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Math Help - examples of a binary relations

  1. #1
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    examples of a binary relations

    Is there any certain ways of thinking of these examples? Or just guess and try?

    a) Reflexive and symmetric, but not transitive
    = abs(a-b) < 1... right? any other example?

    b) Reflexive but neither symmetric nor transitive
    = a-b < 1, any other type?

    c) Symmetric, but neither reflexive nor transitive
    = abs(a-b) = 1, any other?

    d) Transitive, but neiter reflexive nor symmetric
    = a < b

    any other examples will be appreciated, thanks
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  2. #2
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    Quote Originally Posted by ninano1205 View Post
    Is there any certain ways of thinking of these examples? Or just guess and try?

    a) Reflexive and symmetric, but not transitive
    = abs(a-b) < 1... right? any other example?

    b) Reflexive but neither symmetric nor transitive
    = a-b < 1, any other type?

    c) Symmetric, but neither reflexive nor transitive
    = abs(a-b) = 1, any other?

    d) Transitive, but neiter reflexive nor symmetric
    = a < b

    any other examples will be appreciated, thanks
    a) Let R={(a,b): a,b are real Nos and |a-b|<1}

    For R to be reflexive we must have :

    For all a belonging to reals,then (a,a)εR ..or |a-a|<1,which is correct since 0<1

    For R to be symmetric we must have:

    For all a,b reals and (a,b)εR ,then (b,a)εR.But (a,b)εR===> |a,b|<1 which implies that |b,a|<1 ,hence (b,a)εR,SINCE |a-b|=|b-a|

    And finally for R TO BE not transitive we must prove that:

    There exist x,y,z reals such that (x,y)εR and (y,z)εR BUT ~(x,z)εR ( (x,z) does not belong to R).

    This is the negation of the definition for R being transitive in the real Nos:


    ......for all x,y,z reals (x,y)εR and (y,z)εR ,then (x,z)εR.........................


    So let x=8,....y=8,5..........z=9........................ .and:

    |x-y|=|8-8,5|=0,5<1 .........hence (x,y)εR........................................... ...................................

    |y-z|=|8,5-9|= 0,5<1...........hence (y,z)εR........................................... ...........................................

    But..... |x-z|=|8-9| = 1\geq 1.................................................. ..................................1

    Now for (x,z) not to belong to R we need to have ~|x-z|<1 ,i.e |x-z| must not be less to....1

    From the inequalities of real Nos ,if ~|x-z|<1 ,then |x-z|\geq 1

    Which is exactly what we have proved in (1)............................................... ..................................................

    Hence R is not transitive


    YOU CAN do the rest of problems more or less in the same way.

    To understand what you really doing i suggest you write first the definitions when a relation R in real Nos is reflexive,symmetric, transitive,and then take their negations
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