I'm taking a discrete mathematics course in my computer science studies and came across the following:
Given the relation: $$R = \{(a, b)\in\Bbb Z\ \times\Bbb Z\ | \exists z\in\ \Bbb Z\ : a - b = z\cdot p \}$$
with $p\in \Bbb N_0\;$,
prove whether or not the relation is reflexive, symmetrical, antisymmetrical or transitive.
As I understand the question I only have to provide one z that satisfies this condition for each property. Assuming that is the case I found the following answers:
The relation is:
reflexive, since $z = 0$ satisfies $a - a = z \cdot p$
$^1$symmetrical, since $a-b = z \cdot p \implies b - a = z \cdot p$ is true for $z = 0$
anti-symmetrical, since $ a- b = z \cdot p \land b-a = z \cdot p \implies a = b$ is true for $z = 0$
transitive, since $a-b = z\cdot p \land b -c = z\cdot p\implies a = c$ is true for $z = 0$
However, I think I commit some very basic fallacies, because I completely ignore p. Additionally, if we assume that R is not transitive, would I have to prove it for every z?
Edit: Forgot to ask a question:
Are these answers correct and if not why?
The set of evens is the set of integers such that there exists an integer $k$ such that $2k$. Your argument is analogous to saying the set of evens has the "all of them are squares" property because $4$ is a square.
Since there are evens other than the one ($4$) with $k=2$ and ordered pairs in the relation other than the ones with $z=0$, these arguments are generally invalid for general properties that say something holds for every two or three pairs in the relation (but reflexivity is okay); you were right to be suspicious.
For each property, you have to either show that it holds for all pairs in the relation or find a counterexample (which may only involve one or two $z$s). As a hint, your answers for some properties might depend on the value of $p$.