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Sets and Relations (just needs checking please)

  1. Oct 27, 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

    As p is an equivalence relation there are associated equivalence classes.

    (iv) Find all the ordered pairs in the equivalence class of (2,6). Why could this equivalence class be identified with the integer -4?

    (v) Give the equivalence classes (as sets of ordered pairs) defined by p for each of the integers: 0, -1 and +1

    (vi) Consider two general ordered pairs, (a,b) and (c,d). If addition is defined by (a,b) + (c,d) = (a+c, b+d) and multiplication is defined by (a,b) x (c,d) = (ac+bd, ad+bc), show that these definitions provide a way of demonstrating that (+1) + (-1) = 0 and (-1) x (-1) = (+1)


    (iv) The ordered pairs of the equivalence class (2,6) are infinite i.e. (4,8) or (6,10) so to find the general term lets write the condition as follows: a - b = c - d (so that the values of one pair appear on the left and the values of the other on the right). Here a - b = 2 - 6 = -4. Therefore any pair (c,d) with c-d = -4 (or d-c = 4) is also related.

    (v) The set of ordered pairs are:

    0 ----------> (a,a) where b=a

    -1 ---------> (a-1, a) where b=a+1 or should I write like this instead (a, a+1)? Not sure let me know

    +1 ---------> (a+1, a) where b=a-1 or should I write like this instead (a, a-1)? Not sure let me know

    (vi) Lets consider 2 pairs say (2,3) and (6,7) which are as seen previously have an integer of -1 then if the multiplication of two orderd pairs is defined by (a,b) x (c,d) = (ac+bd, ad+bc) then

    (2,3) x (6,7) = (12+21,14+18) = (33,32) which is has as seen in the previous question an integer of +1

    Hence -1 x -1 = +1

    Again, if we choose 1 pair say (33, 32) with an integer of +1 and the pair (5,6) with an integer value of -1 then the addition of two ordered pairs is defined by (a,b) + (c,d) = (a+c, b+d) and we get

    (33,32) + (5,6) = (33+5, 32+6) = (38,38) which has an integer value of 0

    Hence +1 + -1 = 0

    Please correct any of my mistakes anyone... many thanks
  2. jcsd
  3. Oct 28, 2005 #2


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    Science Advisor

    Yes, those are correct. That is, by the way, a standard way of defining the integers given the natural numbers.
    If (a,b), with a> b, is in an equivalence class, then all members of the equivalence class, (x,y) have x-y= a- b and we associate that "integer" (equivalence class) with the natural number a-b.
    If (a,b), with a< b, is in an equivalence class, then all members of teh equivalence class, (x,y) have y-x= b-a and we associate that "integer" with
    If an equivalence class contains (a,a), then all members of the equivalence class (x,y) have x= y and we associate that "integer" with 0.
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