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Math Help - Am I doing this right? (negation of this proposition)

  1. #1
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    Am I doing this right? (negation of this proposition)

    I need to do the negation of y∀x(x y +1) using generalized de morgan's law to simplify the expression. This is what I have;

    (∃y∀x(x ≥ y +1)) ≡ ∀y(∀x(x≥y +1)) ≡ ∀y∃x(x ≥ y +1) ≡ ∀y∃x(x < y +1)

    I'm pretty sure I've done it all right except I'm not sure about the last statement in particular the < sign and the +1.




    Okay that's my main problem that I would love to get some help on but if someone could also tell me if the following is correct.

    For y∀x(x y +1) I need to prove whether the statement is true or not if the domain of discourse is Z+ which I'm pretty sure I have the right answer for (this isn't what I want clarify) but the question is I need to state a pair of domains (one for x, one for y) for which the proposition is true. Would I be correct with the following?~

    5 for x, 4 for y as 5 ≥ 4 + 1 ≡ 5 ≥ 5 is true.

    Thanks for the help!
    Last edited by Herblore; April 5th 2013 at 02:02 AM.
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  2. #2
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    Re: Am I doing this right? (negation of this proposition)

    Quote Originally Posted by Herblore View Post
    (∃y∀x(x ≥ y +1)) ≡ ∀y(∀x(x≥y +1)) ≡ ∀y∃x(x ≥ y +1) ≡ ∀y∃x(x < y +1)
    This is correct.

    Quote Originally Posted by Herblore View Post
    the question is I need to state a pair of domains (one for x, one for y) for which the proposition is true. Would I be correct with the following?~

    5 for x, 4 for y as 5 ≥ 4 + 1 ≡ 5 ≥ 5 is true.
    A domain is a set of values, not an individual value. You can consider domains that are singletons, though. An alternative would be {0, 1, ...} for y and {1, 2, ...} for x.
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  3. #3
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    Re: Am I doing this right? (negation of this proposition)

    Really appreciate the help, I'm quite surprised that ≡ ∀y∃x(x < y +1) is right I thought it might be something like ≡ ∀y∃x(x < y -1).

    Anyway for the second part the question is worded exactly "(c) State a pair of domains (one for x, one for y) for which the proposition is true." and I have to use the domain of discourse Z+ so I can't use 0 but you're telling me that domains is in fact a pair of sets I need to state {1, 2 ...} for y and {2, 3 ...} for x. Is using "..." correct for demonstrating the pattern +1 every time?

    Thanks again.

    Edit: P.S. I'm a complete math noob, I didn't even do math from year 11 in high school and this is the only math course I need to do in my degree so I apologize if what I'm asking is obvious.
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  4. #4
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    Re: Am I doing this right? (negation of this proposition)

    Quote Originally Posted by Herblore View Post
    Really appreciate the help, I'm quite surprised that ≡ ∀y∃x(x < y +1) is right I thought it might be something like ≡ ∀y∃x(x < y -1).
    Yes, the negation of a\ge b is a<b.

    Quote Originally Posted by Herblore View Post
    Anyway for the second part the question is worded exactly "(c) State a pair of domains (one for x, one for y) for which the proposition is true." and I have to use the domain of discourse Z+ so I can't use 0 but you're telling me that domains is in fact a pair of sets I need to state {1, 2 ...} for y and {2, 3 ...} for x. Is using "..." correct for demonstrating the pattern +1 every time?
    Usually yes, though it depends on the instructor. To be more precise, \{1,2,\dots\} =\mathbb{Z}^+ and \{2,3,\dots\}= \{n\in\mathbb{Z}^+\mid n>1\}.
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