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Math Help - [SOLVED] Disprove Relation - Can somebody check my work

  1. #1
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    Question [SOLVED] Disprove Relation - Can somebody check my work

    Disprove R on Q is either reflexive, symmetric, or transitive. x, y are elements in Q. xRy iff x = yk -1 for some integer k.
    In order for to disprove this relation, I think I should show that it is

    • NOT reflexive
    • NOT symmetric
    • NOT transitive

    REFLEXIVE:
    Suppose xRx.
    x = xk -1.
    1 = k - 1/x
    1 + 1/x = k
    k = (1+x)/x, since k is rational, R is not reflexive.

    SYMMETRIC:
    Suppose xRy.
    x = yk-1.
    Suppose yRx.
    y = xm-1.
    Plugging in for x, and solving for y
    y = (yk-1)m-1
    y = y(km - m) - 1
    1 = (km-m) - 1/y
    (km-m) = (y + 1)/y, once again, (km-m) is rational, so R is not symmetric.

    TRANSITIVE:
    Suppose xRy.
    x = yk-1.
    Suppose yRz
    y = zn-1
    Plugging in for y and solving for x
    x = (zn-1)k - 1
    x = z(nk - k) - 1
    x/z = (nk - k) -1/z
    x/z + 1/z = (nk - k)
    (x+1)/z = (nk - k), so again, (nk - k) is rational, so R is not transitive

    This is the only way that I could come up to disprove the relation. Please correct me if I am wrong on any work and show how I can correct with an explanation. Much thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by spearfish View Post
    Disprove R on Q is either reflexive, symmetric, or transitive. x, y are elements in Q. xRy iff x = yk -1 for some integer k.
    In order for to disprove this relation, I think I should show that it is

    • NOT reflexive
    • NOT symmetric
    • NOT transitive

    REFLEXIVE:
    Suppose xRx.
    x = xk -1.
    1 = k - 1/x
    1 + 1/x = k
    k = (1+x)/x, since k is rational, R is not reflexive.
    k is a NON-INTEGER rational is what you want to say. also, consider the case where x = 0. it is trivial, but if you are dividing by x, you must consider it.

    SYMMETRIC:
    Suppose xRy.
    x = yk-1.
    Suppose yRx.
    y = xm-1.
    Plugging in for x, and solving for y
    y = (yk-1)m-1
    y = y(km - m) - 1
    1 = (km-m) - 1/y
    (km-m) = (y + 1)/y, once again, (km-m) is rational, so R is not symmetric.
    again, say k is a non-integer rational, clearly. and consider the case y = 0.

    TRANSITIVE:
    Suppose xRy.
    x = yk-1.
    Suppose yRz
    y = zn-1
    Plugging in for y and solving for x
    x = (zn-1)k - 1
    x = z(nk - k) - 1
    x/z = (nk - k) -1/z
    x/z + 1/z = (nk - k)
    (x+1)/z = (nk - k), so again, (nk - k) is rational, so R is not transitive

    This is the only way that I could come up to disprove the relation. Please correct me if I am wrong on any work and show how I can correct with an explanation. Much thanks.
    now this one is a little shaky. how do we know that (x + 1)/z is not an integer? can z = x + 1? and of course, consider the case z = 0 since you divide by it
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  3. #3
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    Awsome, I 'll get to work on that transitive part.
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  4. #4
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    Hello spearfish
    Quote Originally Posted by spearfish View Post
    Disprove R on Q is either reflexive, symmetric, or transitive. x, y are elements in Q. xRy iff x = yk -1 for some integer k.
    All you really need to do here is to find a single counter-example in each case.

    R is reflexive \Rightarrow\forall x, \, xRx. But when x = 2

    2 = 2k-1 \Rightarrow k = \tfrac32

     \Rightarrow R is not reflexive.

     R is symmetric \Rightarrow (\forall x,y,\,xRy \Rightarrow yRx)

    Now x=5, y = 2 \Rightarrow k = 3

    So 5R2. But x = 2, y = 5 \Rightarrow k = \tfrac35

     \Rightarrow R is not symmetric.

    And R is transitive \Rightarrow (\forall x, y, z, \,xRy \wedge yRz \Rightarrow xRz)

    But 7R4 \,(k=2) \wedge 4R5 \,(k=1). But 7 = 5k-1 \Rightarrow k = \tfrac85

    So R is not transitive.

    Incidentally, the algebra goes a bit wrong here:
    x = (zn-1)k - 1
    x = z(nk - k) - 1
    Grandad
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  5. #5
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    Silly me! Thanks for the tip and for pointing out the error Grandad!
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