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Math Help - Question on Relations

  1. #1
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    Question on Relations

    Question 1:
    D on Z(all integers): mDn iff m|n

    My thoughts:
    Why is 0D0 not reflexive as stated by the solution? 0|0 --> 0=n.0
    So that means it does not matter what value n is, the answer will be still be 0. So shouldn't it be reflexive?


    Question 2:
    P on N(natural numbers): mPn iff There exists p, p|m,p|n and p is prime.

    My thoughts:
    I am not sure how to apply the transitive rules on this where if xPy and yPz then xPz.
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  2. #2
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    Hello,

    Another way to say that 0|0 is that 0/0=an integer, and 0/0 is ... i'll say undetermined...

    I am not sure how to apply the transitive rules on this where if xPy and yPz then xPz.
    You can find out a counter example.

    The relation here is :

    If xPy <=> p|gcd(x,y), with p prime
    and yPz <=> p'|gcd(y,z), with p' prime

    Does p'', prime, exist such as p''|gcd(x,z) ?

    Counter example :

    3P6 because 3, prime, divides 3 and 6.
    6P14 because 2, prime, divides 6 and 14.

    Does a prime number divide 3 and 14 ? The answer is no, because they are coprime (1 is not a prime number !)
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  3. #3
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    Thanks for the reply!

    Then why is it not reflexive for question 2? It works out to be like this right?
    xPx <=> p|gcd(x,x). Then shouldn't it be reflexive?
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  4. #4
    Moo
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    I would have said that it's reflexive too , because there always exists a prime integer which divides a natural integer...
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    Quote Originally Posted by Moo View Post
    Hello,

    Another way to say that 0|0
    0 does not divide 0. But 0|0 is true. Because a|b if and only if there exists c so that a=bc. This is definitely true for 0|0.
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  6. #6
    Moo
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    But dividing by 0 is not true ? How do we know if it can be a counterexample or not ? (just asking, don't screw up )
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    Quote Originally Posted by Moo View Post
    I would have said that it's reflexive too , because there always exists a prime integer which divides a natural integer...
    What prime divides 1?
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  8. #8
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    Oh yeah
    None.

    Sorry ~
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  9. #9
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    Quote Originally Posted by Moo View Post
    But dividing by 0 is not true ? How do we know if it can be a counterexample or not ? (just asking, don't screw up )
    What are you talking about? Remember 0|0 if and only if there exists a c\in \mathbb{Z} such that 0 = 0\cdot c. Clearly, such a c exists. Thus, 0|0.
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  10. #10
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    Just going the wrong way
    Forgive me
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