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If you have three natural numbers a, b and c.

How can you prove that for (bc), if a does not divide b and it does not divide c, then it also does not divide (bc)?

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    $\begingroup$ You can't. $4 \nmid 6$ and $4 \nmid 10$, but $4 \mid 60$. $\endgroup$ – Daniel Fischer Feb 1 '17 at 14:23
  • $\begingroup$ ok. Now let's add that all three numbers are prime numbers. Would the statement then be true and if yes why? $\endgroup$ – Kevin Wu Feb 1 '17 at 14:26
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    $\begingroup$ It suffices to require that $a$ is prime. Depending on context, that can be essentially the definition of prime, or Euclid's lemma. $\endgroup$ – Daniel Fischer Feb 1 '17 at 14:28
  • $\begingroup$ What if b and c are prime, how can you prove that a, also prime, does not divide the product of b and c? $\endgroup$ – Kevin Wu Feb 1 '17 at 14:44
  • $\begingroup$ What is the definition of a prime you're using? $\endgroup$ – Daniel Fischer Feb 1 '17 at 14:47

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