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)
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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