1
$\begingroup$

Basically, prove or disprove that $ab|c \implies a|c$.

It seems like it should, but I don't know how to go about proving it.

$\endgroup$
  • 3
    $\begingroup$ If $ab \mid c$ then we can write $c = abn$ for some integer $n$. Now let us add some suggestive parentheses: $c = a(bn)$. What can you conclude? $\endgroup$ – Bungo Mar 25 '16 at 1:53
  • 1
    $\begingroup$ just write down the definition of dividing. $\endgroup$ – Melquíades Ochoa Mar 25 '16 at 1:55
5
$\begingroup$

ab|c means there exists an integer d such that (ab)d=c

So consider the integer bd now

a(bd)=c hence a divides c

$\endgroup$
2
$\begingroup$

Hint: Doesn't "divides" really mean "is a factor of"? If so, then aren't factors of the divisor also factors of the dividend?

Example: $100$ is a factor of $1000$. Because $100=4\times 25$, you immediately know that $4$ is a factor of $1000$, and $25$ is a factor of $1000$.

$\endgroup$
1
$\begingroup$

Hint:

Divisibility is a transitive relation.

$\endgroup$
0
$\begingroup$

By definition, $ab|c \Leftrightarrow c = abk$ for some $k \in \mathbb{Z}$. Similarly, $a|c \Leftrightarrow c = am$ for some $m \in \mathbb{Z}$. Let $m = bk$ and you are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.