# Sets and Relations (just needs checking please)

1. Oct 27, 2005

### Natasha1

MY WORK FOLLOWS BELOW THE QUESTIONS

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)

HERE IS MY WORK

(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. Oct 28, 2005

### HallsofIvy

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
-(b-a).
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.