Function and equivalence relations question

Question

1. Let A and B be subsets of the set Z of all integers, and let F denote the set of all functions f : A to B. Define a relation R on F by: for any f,g element of F, fRg if and only if f - g is a constant function; that is, there is a constant c so that f(x) - g(x) = c for all x element of A.

   (a) Prove that R is an equivalence relation on F.
   (b) Suppose that A = B = Z. Let Iz be the identity function on Z so Iz(x) = x 
       for all x element of Z. Find a function f (6)= Iz which belongs to the
       equivalence class [Iz].
   

   Assume that A={1,2,3} and B={1,2,...,n} where n >= 2 is a fixed integer

   (c) Let f1 element of F be defined by: f1(1) = 2, f1(2) = n, f1(3) = 1.
       (As a set ofordered pairs, f1 = {(1,2), (2,n), (3,1)}
       Suppose that g element of F is arbitrary so that gRf1. 
       Prove that g = f1, and thus the equivalence class [f1] is just {f1}.
   

Answer 1

HINTS: For (a):

• Reflexivity: What is $f(n)-f(n)$? Does it depend on $n$?
• Symmetry: If $f(n)-g(n)=k$, what is $g(n)-f(n)$?
• Transitivity: If $f(n)-g(n)=k$ and $g(n)-h(n)=\ell$, what is $f(n)-h(n)$?

For (b): You want a function $f:\Bbb Z\to\Bbb Z$, different from $I_{\Bbb Z}$, such that $f(n)-I_{\Bbb Z}(n)$ is a constant that does not depend on $n$. Note that $f(n)-I_{\Bbb Z}(n)=f(n)-n$. Can you find a function $f$ such that $f(n)-n$ is always $2$, say? Or $7$? Or any other fixed integer?

Answer 2

Ok, if f belongs to the equivalence class [Iz], f(x)-x=c, c being a constant. So f(x)=x (mod c). Since congruence is an equivalent relation, I would say that any function f, where f(x) - Iz(x) is equal to a multiple of c, that belongs to the equivalent class [Iz]. I'm really not sure about that :)