$a\Big | b,\; b = ak.$ $a\Big|c, c = al,$

So do I multiply $b$ and $c$ to get $a(kl)$ to prove that $bc = a$ multiplied by some integer $kl$ closed under multiplication?

  • 3
    $\begingroup$ Welcome to stackexchange. This is not true. $6$ divides $3 \times 4$ but divides neither factor. Make sure you have correctly stated what you have been asked to prove. edit the question to clarify, don't use comments. And use mathjax: math.meta.stackexchange.com/questions/5020/… (as some editors have done). $\endgroup$ – Ethan Bolker Mar 13 at 1:56
  • $\begingroup$ Need to assume $a$ is prime to show this, then it's a fairly well known property of primes. $\endgroup$ – coffeemath Mar 13 at 2:39

$6 \mid 2\cdot 3, 6 \not \mid 2, 6 \not \mid 3 $.


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