I have been stuck on this question for quiet a bit of time. I'm not sure how to prove a relation $R$ is a partial order of $X$.
Let $\mathbb{Z}$ be the set of integers and consider the set $X = \mathbb{Z} \times \mathbb{Z}$ Consider the realtion $R$ on $X$ defined by
$$ (x,y)R(z,w) $$
if: $$ x < z $$ or $$ x = z \land y \le w$$
At most I know how to show the relation $R$ is reflexive.
If $x = z$ and $y \le w$ $\Rightarrow x = y \Rightarrow R$ is reflexive.
I can't seem to be able to wrap my head around how one would prove that $R$ is antisymmetric or transitive. Any help or advice would be really appreciated.
With transitivity you have the initial hypothesis: $$ (x,y) R (z,w) \land (z,w) R (a,b) $$ There are a total of four scenarios in which the hypothesis can be satisfied as per the relation.
You must prove under this hypothesis and in each scenario that $(x,y) R (a,b)$.
For example, let's look at one of the scenarios for: $$ (x,y) R (z,w) \land (z,w) R (a,b) $$ One case, might be that $x<z$ and $z<a$, in which case by transitivity of $<$, you have that $x<a$ and therefore, $(x,y) R (a,b) \checkmark$
I'll leave it to you to examine the other scenarios under which the hypothesis could be satisfied.
For antisymmetry, the hypothesis is that two ordered pairs are related symmetrically: $$ (x,y) R (z,w) \land (z,w) R (x,y) $$ It is not that case $x<z \land z<x$, so assume $x=z$. This, in turn, yields $y\le w$ and $w\le y$ which indeed shows $w=y$.