If a divides bc, then a divides b or a divides c

[Dustinsfl]

If $$a|bc$$, then $$a|b$$ or $$a|c$$.

$$a|bc\Leftrightarrow am=bc\Leftrightarrow a\left(\frac{m}{b}\right)=c$$

Hence, $$a|c$$

 

[LCKurtz]

If $$a|bc$$, then $$a|b$$ or $$a|c$$.

$$a|bc\Leftrightarrow am=bc\Leftrightarrow a\left(\frac{m}{b}\right)=c$$

Hence, $$a|c$$

That statement is false as stated. Let b = 6, c = 10. Then a = 4 divides bc = 60 but a doesn't divide either b or c.

 

[ocohen]

Your proof is false because you are dividing by b. This is incorrect since b could be 0.

 

[cap.r]

This is only true if b or c is prime. As a side note, your proof is false because you are dividing by b. This is incorrect since b could be 0.

he also doesn't know whether m/b will be an integer or not.

I am guessing you are reading an introductory number theory book. you should refrain from dividing as much as possible in your proofs. modding out is acceptable later when you start doing it, but don't divide unless you know that the denominator is NOT zero and the quotient is an integer.