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Math Help - Negations Project

  1. #1
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    Negations Project

    Hey all, I need to negate the following statements, can I get feedback on if this is correct?

    Every cubic polynomial has a real root.
    Negation: There exists a cubic polynomial that has no real root.

    G is normal and H is regular.
    Negation: G is not normal or H is not regular.

    There does not exist 0 such that for all x, x+0=x
    Negation: For no 0 does there exist x, where x+0 =/ x

    The newspaper article was neither accurate nor entertaining.
    Negation: The newspaper article was accurate and entertaining.

    If gcd(m,n) is odd, then m or n is odd.
    Negation: Gcd(m,n) is odd and m is not odd and n is not odd.

    H/N is a normal subgroup of G/N if and only if H is a normal subgroup of G.
    Negation: H/N is a normal subgroup of G/N and H is not a normal subgroup of G or H is a normal subgroup of G and H/N is not a normal subgroup of G/N.

    For each epsilon > 0 there exists N in the Natural Numbers such that for all n >= N, the abs|An - L| < epsilon

    Negation: There exists epsilon > 0 for each N in the Natural Numbers such that there exists n >= N, |An - L| >= epsilon


    Any help would be appreciated!
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  2. #2
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    There does not exist 0 such that for all x, x+0=x
    Negation: For no 0 does there exist x, where x+0 =/ x
    The original statement is equivalent to "It is not the case that there exists 0 such that for all x, x+0=x", so, to form negation, you just remove "It is not the case that".

    The newspaper article was neither accurate nor entertaining.
    Negation: The newspaper article was accurate and entertaining.
    Should be "accurate or entertaining".

    For each epsilon > 0 there exists N in the Natural Numbers such that for all n >= N, the abs|An - L| < epsilon

    Negation: There exists epsilon > 0 for each N in the Natural Numbers such that there exists n >= N, |An - L| >= epsilon
    You are making this more complicated than necessary. "There exists epsilon > 0 for each N" means "For each N there exists an epsilon > 0 (which may depend on N)", and it's a wrong way to start the negation. Rather, you keep the same order of variables: epslion, N, n, but change the quantifiers into the dual ones. "There exists an epsilon > 0 such that for each N in the Natural Numbers, there exists an n >= N such that |An - L| >= epsilon".

    The rest seems good to me.
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