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Math Help - logic problems

  1. #1
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    logic problems

    trying to figure out the following statements' true or false, and why?
    figured out the first one then stuck on the rest. thanks

    1.∀x ∈R, ∃y ∈ R such that x.y=1. false x=0, 0.y≠0
    2∀x ∈Z+ and ∃y ∈ Z+ ,∃z ∈ Z+ such that z=x-y
    3. ∀x ∈Z and ∃y ∈ Z ,∃z ∈ Z such that z=x-y
    4. ∃u ∈R+ such that ∀v ∈ R+, u.v<v
    5. ∀x ∈R+ , ∃u ∈ R+ such that u.v<v
    Last edited by demarkus; August 2nd 2011 at 01:06 PM. Reason: spelling
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  2. #2
    Super Member TheChaz's Avatar
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    Re: logic problems

    Quote Originally Posted by demarkus View Post
    trying to figure out the following statements' true of false, and why?
    figured out the first one then stuck on the rest. thanks

    1.∀x ∈R, ∃y ∈ R such that x.y=1. false x=0, 0.y≠0
    2∀x ∈Z+ and ∃y ∈ Z+ ,∃z ∈ Z+ such that z=x-y
    3. ∀x ∈Z and ∃y ∈ Z ,∃z ∈ Z such that z=x-y
    4. ∃u ∈R+ such that ∀v ∈ R+, u.v<v
    5. ∀x ∈R+ , ∃u ∈ R+ such that u.v<v
    4. Write u.v < v as
    u.v - v < 0
    v.(u - 1) < 0
    Is this product always negative?
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  3. #3
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    Re: logic problems

    I suggest offering some of your own thoughts on 2-5. I'll offer mine below the fold.

    Spoiler:

    What does your notation "x.y" mean? Is that multiplication? If so, you are correct. Every real number besides zero has a unit: i.e., another number for which its product equals 1.

    The second says that every positive x can be written as the sum of two other positive numbers. If 0 is not included in Z+, then this is false because what would 1 equal? At best, 1 + 0, but that is excluded. If it is included, we can write 0 + 0 = 0, (0 = 0 - 0, correctly), 0 + 1 = 1 (1 = 1 - 0, correctly), and the rest are obviously true.

    I assume 2 does not include 0, so 3 would. Per the above discussion, clearly 3 is true.

    4 is saying there is a positive real u which multiplied by v is less than v. Well, first thing that comes to mind is a fraction. If you're always normalizing v by some fraction of v, then you're going to have a number less than v. This is exactly what 4 states. The reason we want positive real numbers is because with negative multiplication we can end up on one side of the real number line or the other, which can pose problems, clearly.

    You need to restate 5. I assume you mean "for all v, there is a u ...", not "for all x," right? The answer should be obvious from the above discussion, just think about the quantifiers.
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  4. #4
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    Re: logic problems

    Quote Originally Posted by demarkus View Post
    trying to figure out the following statements' true of false, and why?
    1.∀x ∈R, ∃y ∈ R such that x.y=1. false x=0, 0.y≠0
    2∀x ∈Z+ and ∃y ∈ Z+ ,∃z ∈ Z+ such that z=x-y
    3. ∀x ∈Z and ∃y ∈ Z ,∃z ∈ Z such that z=x-y
    4. ∃u ∈R+ such that ∀v ∈ R+, u.v<v
    5. ∀x ∈R+ , ∃u ∈ R+ such that u.v<v
    What if in #2, x=1~?

    In #4 are you sure that it is <\text{ and not }\le~?
    If #4 is correct, what if v=1~?.

    You do the others.
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