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Question: Let $m, n, k \in \mathbb{Z}$ with $\operatorname{gcd}(m, n) = 1 = \operatorname{gcd}(n, k)$

Determine if it is true that necessarily $\operatorname{gcd}(m, k) = 1$.

Answer: The statement is FALSE.

Proof : [By Contradiction]

Let $m=3, n=5, k=6$.

The $\operatorname{gcd}(3,5)=1=\operatorname{gcd}(5,6)$ , then the $\operatorname{gcd}(3,6)\neq1$ .

Hence, the statement is FALSE.

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    $\begingroup$ Your proof is ok. You could even set $k=m$ ! $\endgroup$ – Evariste Aug 23 '16 at 11:32
  • $\begingroup$ It's not exactly a proof by contraduction (= reductio ad absurdum). It's only a counter-example (which is well enough, anyway). $\endgroup$ – Bernard Aug 23 '16 at 11:42
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Your proof is correct. Another example would be $m=k=2$ and $n=1$.

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