Answer:
Step-by-step explanation:
REcall that given a set A, * is a equivalence relation over A if
- for a in A, then a*a.
- for a,b in A. If a*b, then b*a.
- for a,b,c in A. If a*b and b*c then a*c.
Consider A the set of all ordered pairs of positive integers.
- Let (a,b) in A. Then a+b = a+b. So, by definition (a,b)R(a,b).
- Let (a,b), (c,d) in A and suppose that (a,b)R(c,d) . Then, by definition a+d = b+c. Since the + is commutative over the integers, this implies that d+a = c+b. Then (c,d)R(a,b).
- Let (a,b),(c,d), (e,f) in A and suppose that (a,b)R(c,d) and (c,d)R(e,f). Then
a+d = b+c, c+f = d+e. We have that f = d+e-c. So a+f = a+d+e-c. From the first equation we find that a+d-c = b. Then a+f = b+e. So, by definition (a,b)R(e,f).
So R is an equivalence relation.
[(a,b)] is the equivalence class of (a,b). This is by definition, finding all the elements of A that are equivalente to (a,b).
Let us find all the possible elements of A that are equivalent to (2,4). Let (a,b)R(2,4) Then a+4 = b+2. This implies that a+2 = b. So all the elements of the form (a,a+2) are part of this class.
