I was given the following question to determine if it was reflexive, symmetric and transitive. Just wanted to make sure my explanations are acceptable or if they can be made stronger. Question: Suppose $\sim$ is defined on ordered pairs of integers as follows: $(a,b)\sim (c,d)$ iff there is an integer $u$ such that $c=au$ and $d=bu$.
Reflexive: $(a,b)\sim (a,b)$, then $a=au$ , $1=u$ so $a=a(1)$ and $b=b(1)$ therefore it is reflexive.
Symmetirc: $(a,b)\sim (c,d)$ then ($c,d)\sim (a,b)$
$c=au$ and $d=bu$, $a=cu$ and $b=du$
In each I divided by $u$, getting for instance $(1/u) c=a$, where $1/u$ is not a integer....(stuck with finishing it up)... However I know it is not symmetric
Transitive: $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)$ then $(a,b)\sim (e,f)$
$c=au$ and $d=bu$, $e=ck$ and $f=dk$, so, $e=auk$ and $f=buk$
So, $e=ck$ so $e=auk$ we substitute for $e$ getting $ck=auk$, cancel out our $k$....getting $c=au$ repeat the same thing $f=buk$ given $f=dk$ getting $d=bu$.
It's transitive.
Please forgive the disorder, any help or tips would be appreciative.
To prove reflexive, you want to show that $(a,b)~\sim(a,b)$:
Since $a=a(1)$ and $b=b(1)$ and $1 \in \mathbb{Z}$, it is reflexive.
For symmetric, if we have $(a,b) \sim (c,d)$, we want to check whether we have $(c,d) \sim (a,b)$ all the time.
An explicit example helps, for instance, $(1,1) \sim (0,0)$ as $0=0(1)$. However, the reverse direction is not true, we cannot find integer $u$ such that $1=u(0)$.Hence it is not symmetric.
For transitivity, if $(a,b) \sim (c,d)$ and $(c, d) \sim (e, f)$,
Hence $e=(uv)a$ and $f=(uv)b$, since $uv \in \mathbb{Z}$,
we have $(a,b) \sim (e,f)$. It is transitive.