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I'm trying to prove this by contradiction but I'm really struggling to find the contradiction position for my proof. The idea is to assume the statement is false and show the idea leads to a contradiction, then conclude that it must be true. Should I start with "Suppose that $∤b$ and $() \mid ()$"?

Bill Dubuque
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Oscar
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  • may be start with $a \mid b$ and $c \nmid d$ but $ac \mid bd$ – Math Lover Oct 13 '20 at 18:33
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    A proof by condradiction would be to assume $a|b$ and $b|c$ and $ab\not \mid cd$. But I don't think that the $ab\not \mid cd$ will help you much and you might as well do a direct prove (which is easy). You could do a prove by contrapositve which assumes $ab\not\mid cd$ and conclude that either $a\not \mid b$ or $c\not \mid d$. But taking the negative statement that $ab\not \mid cd$ is really hard to draw conclusions from (because it doesn't tell you any thing that is, just want isn't) and drawing conclusions from $a|b$ and $c|d$ is pretty easy. I strongly advice a direct proof. – fleablood Oct 13 '20 at 21:52
  • This question is clearly different from the dupicated propose. The main question is totally different. Voting to reopen. – L F Oct 13 '20 at 22:06
  • @LuisFelipe Contrapositive statements are equivalent, not "totally different". – Bill Dubuque Oct 14 '20 at 00:50
  • TOTALLY DIFFERENT bro, because he asking for one way of proof, not its equivalence – L F Oct 14 '20 at 05:11

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No, You want to prove that $c \not\mid d$. If you want proof by contradiction, assume the opposite, i.e., $c\mid d$. Then use that to show that it is not possible, which implies our assumption was wrong.

$\blacksquare $ Solution:(Proof by Contradiction). On the contrary, let's assume that $c\mid d$. Thus we have $k_1, k_2 \in \mathbb{Z}$ such that $$ b = ak_1 \quad \text{and} \quad d = ck_2 $$ This implies $(bd) = (ac) \cdot (k_1k_2) \overset{\text{not.}}{=} (ac) \cdot k \implies ac \mid bd$. A contradiction!

Hence, our assumption was false, i.e., $c \not \mid d$.

Ralph Clausen
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