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Math Help - Proof about relations

  1. #16
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    Re: Proof about relations

    I don't understand this example. Which are the elements of R?
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  2. #17
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    Re: Proof about relations

    Quote Originally Posted by Hartlw View Post
    The best I could come up with was an example:

    A={1.2.3.4,5} R = <
    R[A]=(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4), (3,5),(4,5)
    B={3,4,5,6}
    R[B]= (3,4),(3,5),(3,6),(4,5),(4,6),(5,6)
    R[A]-R[B]= (1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)
    [A-B]={1,2}
    R[A-B]=(1,2)

    so R[A-B] is a subset of R[A]-R[B]
    ??
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  3. #18
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    Re: Proof about relations

    Quote Originally Posted by sebasvargasl View Post
    ??
    Defininitions:

    AXB is the set of all ordered pairs (a,b) with a ϵ A and b ϵ B.

    A relation R from A to B is a subset of AXB.

    If B =A then you have a relation from A to A: R[A], a subset of AXA.

    Example: A = {1,2,3}
    AXA = { (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
    Let the relation R[A] be “less than,” <. Then
    R[A] = {(1,2), (1,3), (2,3)}
    Write R[A] for the relations “equal” and “greater than.”

    Definition: A – B: All elements of A which are not also elements of B. (Draw a Venn diagram with B partially overlapping A and subtract B)
    Examples:
    If B = {2,3,4,5} and A as above,
    A - B = {1}.

    If A and B don’t have elements in common, A-B is empty.
    Last edited by Hartlw; March 18th 2013 at 07:40 AM. Reason: change "subtract A" to "subtract B"
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  4. #19
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    Re: Proof about relations

    Your definition of R[A] is wrong. The elements of R[A] are not necessarily ordered pairs. In order to say things about R[A] first you have to define R. Again, R[A]={y: ∃x(x∈A∧xRy)}. When you state R[A] = {(1,2), (1,3), (2,3)} this implies that for some x (x,(1,3))∈R, therefore R is a ternary relation and binary at the same time, a contradiction.
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  5. #20
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    Re: Proof about relations

    Quote Originally Posted by sebasvargasl View Post
    Your definition of R[A] is wrong. The elements of R[A] are not necessarily ordered pairs. In order to say things about R[A] first you have to define R. Again, R[A]={y: ∃x(x∈A∧xRy)}. When you state R[A] = {(1,2), (1,3), (2,3)} this implies that for some x (x,(1,3))∈R, therefore R is a ternary relation and binary at the same time, a contradiction.
    I DID define R[A], by assuming that’s what you meant by your unintelligible definition of Post #3, and my conclusions from this definition are correct. If I misinterpreted your original definition, why wait till now before explaining it? If you weren’t sure and have been working your way into it, that’s understandable.

    You are now proposing a new problem by giving, for the first time, your new, intelligible, definition of R[A], which is:

    R[A] is the set of all y in xRy.

    I note you don’t define xRy but I know what you mean. For a proper definition, you should define what you mean by xRy.

    I am not a mind-reader. Why didn’t you just say in the first place what your definition of R[A] is instead of waiting till now?

    The more judicious way of doing this would be a new post of the form: If I define R[A] as set of all y in xRy, then show …..
    Last edited by Hartlw; March 18th 2013 at 10:50 AM.
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  6. #21
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    Re: Proof about relations

    Here is new example to go with your new definition of R[A] in post #19.

    Def: R[A] is set of all y of xRy. R of xRy is “less than,” <.

    A={1,2,3,4}
    xRy={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}
    R[A]={2,3,4}

    B={3,4,5}
    xRy=(3,4),(3,5),(4,5)
    R[B]={4,5}

    R[A]-R[B]= {2,3}

    [A-B]={1,2}
    xRy={(1.2)}

    R[A-B]={2}

    so R[A-B] is a subset of R[A]-R[B]
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  7. #22
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    Re: Proof about relations

    And here is new proof of OP to go with new definition of post#19

    x,y ϵ [A-B] → x,y ϵ A and x,y ϵ’ B → (y ϵ xRy → y ϵ A and y ϵ’ B)

    So R[A-B] is a subset of R[A]-R[B]
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  8. #23
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    Re: Proof about relations

    Now you're using xRy to denote a set, which is wrong, xRy is another form to write (x,y)ϵR: a sentential function, not a set. How is "y belongs to R[A] if and only if there exists x belonging to A for which the ordered pair (x,y) belongs to R" unintelligible?. Once again your example is not correct for what you're doing is defining two different relations namely, {(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)} and {(3,4),(3,5),(4,5)}. They have to be one and the same relation to serve as an example. To write R[A] you use the first one, but, when you write R[B] you use the second one, so you're not really doing R[A]-R[B] but, R[A]-R'[B]
    Thanks from emakarov
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  9. #24
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    Re: Proof about relations

    Quote Originally Posted by Hartlw View Post
    And here is new proof of OP to go with new definition of post#19
    x,y ϵ [A-B] → x,y ϵ A and x,y ϵ’ B → (y ϵ xRy → y ϵ A and y ϵ’ B)
    So R[A-B] is a subset of R[A]-R[B]
    This is completely wrong.

    Relpy #13 is the only clear and correct proof in the thread.

    THREAD CLOSED.
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  10. #25
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    Re: Proof about relations

    Quote Originally Posted by Hartlw View Post
    And here is new proof of OP to go with new definition of post#19

    x,y ϵ [A-B] → x,y ϵ A and x,y ϵ’ B → (y ϵ xRy → y ϵ A and y ϵ’ B)

    So R[A-B] is a subset of R[A]-R[B]
    And this is by no means a proof, I'm sorry. In fact one can't prove "R[A-B] is a subset of R[A]-R[B]"
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