# Just needs checking please (Sets and Relations)

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 27, 2005

### hypermorphism

I would write it the second way, since we're dealing with N. If by (a-1,a), you're writing shorthand for the set $\{(a-1, a): a \in \mathbb{N}\}$ then there is no corresponding term for a=0 (or a=1 depending on your definition of N). The next one is fine since adding one to a natural number will never take it out of the set.
This is not enough to show that (-1)x(-1) = (+1). You must show it for an arbitrary element of the equivalence class, not any specific element. This is to show that only the fact that it is in the class (-1) makes it behave this way, and that it is not simply a side effect of the specific pair you picked. You should start off something like "All elements of the class (-1) have the form (a, a+1), so multiplying two arbitrary elements (a, a+1)x(b,b+1) gives us..." and so on. Same for the next question.

Last edited: Oct 27, 2005
3. Oct 28, 2005

### Natasha1

Right then,

All elements of the class (-1) have the form (a, a+1), so multiplying two arbitrary elements (a, a+1)x(b,b+1) gives us

(a, a+1)x(b,b+1) = (ab+(a+1)(b+1), a(b+1)+b(a+1))
= (ab+ab+a+b+1, ab+a+ab+b)
= (2ab+a+b+1, 2ab+a+b) which has an integer of +1 (Should I say it like that??? how can I end this nicely???)

Hence -1 x -1 = +1

If one element of the class (-1) , which has the form (a, a+1) is added to an element of the class (+1) which has the form (a,a-1) gives us

(a, a+1) + (b,b-1) = (a+b, a+1+b-1)
= (a+b, a+b) which has an integer of 0 (Should I say it like that??? how can I end this nicely???)

Hence +1 + -1 = 0

4. Oct 28, 2005

### hypermorphism

Well, the last line is enough, but you could clarify by saying that it is of the form (c+1,c) = (+1) where c = 2ab + a + b. You already established earlier that pairs of that form are in the equivalence class (+1), but you can do it again by just noting that (c+1)-c = +1. The same holds for the next question.