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Thread: My last 2 what should be easy problems are crushing my soul

  1. #1
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    My last 2 what should be easy problems are crushing my soul

    Let x, y, z, be real numbers and let n be an integer. Negate this statement:

    For all of x, For all of y [(x > 0 and y > 0) --> there exists n(nx > y)]


    AND...


    Use the properties of logical equivalence to show that:
    ~((~p and Q) or (~p and ~q)) or (p and q) logically equivalent p




    maybe its because i have been up all night but these are killing me.
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  2. #2
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    Quote Originally Posted by thecleric View Post
    Let x, y, z, be real numbers and let n be an integer. Negate this statement:

    For all of x, For all of y [(x > 0 and y > 0) --> there exists n(nx > y)]
    Try to read this negation: $\displaystyle \left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {x > 0 \wedge y > 0} \right) \wedge \left( {\forall n} \right)\left[ {nx \leqslant y} \right]} \right]$
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  3. #3
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    Your formula $\displaystyle F$ is $\displaystyle F:=\ \forall x\forall y((x>0\wedge y>0)\Rightarrow \exists n(nx>y))$

    $\displaystyle \neg F\equiv \exists x\exists y ((x>0\wedge y>0)\wedge \forall n(nx\leq y))$

    which can be written $\displaystyle \exists x\exists y (x>0\wedge y>0\wedge \forall n(nx\leq y))$


    $\displaystyle \neg((\neg p\wedge q)\vee(\neg p\wedge\neg q))\vee(p\wedge q)$
    $\displaystyle \equiv (\neg(\neg p\wedge q)\wedge\neg(\neg p\wedge\neg q))\vee(p\wedge q)$
    $\displaystyle \equiv ((p\vee\neg q)\wedge(p\vee q))\vee(p\wedge q)$
    $\displaystyle \equiv ((p\wedge(\neg q\vee q))\vee(p\wedge q)$
    $\displaystyle \equiv p\vee(p\wedge q)$
    $\displaystyle \equiv p$
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  4. #4
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    ok i didn't know how to do the symbols but here it is:

    $\displaystyle
    \forall x\forall y [(x>0\wedge y>0) \rightarrow \exists n(nx > y)]
    $

    where $\displaystyle \rightarrow $ is the symbol for implies

    i have to find and simplify the negation of this statement
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  5. #5
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    Did you see my reply above?
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