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Math Help - Need to check my work

  1. #1
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    Need to check my work

    Hi

    I want to negate the following statements. Please check if I am doing it correctly.

    1)There exists p > 0 such that for every x we have f(x+p)= f(x)

    2) For all \varepsilon > 0 there exists \delta > 0 such that
    whenever x and t are in D and satisfy \lvert x-t \rvert < \delta
    , then \lvert f(x) - f(t) \rvert < \varepsilon

    3)For all \varepsilon > 0 there exists \delta > 0 such that
    whenever x \in D and 0 <\lvert x-a \rvert < \delta ,
    then \lvert f(x)-A \rvert < \varepsilon

    Following are my negated statements

    1)For all p> 0 there exists x such that f(x+p)\neq  f(x)

    2)There exists \varepsilon > 0 such that for every
    \delta >0 there exists x and t in D such that \lvert x-t \rvert < \delta and \lvert f(x) - f(t) \rvert \geqslant  \varepsilon

    3)There exists \varepsilon > 0 such that for every
    \delta >0 there exists x\in D such that
    0 <\lvert x-a \rvert < \delta and \lvert f(x)-A \rvert \geqslant \varepsilon

    Please tell me if I am doing it right. I have followed the examples given in my book.
    The book is "A Friendly Introduction to Analysis: Single and Multivariable, second edition" Author- Witold Kosmala
    I am not satisfied with his treatment of the topic there. He discusses these things under the topic of "Proof Techniques", where he also talk talks about methods of proofs, like contradiction , contrapositive.

    If you can suggest a good source on internet ( preferably free) where they discuss
    negation of the mathematical statements, please suggest me.

    thanks

    newton
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  2. #2
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    Yes, you are correct.

    Negating such statements is pretty easy. They usually have the form ∀x (P(x) → Q(x)) or ∃x (P(x) ∧ Q(x)) where Q(x) either is an atomic formula (equality, inequality, etc.) or in turn has the same form. We have the following equivalences.

    (A → B) ⇔ A ∧ B
    (A ∧ B) ⇔ A → B
    ∀x A(x) ⇔ ∃x A(x)
    ∃x A(x) ⇔ ∀x A(x)

    Therefore,

    (∀x (P(x) → Q(x))) ⇔ ∃x (P(x) ∧ Q(x))
    (∃x (P(x) ∧ Q(x))) ⇔ ∀x (P(x) → Q(x))

    Thus, to negate a proposition that starts with a sequence of quantifiers, you switch the quantifiers without changing the restrictions on quantified variables (P(x) above, e.g., x ∈ D, δ > 0 or |x - t| < δ) and negate only the last atomic formula.
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  3. #3
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    Priviet Makarov

    Thanks. That makes sense.... I never had a logic course. Can you suggest some good book on logic for self study ?

    newton
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by issacnewton View Post
    Priviet Makarov

    Thanks. That makes sense.... I never had a logic course. Can you suggest some good book on logic for self study ?

    newton

    Introduction to Logic by Irving-M-Copi.
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  5. #5
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    thanks zarathustra
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