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Math Help - Partial Orders

  1. #1
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    Partial Orders

    Hi,
    A = {1,3,5,7,9,10,12,15,8,21}

    R = {(x,y) | y divided by x}


    Im trying to prove that R is a partial order (reflexsive, antisymmetric and transitive).

    Reflexsive: I said that it is true, because for all x, xRx. That is, every number in A is divided into itself once.

    Antisymmetric: is false. Because if we have (5, 10) and (10, 5) then x is not equal to y. Is that correct? Because we get 2 and 0.5 and thats not equal?

    Transitive:: if we have (5, 10, 15) then thats true but (15, 10, 5) false? Hence not transitive?

    Can somebody help me out on this question?

    I drew a hasse diagram for this but i dont know how to draw one on this forum. If someone can draw me one on the forum so i can check mine it would be of great help?

    My maximal was 12 and 15 and minimal element was 1. Correct?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kurac View Post
    Hi,
    A = {1,3,5,7,9,10,12,15,8,21}

    R = {(x,y) | y divided by x}
    clarify here. do you means y "divides" x?, as in y|x??

    if so, you have certain flaws in your logic. for instance, (5,10) would not be in the relation. etc etc
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  3. #3
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    Thats what confusing me alittle.

    All is said that "y is divided by x". I need prove that its a partial order.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kurac View Post
    Thats what confusing me alittle.

    All is said that "y is divided by x". I need prove that its a partial order.
    that is what it says in words? that makes no sense. it should be y "divides" x, or maybe something like "y divided by x is an integer". just saying y divided by x can mean many things. divides exactly? has a remainder of a certain amount?...etc

    if there was a symbol used, say so. do not type the english if you are not sure how the notation is read
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  5. #5
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    Quote Originally Posted by kurac View Post
    Hi,
    A = {1,3,5,7,9,10,12,15,8,21} R = {(x,y) | y divided by x}
    Im trying to prove that R is a partial order (reflexive, anti-symmetric and transitive).
    Reflexsive: I said that it is true, because for all x, xRx. That is, every number in A is divided into itself once.
    Antisymmetric: is false. Because if we have (5, 10) and (10, 5) then x is not equal to y. Is that correct? Because we get 2 and 0.5 and thats not equal?
    Transitive:: if we have (5, 10, 15) then thats true but (15, 10, 5) false? Hence not transitive?
    Reflexive: I said that it is true, because for all x, xRx. That is, every number in A is divided into itself once.

    Anti-symmetric: is false. Wrong
    Transitive:: if we have (5, 10, 15) then thats true but (15, 10, 5) false? Wrong again.
    You seem to have a most odd understanding of this relation.

    2 divides 4 but 4 does not divide 2.
    So if x divides y and y divides x then x=y: anti-symmetric.

    So if x divides y and y divides z then clearly x divides z: transitive.
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  6. #6
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    Plato, so your saying my relation definition will always only apply too numbers where x and y are the same; for e.g (2,2) (3,3). Hence x and y will divide and y and x will divide?

    I tried drawing a hasse diagram of the partial order R to try get better understanding of it...

    Can someone who knows how to draw these diagrams check if 12 and 15 is the maximal and 1 the minimal?

    Thanks for the help, im not very good at maths but im trying my best!
    Kurac.
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