I am getting these questions from a book called, An Inquiry-Based Introduction to Proofs by Jim Hefferon. However, I do not understand what he means by "Generalise to any divisor.". I thought the divisor would always be 2, given we are talking about an even number.

My solution I have is simply this:
1. $a$ and $b$ are even.
2. $\exists k,m \in \mathbb{Z}: a = 2k$ and $b = 2m$.
3. $a \times b = 2k \times 2m = 2(2km) =$ even.

However, I am not sure if that satisfies the request to "Generalise to any divisor" because I don't really know what that means? Does anyone know? Thanks.

  • $\begingroup$ Replace "even" by "multiple of $2$" then you can generalize to any larger integer. $\endgroup$ – didgogns Jun 15 '17 at 14:37
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    $\begingroup$ You made a calculating error, $2k\times 2m=4km$ $\endgroup$ – User123456789 Jun 15 '17 at 14:37
  • $\begingroup$ $2k \cdot 2m \ne 2(k+m)$ although it is even. $\endgroup$ – NickD Jun 15 '17 at 14:37
  • $\begingroup$ You're right @Daan, thanks for the headsup. $\endgroup$ – Bucephalus Jun 15 '17 at 14:39

Note that in your answer, you should have $2k\cdot 2m=4mk=2(2mk)$, not $2(k+m)$. Otherwise, your proof is correct.

By generalizing to any divisor, it means that it wants you to prove the following:

Suppose $n$ and $m$ are both divisible by $k$. Prove that $n\cdot m$ is divisible by $k$.

  • $\begingroup$ Oh yeah that explanation is what I'm looking for. So if I want to go on and do this for n and m as you have stated, do I edit my original post or I answer my post? $\endgroup$ – Bucephalus Jun 15 '17 at 14:43
  • $\begingroup$ Well, you may accept whichever answer you feel answers your question best, including your own. Editing your own question might make hide the original purpose of it. $\endgroup$ – ervx Jun 15 '17 at 14:49
  • $\begingroup$ Yes, but I mean generally, sometimes someone will give you a hint, and not the answer, and so you have to continue with your solution. Should you edit and append it to your original post? Or is it preferred that you answer your own post with an updated version? $\endgroup$ – Bucephalus Jun 15 '17 at 14:51
  • $\begingroup$ If the hint helps you answer your question, you should probably accept that answer. There is no need to then post a full solution to your question. $\endgroup$ – ervx Jun 15 '17 at 14:53
  • $\begingroup$ ok, thanks @ervx $\endgroup$ – Bucephalus Jun 15 '17 at 14:53

$2p\times 2k=2^2pk$

$2\mid 2^2pk.$ QED

$np\times nk=n^2pk$

$n\mid n^2pk$. QED

  • $\begingroup$ Oh yeah, ok so that's what I should be putting if I were to add it. Thanks @RobertFrost $\endgroup$ – Bucephalus Jun 15 '17 at 14:44
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    $\begingroup$ @Bucephalus yeah, to generalise just means prove for the general case, i.e. for any divisor (in this case $n$), not just $2$. $\endgroup$ – user334732 Jun 15 '17 at 14:45

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