MHB If c|ab and gcd(b, c) = 1 why does c|ac?

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If c divides ab and the greatest common divisor of b and c is 1, then it follows that c divides a. The proof relies on the relationship between the products ab and ac, demonstrating that since c divides ab, it must also divide ac. The discussion highlights some confusion regarding the definition of the divisibility relation. Clarification is sought on the concept of divisibility, indicating a need for a deeper understanding of the theorem. Overall, the theorem is affirmed, emphasizing the importance of grasping the foundational definitions in number theory.
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Theorem: If c|ab and (b, c) = 1 then c|a.

Proof: Consider (ab, ac) = a(b, c) = a. We have c|ab and clearly c|ac so c|a. It's not so clear to me why c|ac. Perhaps I'm missing something really obvious.
 
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MI5 said:
It's not so clear to me why c|ac.
This is because ac = a * c, i.e., by the definition of the | (divides) relation.
 
Evgeny.Makarov said:
This is because ac = a * c, i.e., by the definition of the | (divides) relation.
I still don't understand I'm afraid. Could you say bit more please?
 
MI5 said:
I still don't understand I'm afraid. Could you say bit more please?
I need to be sure you know the definition of the relation denoted by |. Could you write this definition?
 
WOW! All I can say is thanks. (Giggle)
 

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