# Ordered pairs

• Oct 27th 2005, 05:56 PM
Natasha
Last 3 questions for Professor Jameson
Lets see if you are a real professor or a simple lecturer... lol, just kidding ;)

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)
• Oct 27th 2005, 06:04 PM
Jameson
In about 13 years I hope to be a full professor, but for now I remain a humble student. To be honest, I'm not seeing the answer right now. I'm terribly sorry. If you some thoughts to contribute I'll look at them and see if they spark anything but between school and running my own math help website, I'm exhausted.

Best,
Jameson
• Oct 27th 2005, 06:18 PM
Natasha
Right then....

(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 (a, a+1)?

+1 ---------> (a+1, a) where b=a-1 or should I write (a, a-1)?

(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 :)
• Oct 27th 2005, 06:22 PM
Jameson
I can tell you I'm probably the only person coming on this site tonight. If you need help, I'd suggest www.physicsforums.com. Very smart people there. But you better show your work or they'll eat you alive.

Goodnight and goodluck,
Jameson