Yes this is a homework problem but I have attempted to solve it and my work is below, also this is my first question here so I'm sorry for any mistakes:
Question:
Context:
Let $A$ and $B$ be subsets of $\Bbb{Z}$, and let $F = \{f : A\to B\}$. Define a relation $R$ on $F$ by: for any $f,g\in F$, $fRg$ 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 $x\in A$.
Assume that $A=\{1,2,3\}$ and $B=\{1,2,\ldots,n\}$ where $n \geq 2$ is a fixed integer
Actual question:
Prove that for all $g,h \in F$ so that $gRh$, if there exists $a,b \in A$ so that $g(a) = 1$ and $h(b) = 1$ then $g = h$.
Attempted solution:
By definition $g(x) - h(x) = c$, from which we can write that $g(a) - h(a) = c$ and $g(b) - h(b) = c$. Now if we plug in the initial values: $1 - h(a) = c$ and $g(b) - 1 = c$. Next I equated them: $1 - h(a) = g(b) - 1$. And this is where I'm stuck, I know that in order for that to be true: $c = 0$, but I know that starting a proof with what I want to prove is incorrect so I can't use that in my reasoning. And where I am right now seems a bit useless for me...
I'm not asking for answers, any insight will be helpful, thank you!
Thanks! (And sorry if I made it complicated :-/
You just need to employ the properties of your range $B$. From your arguments, we have $1-h(a) = c$ and $g(b) -1 = c$. Additionally, we know that $h(a) \geq 1$, and $g(b) \geq 1$. Use these facts to conclude that $c=0$ indeed.
In more detail, since $h(a) \geq 1$, we have $c = 1-h(a) \leq 0$, and ...