correctly.

    It also seems like one could read this as "the set of solutions for the second equation must be contained within the set of solutions for the first equation". And since the empty set is contained within all sets (including x=-3), then the statement would be true.

    First week of my proofs class and I'm already stumped. Any help?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Consider the well formed formulas ( with domain \mathbb{R} )

    \mathcal{A}(x):\; ( \forall{x)}(x^2+32=0),\quad \mathcal{B}(x):\; ( \forall{x)}(x^2+32=5)

    then,

    \mathcal{A}(x)\rightarrow \mathcal{B}(x)

    is satisfied, if and only if is satisfied:

    \vee \mathcal{B}(x)" alt="(\;\sim \mathcal{A}(x)\ \vee \mathcal{B}(x)" />


    Fernando Revilla
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  3. #3
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    There aren't any real numbers that satisfy \displaystyle x^2 + 32 = 0.

    Since you are going from something false, any statement you make from this is true.
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  4. #4
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    To summarize, what everyone else said, the statement is True because

    F\rightarrow T = T
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  5. #5
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    Quote Originally Posted by FernandoRevilla View Post
    Consider the well formed formulas ( with domain \mathbb{R} )

    \mathcal{A}(x):\; ( \forall{x)}(x^2+32=0),\quad \mathcal{B}(x):\; ( \forall{x)}(x^2+32=5)

    then,

    \mathcal{A}(x)\rightarrow \mathcal{B}(x)

    is satisfied, if and only if is satisfied:

    \vee \mathcal{B}(x)" alt="(\;\sim \mathcal{A}(x)\ \vee \mathcal{B}(x)" />
    To be precise, one has to remove quantifiers from \mathcal{A}(x) and \mathcal{B}(x). The question is not about the formula (\forall x.\,x^2+32=0)\to(\forall x.\,x^2+32=5), but about \forall x\,(x^2+32=0\to x^2+32=5).
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by emakarov View Post
    To be precise, one has to remove quantifiers from \mathcal{A}(x) and \mathcal{B}(x)
    Of course. Too quickly I read another statement form in the OP.


    Fernando Revilla
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  7. #7
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    Thank you to everyone who responded, it is very much appreciated. I've run into a number of other questions, but I will post them in a new thread. Thanks again.
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