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

MI5

New member
Sep 8, 2013
8
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.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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.
 

MI5

New member
Sep 8, 2013
8
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?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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?
 

MI5

New member
Sep 8, 2013
8
WOW! All I can say is thanks. (Giggle)