Proof with Error, if a|b and b|c then a|c since b =ak and c=bk so c=bk=(ak)k=ak^2

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Having trouble figuring out the issue with the question in my textbook

Here is a sample proof that contains an error. Explain why the proof is not correct.

Theorem: If a\b and b\c, then a\c.

Proof: Since a|b, b = a*k.

Since b|c, c = b*k.

Therefore c = bk = (ak)k = ak*k = a(k^2)

Therefore a|c

From my understanding (ak)k = ak^2 checks out and properties of divisibility rules state that a|b means that b is some multiple of a. Is the issue with the proof that there is no "assumption" being made at the start or some kind of notation error?

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No. The error lies in the fact that the number $k$ such that $b=a\times k$ doesn't have to be such that $c=b\times k$. From the fact that $b\mid c$, what follows is that $c=b\times k'$ for some integer $k'$ (which doesn't have to be $k$). Therefore $c=a\times(k\times k')$. It follows that $a\mid c$.