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Thread: mth202 discrete mathmatics second assignment solution fall 2010

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    Icon51 mth202 discrete mathmatics second assignment solution fall 2010

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    Maximum Marks: 15
    Due Date: November 22, 2010

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    Question 1)Prove that (Bc (Bc A)) c = (Bc (Bc Ac)) c by using Membership Table.
    Question 2)Let A= {0, 1, 2} be a set, construct AA and determine weather R = AA is partially ordered relation, justify your answer.
    (Hint: check each condition Reflexive, Anti-symmetric and Transitive)


    solution will be updated soon.


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    Icon7 mth202 final solution assignment 2

    First question:

    Second question answer:

    Let A= {0, 1, 2} so we can determined AA as under
    {(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), A A = (2,0), (2,1), (2,2)}
    Here R = AA so R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    For Reflexive:
    R is reflexive if and only if, for all a A , (a,a)R
    A= {0, 1, 2}
    A A = {(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    We can see that {(0,0), (1,1), (2, 2)}R so R is Reflexive
    For Anti-symmetric:
    R is anti-symmetric if and only if, for all b A if
    (a,b)R and (b, a)R then a = b
    aRb and bRa then a = b
    A= {0, 1, 2}
    A A = {(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    We can see that {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    As we can see that,
    (0,1)R and (1,0)R then 0=0 [As a = b]
    (2,1)R and (1, 2)R then 2=2 [As a = b]
    all pairs are available such that (a,b)R and (b, a)R so R is Anti-symmetric
    For Transitive:
    R is transitive if and only if, for all a,b, c A if
    (a,b)R and (b, c)R then (a, c)R
    That is aRb and bRc then aRc
    A= {0, 1, 2}
    A A = {(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}
    As we can see
    (0, 2)R and (2,1)R then (1,0)R
    (0,1)R and (1, 2)R then (0, 2)R and so on.
    all pairs are available such that (a,b)R and (b, c)R then (a, c)R so R is Transitive

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