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Sets and Relations (just needs checking please)

  1. Oct 28, 2005 #1
    Let AxB be the set of ordered pairs (a,b) where a and b belong to the set of natural numbers N.

    A relation p: AxB -> AxB is defined by: (a,b)p(c,d) <-> a+d = b+c

    (i) Is (2,6) related to (4,8)? Give three ordered pairs which are related to (2,6)

    ANSWER:

    Yes (2,6)p(4,8) as 2+8 = 6+4 = 10

    (2,6)p(2,6) = 2+6 = 6+2 = 8
    (2,6)p(6,10) = 2+10 = 6+6 = 12
    (2,6)p(12,16) = 2+16 = 6+12 = 18

    OR should it be written like this as the question stipulates ordered pairs related to (2,6) i.e. p(2,6)

    (2,6)p(2,6) = 2+6 = 6+2 = 8
    (6,10)p(2,6) = 6+6 = 2+10 = 12
    (12,16)p(2,6) = 6+12 = 2+16 = 18

    ANY SUGGESTIONS???

    (ii) Is (2,6) related to (a, a+4)? Justify your answer

    ANSWER

    A = 2, B= 6, C= a and d = a+4

    as (a,b)p(c,d) <-> a+d = b+c

    Then a+d = 2+a+4 = 6+a and b+c = 6+a

    Therefore a+d = b+c and (a,b)p(c,d)
    Hence (2,6) is related to (a, a+4)

    CORRECT, ANY IMPROVEMENT?


    (iii) Prove (i.e. using general letters) that p is an equivalence relation on the set AxB

    An equivalence relation on a set X is a binary relation on X that is reflexive, symmetric and transitive.

    a) This set AxB is reflexive as any ordered pair is related to itself i.e. (a,b)p(a,b) <-> a+b = b+a
    i.e. (2,6)p(2,6) = 2+6 = 6+2 = 8

    b) This set AxB is symmetric as if we take any 2 ordered pairs (a,b), (c,d) then (a,b)p(c,d), hence a+d = b+c. And then (c,d)p(a,b), hence c+b = d+a.

    i.e. (2,6)p(10, 14) we get 2+14 = 6+10 = 16
    and (10,14)p(2,6) we get 6+10 = 2+14 = 16

    c)This set AxB is transitive as if we take 3 ordered pairs (a,b), (c,d) and (e,f) we have (a,b)p(c,d), hence a+d = b+c and (c,d)p(e,f), then c+f = d+e, which gives (a,b)p(e,f) hence a+f = b+e

    i.e. (2,6), (10,14) and (16,20)

    (2,6)p(10,14) we get 2+14 = 6+10 = 16
    and (10,14)p(16,20) we get 10+20 = 14+16 = 30

    Then (2,6)p(16,20) we get 2+20 = 6+16 = 22

    ANY SUGGESTIONS FOR IMPROVEMENT?
     
  2. jcsd
  3. Oct 28, 2005 #2
    There doesn't seem to be a logical implication here. How do the previous lines imply the last phrase, that (a,b)p(e,f) ?
     
  4. Oct 28, 2005 #3

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    (i) and (ii) look good. Incidentally, ApB is read as "A is related to B". (at least when p doesn't need to be stated explicitly)


    For (iii), I think you're close, but not quite there conceptually.

    First off, you made a notational mistake: your goal is to prove that p is reflexive, symmetric, and transitive.

    Secondly, why are you saying AxB instead of NxN?

    Finally, you're slightly off in how to prove this. For example, in part (b), you never proved p is symmetric. All you've said is:

    (a,b) p (c,d) implies a+d=b+c
    (c,d) p (a,b) implies c+b=d+a

    (Actually, you also said that (a,b)p(c,d) for any two ordered pairs.)

    In particular, you've not said that (a,b)p(c,d) if and only if (c,d)p(a,b), which is what you're trying to prove.
     
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