Proof (Divisibility): If $a \mid b$ and $b\mid c$ then $a \mid c$ Ok, here is what I have for the proof of this conjecture.  Let me know if I'm on the right path? all input appreciated.
There exist integers $j$, $k$, and $m$, such that, 
$b = aj $ and $ c = ajk.$  Then $c = ajk $  (substituting $aj$ for $b$)
let $m = jk$, then $c = ma, => a|c.$ 
 A: Yes, $b=aj$ and $c=bk=ajk=(jk)a$. Since $jk \in \mathbb{Z}$, we have $a|c$.
A: Note that,
$$a|b \implies b=ma \,,$$
and 
$$ b|c \implies c =nb \implies c = nma \implies c = q a \implies a|c \,,$$
for $m,n,q \in \mathbb{Z}$ and $q=nm.$ 
A: (In this answer all variables are integers, i.e., elements of $\mathbb{Z}$.)
By the definition of divisibility we are given that $\;n * a = b\;$ and $\;m * b = c\;$ for some $\;n\;$ and $\;m\;$.  Now we are asked to find a $\;k\;$ which makes $\;k*a = c\;$:
\begin{align}
& k*a = c \\
\equiv & \;\;\;\;\;\text{"use the only fact we know about $\;c\;$"} \\
& k*a = m*b \\
\equiv & \;\;\;\;\;\text{"use the other fact"} \\
& k*a = m*n*a \\
\Leftarrow & \;\;\;\;\;\text{"weaken using Leibniz' rule -- to achieve our goal"} \\
& k = m*n \\
\end{align}
Therefore we have found such a $\;k\;$, and hence proved $\;a|c\;$.
A: Since you given a/b, this means b=ka for some k-interger. Also b/c implies that c=lb for some interger l. now, substituting b-ka into c=lb gives
              c=lka.....>>>c=lk(a)>>>>we conclude a can divide c

Hence a/c
N:B_ the / is not a mathematical symbol for "divides" but rather use |...its just that i like it...lol
