Proving simple property of divisibility Want to confirm my proof for below problems on divisibility :
(a) If $a|b$ and $b|c$ then $a|c$
$\Rightarrow$ Given $b = ae$, and $c = bf$ for $e, f \in \mathbb{N}$. And can easily take case of negative integers as : $b = ae.(-1)$, and $c = bf(-1)$ for $e, f \in \mathbb{N}$. So, $c = aef$, hence a|c.
(b) if $a|b$ and $c|d$ then $ac|bd$
$\Rightarrow$ Given $b = ae$, $d = cf$ for $e, f \in N$. So, $bd = acef$; hence $ac|bd$
(c) if  $m\ne$0, then $a|b$ iff $ma|mb$
$\Rightarrow$ Proving the if part, i.e. $a|b$ if $ma|mb$. If $ma|mb$, then can eliminate common factor $m$ from both sides. As $m\ne 0$ , so if it is positive or negative, this automatically leads to $a|b$.
Proving the iff part, i.e. $ma|mb$ if $a|b$. Assume $b = ea$ for e $\in N$. 
Now, can multiply both sides by non-zero m (positive or negative) to still get the same result.
(d) Assume that $a,b,d,x$ and $y$ are integers. If $d|a$ and $d|b$, then $d|(ax+by)$. 
$\Rightarrow$ Given $a=md$ and $b=nd$, where $d\ne 0$ as $0$ divides none, but is divided by all. Trivially, $a$ can be $0$, and then $m=n=0$. Otherwise, $ax + by = (md)x + (nd)y = d(mx+ny)$, so $d|ax+by$.

Globally, it is correct. A few notes:
a) There is no need to treat negative integers as a special case.
c) You write that you can “can eliminate common factor $m$ from both side”. Well, that's what you are supposed to prove, right? It's easy, of courts: if there is an integer $c$ such that $cma=mb$, the, dividing both sides by $m$; you get that $ca=b$. Therefore, $a\mid b$.
For the proof in the opposite direction, the fact that $m\neq0$ is irrelevant.
d) There's no reason to treat the case $a=0$ as a special case.
