Let R be the relation defined on the set of ordered pairs $Z_+ ×Z_+$ of positive integers defined by $(a, b)R(c, d) ⇔ a + d = b + c$. Prove that the function $f : Z_+ ×Z_+/R → Z$, defined by $f([a, b]) = a−b$, is well-defined and a bijection.
Proof: (For Well-defined function): Let $(a,b)R(c,d)$
$\implies a+d = b+c$
$\implies a-b = c-d$ -----(1)
Since $(a,b)R(c,d) \implies [(a,b)]=[(c,d)]$
Now consider $f[(a,b)]$ and $ f[(c,d)]$
By definition, $f[(a,b)] = a-b $ and $ f[(c,d)] = c-d$
By (1), $\implies a-b = c-d$
Hence, the function is well-defined.
Proving Bijection:
1-1:
Let $f[(a,b)] = f[(c,d)]$
$\implies a-b = c-d$ $\implies a+d = b+c$
$\implies (a,b)R(c,d)$ $\implies [(a,b)]=[(c,d)]$
Hence, f is injective.
Can anyone please verify this? Also, I am unsure as to how to prove this is surjective, any hints?
Thank you.
Your proof seems to have the right ingredients, but it is poorly written, in my opinion.
Let just give a tide up version, and answer your last question.
You want to prove that $f$ is well defined. It is clear that each element has an image, so your task is to prove that if $[(a,b)] = [(c,d)]$ then $f([(a,b)]) = f([(c,d)])$.
If $[(a,b)] = [(c,d)]$, then $(a,b)R(c,d)$, and so $a+d = b+c$; from here it follows that $a-b=c-d$, that is, $f([(a,b)]) = f([(c,d)])$.
To show it is injective, if $f([(a,b)]) = f([(c,d)])$, then $a-b=c-d$, whence $a+d=b+c$, yielding $(a,b)R(c,d)$ and $[(a,b)]=[(c,d)]$.
It is also surjective.
For $k \in \mathbb Z$, if $k \leq 0$, then $k = f([(1,1-k)])$ (notice that $1-k > 0$); if $k > 0$, then $k=f([(k+1,1)])$.