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Math Help - Proving an equivalence relation.

  1. #1
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    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.
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  2. #2
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    Re: Proving an equivalence relation.

    So, what seems to be the problem?
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    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?
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    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.
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    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.
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    Re: Proving an equivalence relation.

    Quote Originally Posted by hcuk View Post
    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)~?
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  7. #7
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    Re: Proving an equivalence relation.

    Quote Originally Posted by hcuk View Post
    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.

    Quote Originally Posted by hcuk View Post
    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.

    Follow Plato's hint on transitivity.

    You may notice that this problem requires proving propositions as well as using assumptions of the form "for all x, A(x)", "there exists an x such that A(x)" and "A implies B". I recommend going over how to do this again.
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