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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 $c = ma, => a|c.$

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Upvoted for showing work, though I feel like I've seen this question 3x on this site! – The Chaz 2.0 Nov 12 '12 at 4:43
up vote 3 down vote accepted

Yes, $b=aj$ and $c=bk=ajk=(jk)a$. Since $jk \in \mathbb{Z}$, we have $a|c$.

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Note that,

$$a|b \implies b=ma \,,$$


$$ b|c \implies c =nb \implies c = nma \implies c = q a \implies a|c \,,$$

for $m,n,q \in \mathbb{Z}$ and $q=nm.$

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(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 $\;a|c\;$.

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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

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