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Prove by contradiction if $a\mid b$ then $ac\mid bc $ for $a,b,c\in\Bbb Z$

I am having considerable difficulty with this problem. Could someone give me a step-by-step solution? Even though the problem seems simple, I cannot solve it. I am new to discrete maths: I only know direct proof and contraposition (I am trying to work on contradiction). All I know is that $a\mid b$ if for some other number $c$, we have $b = a\times c$.

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    $\begingroup$ What are your thoughts about the problem? Can you give a proof that does not use contradiction? Do you know the meaning of the statement $a|b$? Give us something to work with. $\endgroup$
    – littleO
    Jun 26, 2018 at 1:03
  • $\begingroup$ Welcome to math.SE! I can relate to the fact that you are stuck on a math problem; it happens to me all the time and it can be frustrating, I know. However, regardless of your confusion, you will get more useful answers on this and future questions if you provide more context. A good first step would be to post what you have tried so far into the body of the post. Whether you have but a few words to say, or several paragraphs, you will get a better answer this way :) $\endgroup$
    – Crosby
    Jun 26, 2018 at 1:04
  • $\begingroup$ With my new edit, can you help ? $\endgroup$ Jun 26, 2018 at 1:29

2 Answers 2

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Start with not ac | bc. We wish to show not a | b.
Assume a|b. Easily prove ac | bc, a contradiction
from the starting premise. Thus not a|b.

It is pointless to prove this by contradiction
as a direct proof is easy and simple.

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  • $\begingroup$ Why is a contradiction ? $\endgroup$ Jun 26, 2018 at 12:47
  • $\begingroup$ @HenriqueN.Mendes. Proof by contradiction is a red herring because a direct proof in much easier. Certainly you can prove a|b implies ac|bc directly. $\endgroup$ Jun 26, 2018 at 20:46
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Assume $c>0 $, suppose on contrary $\forall k\in \mathbb Z$ $$ac\neq k bc.$$ Let $ac>kbc$. Then $a>kb$. similarly for $ac<kbc$. consider the other case also $c<0$. In each case we get $a $ not divides $b$.

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