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Math Help - 4 Relations Problems!

  1. #1
    anu
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    4 Relations Problems!

    1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) E R if and only if

    a) x + y = 0.
    b) x = y.
    c) x - y is a rational number.
    d) x = 2y.
    e) xy >= 0.
    f) xy = 0.
    g) x = 1.
    h) x = 1 or y = 1.



    2) Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relations from {1, 2, 3} to {1, 2, 3, 4}. Find R1 - R2.

    Would the answer of # 2 be or {} ?



    3) Show that the closure of a relation R with respect to a property P, if it exists, is the intersection of all the relations with property P that contain R.



    4) Use Algorithm 1 to find the transitive closures of these relations on {a, b, c, d, e}.

    a) {(a, b), (a, c), (a, e), (b, a), (b, c), (c, a), (c, b), (d, a), (e, d)}


    Thank You very much for your help!
    Anu

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anu View Post
    1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) E R if and only if

    a) x + y = 0.
    i will illustrate what you need to do with this one. the rest of the questions in this section are just applying the definitions again.

    Definition: A relation R defined on a set A is called reflexive if xRx for every x \in A (that is, every element in the set is related to itself under the relation).

    since x + x = 0 is not always true, each element in the real numbers does not relate to itself under this relation. thus this relation is not reflexive


    Definition: A relation R defined on a set A is called symmetric if whenever xRy, then yRx for all x,y \in A

    we are dealing with real numbers, thus we have the property of commutativity of addition.

    so if x + y = 0 then y + x = 0 thus we have xRy \implies yRx as desired. this relation is symmetric


    Definition: A relation R defined on a set A is called anti-symmetric if whenever xRy and yRx then x = y for all x,y \in A

    now if x + y = 0 and y + x = 0 it does not imply that x = y

    case in point, consider x = 1 and y = -1

    thus, the relation is not anti-symmetric


    Definition: A relation R defined on a set A is called transitive if whenever xRy and yRz then xRz for all x,y,z \in A

    now if x + y = 0 and y + z = 0 it means that x = z. but we know that the relation is not reflexive, and therefore x does not relate to itself. so the relation is not transitive.

    case in point: x = z = 1 and y = -1

    then we have xRy and yRz (that is, x + y = 1 - 1 = 0, and y + z = -1 + 1 = 0), but x \not{R}z since x + z = 1 + 1 = 2. this is an implication, so if we have a true statement implying a false statement, the whole implication is false. thus our conclusion follows


    try the rest. as you see, i just relied solely on the definitions, nothing fancy
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anu View Post
    2) Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relations from {1, 2, 3} to {1, 2, 3, 4}. Find R1 - R2.

    Would the answer of # 2 be or {} ?
    i don't recall doing this in class when we were doing relations (that is, to subtract one relation from another). i suppose we treat them as sets though. so i interpret R1 - R2 to mean we take the set R1 and throw out all the elements that appear in R2 (is that correct?) thus, under this assumption, the answer would be \emptyset NOT \{ \emptyset \}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anu View Post

    3) Show that the closure of a relation R with respect to a property P, if it exists, is the intersection of all the relations with property P that contain R.
    i have no idea what to do here


    4) Use Algorithm 1 to find the transitive closures of these relations on {a, b, c, d, e}.

    a) {(a, b), (a, c), (a, e), (b, a), (b, c), (c, a), (c, b), (d, a), (e, d)}
    what is this "Algorithm 1" you are referring to? that sounds like it's something specific to your text or your class
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  5. #5
    anu
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    Quote Originally Posted by Jhevon View Post
    what is this "Algorithm 1" you are referring to? that sounds like it's something specific to your text or your class
    Ignore!!! sorry
    Last edited by anu; October 24th 2007 at 10:39 PM.
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  6. #6
    anu
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    Quote Originally Posted by Jhevon View Post
    i will illustrate what you need to do with this one. the rest of the questions in this section are just applying the definitions again.

    Definition: A relation R defined on a set A is called reflexive if xRx for every x \in A (that is, every element in the set is related to itself under the relation).

    since x + x = 0 is not always true, each element in the real numbers does not relate to itself under this relation. thus this relation is not reflexive


    Definition: A relation R defined on a set A is called symmetric if whenever xRy, then yRx for all x,y \in A

    we are dealing with real numbers, thus we have the property of commutativity of addition.

    so if x + y = 0 then y + x = 0 thus we have xRy \implies yRx as desired. this relation is symmetric


    Definition: A relation R defined on a set A is called anti-symmetric if whenever xRy and yRx then x = y for all x,y \in A

    now if x + y = 0 and y + x = 0 it does not imply that x = y

    case in point, consider x = 1 and y = -1

    thus, the relation is not anti-symmetric


    Definition: A relation R defined on a set A is called transitive if whenever xRy and yRz then xRz for all x,y,z \in A

    now if x + y = 0 and y + z = 0 it means that x = z. but we know that the relation is not reflexive, and therefore x does not relate to itself. so the relation is not transitive.

    case in point: x = z = 1 and y = -1

    then we have xRy and yRz (that is, x + y = 1 - 1 = 0, and y + z = -1 + 1 = 0), but x \not{R}z since x + z = 1 + 1 = 2. this is an implication, so if we have a true statement implying a false statement, the whole implication is false. thus our conclusion follows


    try the rest. as you see, i just relied solely on the definitions, nothing fancy
    I kinda know all this but still confused!
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anu View Post
    I kinda know all this but still confused!
    type up your solution for (b) and we'll see how you do
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  8. #8
    anu
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    Quote Originally Posted by Jhevon View Post
    type up your solution for (b) and we'll see how you do
    b) x = +-y

    not reflexive b/c x cannot equal both + y and -y, is symmetric b/c x would equal y or -y and then y or -y would equal x; is not antisymmetric and it is transitive

    am i right?
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anu View Post
    b) x = +-y

    not reflexive b/c x cannot equal both + y and -y, is symmetric b/c x would equal y or -y and then y or -y would equal x; is not antisymmetric and it is transitive

    am i right?
    check reflexive again. is x = \pm x?
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  10. #10
    anu
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    Quote Originally Posted by Jhevon View Post
    check reflexive again. is x = \pm x?
    so it is reflexive, to check if it is reflexive we just plug in x for the y?

    Quote Originally Posted by anu View Post
    1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) E R if and only if

    a) x + y = 0.
    b) x = y.
    c) x - y is a rational number.
    d) x = 2y.
    e) xy >= 0.
    f) xy = 0.
    g) x = 1.
    h) x = 1 or y = 1.
    a) symmetric
    b) reflexive, symmetric, transitive

    c) wait is 0 a rational number? i think yes so it is reflexive, antisymmetric b/c x or y can be not rational numbers and still have a result of a rational number.

    d) antisymmetric?

    e) reflexive, antisymmetric b/c xy >= 0 but x and y individually can be less than zero, right?

    f) reflexive, antisymmetric b/c x or y has to be zero but not the other?

    g) none

    h) symmetric?
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anu View Post
    so it is reflexive, to check if it is reflexive we just plug in x for the y?
    no. to check reflexivity, you just check if any element relates to itself. any element x relates to itself in the case of x = +/- y


    yes, 0 is a rational number
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  12. #12
    anu
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    Quote Originally Posted by anu View Post
    1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) E R if and only if

    a) x + y = 0.
    b) x = y.
    c) x - y is a rational number.
    d) x = 2y.
    e) xy >= 0.
    f) xy = 0.
    g) x = 1.
    h) x = 1 or y = 1.
    Quote Originally Posted by anu View Post
    a) symmetric
    b) reflexive, symmetric, transitive

    c) wait is 0 a rational number? i think yes so it is reflexive, antisymmetric b/c x or y can be not rational numbers and still have a result of a rational number.

    d) antisymmetric?

    e) reflexive, antisymmetric b/c xy >= 0 but x and y individually can be less than zero, right?

    f) reflexive, antisymmetric b/c x or y has to be zero but not the other?

    g) none

    h) symmetric?

    So how did i do for the above?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anu View Post
    c) x - y is a rational number.
    x - x = 0 which is rational, so it is reflexive

    if x - y is rational, then y - x is rational, so it is symmetric

    if x - y is rational and y - x is rational then it doesn't necessarily mean x = y. case in point x = 1, y = 2

    so it is not anti-symmetric

    it is transitive

    (note here that x, y and z themselves do not have to be rational. for instance x = 1 + \sqrt{2}, y = z = \sqrt{2} fulfills the conditions)

    d) x = 2y.
    x is not equal to 2x in general, so it's not reflexive.

    if x = 2y then y = (1/2)x so it is not symmetric

    if x = 2y and y = 2x then x = -y so it would seem not to be anti-symmetric. but when you think about it, the only way this would work is if x = y = 0. so i'd say it is anti-symmetric

    if x = 2y and y = 2z then x = 4z so it is not transitive.

    e) xy >= 0.
    xx = x^2 >= 0 so it is reflexive

    xy = yx, so if xy >= 0 then yx >= 0 so it is symmetric

    if xy >=0 and yx >= 0 it obviously does not mean that x = y so it is not antisymmetric (come up with your own counter-example here. it is good practice to come up with reasons for why something does not fulfill the criteria, we do that by finding a counter-example)

    if xy >= 0 and yz >= 0 then xz >= 0 (why is this?) so it is transitive

    f) xy = 0.
    obviously this is not reflexive

    if xy = 0 then yx = 0 so it is symmetric

    it is obviously not anti-symmetric

    it is not transitive. if xy = 0 and yz = 0 it does not mean xz = 0

    consider x = z = 1, y = 0

    g) x = 1.
    the relation is {(1,1)} it is reflexive, symmetric, antisymmetric and transitive (recall we have implications for the definitions here, the conditions can be fulfilled vacuously)

    h) x = 1 or y = 1.
    again, this seems to fulfill the conditions vacuously


    now i'm tired, i'm going to bed
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  14. #14
    anu
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    Thanks a Million Jhevon
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  15. #15
    soul_musique
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    Can anyone solve this problem for me ??? Thanks.

    3) Show that the closure of a relation R with respect to a property P, if it exists, is the intersection of all the relations with property P that contain R.
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