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Math Help - 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: \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 F is F:=\ \forall x\forall y((x>0\wedge y>0)\Rightarrow \exists n(nx>y))

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

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


    \neg((\neg p\wedge q)\vee(\neg p\wedge\neg q))\vee(p\wedge q)
    \equiv (\neg(\neg p\wedge q)\wedge\neg(\neg p\wedge\neg q))\vee(p\wedge q)
    \equiv ((p\vee\neg q)\wedge(p\vee q))\vee(p\wedge q)
    \equiv ((p\wedge(\neg q\vee q))\vee(p\wedge q)
    \equiv p\vee(p\wedge q)
    \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:

    <br />
\forall x\forall y [(x>0\wedge y>0) \rightarrow \exists n(nx > y)]<br />

    where  \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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