# Thread: Proving an equivalence relation.

1. ## Proving an equivalence relation.

Let A and B be subsets of the set Z of all integers, and let F denote the set of all functions f : A--> B. Define a relation R on F by: for any f,g 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 A.

I need to prove that R is an equivalence relation on F.

2. ## Re: Proving an equivalence relation.

So, what seems to be the problem?

3. ## Re: Proving an equivalence relation.

Ok so i believe i have to prove that it is reflexive symmetric and transitive. How would i prove that this is reflexive?

4. ## Re: Proving an equivalence relation.

Please confirm that you know what "reflexive" is. Because if you do, the fact that R is reflexive should be obvious to you. (You know that y - y = 0 for all y, don't you?) On the other hand, if you don't know what "reflexive" means, asking a question that hides this lack of knowledge on a forum is not the best idea. Instead, you should read your textbook or lecture notes or look it up in Wikipedia, MathWorld or a similar site.

5. ## Re: Proving an equivalence relation.

Alright i hope this is correct:

Reflexive: fRf = f(x)-f(x) for x (element of) A
=0 which is a constant. Therefore reflexive

Symmetric: There exists f,g (element of) F if fRg, then gRf
gRf = g(x) - f(x)
=-(f(x) - g(x))
=-C

Still a little confused on transitivity.

6. ## Re: Proving an equivalence relation.

Originally Posted by hcuk
Alright i hope this is correct:

Reflexive: fRf = f(x)-f(x) for x (element of) A
=0 which is a constant. Therefore reflexive

Symmetric: There exists f,g (element of) F if fRg, then gRf
gRf = g(x) - f(x)
=-(f(x) - g(x))
=-C

Still a little confused on transitivity.
If $f\mathcal{R}g~\&~g\mathcal{R}h$ what can be said of $(f-g)+(g-h)~?$

7. ## Re: Proving an equivalence relation.

Originally Posted by hcuk
Reflexive: fRf = f(x)-f(x) for x (element of) A
=0 which is a constant. Therefore reflexive
The idea is correct but should be written differently. Equality = is mostly used between numbers, vectors, functions, matrices and other math objects. On the other hand, fRf is a proposition, i.e., something that can be either true or false. It is customary to write "A ⇔ B" or "A iff B" to indicate that propositions A and B are equivalent. So here I would write:

R is reflexive iff for all f ∈ F, fRf. Fix an arbitrary f ∈ F. Then fRf iff there exists a c such that for all x ∈ A, f(x) - f(x) = c. Consider c = 0 and fix an arbitrary x ∈ A. Indeed, f(x) - f(x) = 0, as required.

Originally Posted by hcuk
Symmetric: There exists f,g (element of) F if fRg, then gRf
gRf = g(x) - f(x)
=-(f(x) - g(x))
=-C
First, it's "for all f, g" and not "there exists "f, g". I would write:

R is symmetric iff for all f, g ∈ F, fRg implies gRf. Fix arbitrary f and g ∈ F and assume fRg. This means that there exists a c such that for all x ∈ A, f(x) - g(x) = c. We need to show gRf, i.e., there exists a c' such that for all x ∈ A, g(x) - f(x) = c'. Take c' = -c and fix an arbitrary x ∈ A. Then g(x) - f(x) = -(f(x) - g(x)) = -c, as required.