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

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

Yadati Kiran
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Youcantseemenow
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    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: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference (as some editors have done). – Ethan Bolker Mar 13 '19 at 01:56
  • Need to assume $a$ is prime to show this, then it's a fairly well known property of primes. – coffeemath Mar 13 '19 at 02:39

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

marty cohen
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