1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)


    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


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


    (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

  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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Sets and Relations (just needs checking please)