The question is If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$.

My attempt is that $b \equiv 0 \pmod a$ can be written $a\mid b-0 = a\mid b$ and the same with $c \equiv 0 \pmod b$ can be written $b\mid c-0 = b\mid c$ then the $c \equiv 0 \pmod a$ can also be written $a\mid c-0 = a\mid c$.

Now all I have to show is that $a\mid b$ and $b\mid c$ then $a\mid c$ and to show $a\mid c$ I can write $c= ak$ for some integer $k$.

Now I start my proof
let $a\mid b$ be $b=a\cdot r$ for some integer $r$ (1)
let $b\mid c$ be $c=b\cdot s$ for some integer $s$ (2)

Sub (1) into (2) and get
$c=(ar)\cdot s$
$c=a\cdot(r\cdot s)$
so $c=a\cdot k$
therefore the statement $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$ is true.

I'm wondering if this is a good/correct proof? any input would be appreciated.


You did not define what $k$ is, so no, your proof is not correct. You simply state

So $c=ak$.

What is $k$? How did you prove that $c=ak$?

  • $\begingroup$ @Ganymede The OP asked "is this proof correct?". I answered him (the answer is no) and explained to him where the answer is wrong. This should motivate him to improve his answer. On the other hand, you actually wrote a mathematically false thing in your answer, thus I downvoted it. Now you fixed your answer and I removed the downvote. I don't see why you should downvote my answer. $\endgroup$ – 5xum Aug 26 '15 at 7:52
  • $\begingroup$ He did define what $k$ is. It is the integer for which $c=ak$. That comes directly from the definition of $a\mid c$. $\endgroup$ – celtschk Aug 26 '15 at 7:55
  • $\begingroup$ @celtschk If I have a proof in which I want to prove the statement: "there exists such a $k$ that $c=ak$", and I say "let $k$ be the number such that $c=ak$", then the proof is incorrect. You cannot assume that the thing you are trying to prove exists. $\endgroup$ – 5xum Aug 26 '15 at 8:00
  • $\begingroup$ @celtschk The proof is almost entirely correct, but is missing one crucial line: "then, if we set $k$ to be $k=*$, we can see that $c=ak$". Without that line (with something written in place of the asterix), the proof is incorrect. $\endgroup$ – 5xum Aug 26 '15 at 8:01
  • $\begingroup$ He rewrites the statement "$a\mid c$" he wants to prove into "$c=ak$ for some $k$". That's nothing but another way to state the same. The statement "$c=ak$ for some $k$" is the statement he's going to proof. Note that he writes "Now I start my proof" below that line, so that line is not even part of the proof, so how can it make the proof wrong? $\endgroup$ – celtschk Aug 26 '15 at 8:05

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