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

  • MHB
  • Thread starter MI5
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In summary, the theorem states that if c divides ab and b and c are coprime, then c also divides a. This can be proven by considering the fact that ac = a * c, and by the definition of the | (divides) relation, c must also divide ac.
  • #1
MI5
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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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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
WOW! All I can say is thanks. (Giggle)
 

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

How does the statement "c|ab" relate to the statement "gcd(b, c) = 1"?

The statement "c|ab" means that c is a common factor of both a and b, while the statement "gcd(b, c) = 1" means that b and c have no common factors other than 1. This suggests that c is a prime factor of a, since it is the only factor that is common between a and b.

Why is it necessary to specify that gcd(b, c) = 1 in this statement?

The condition gcd(b, c) = 1 is necessary in order to show that c is a prime factor of a. If gcd(b, c) was not equal to 1, then c would not be a prime factor of a and the statement would not hold true.

Can you give an example of when c|ab and gcd(b, c) = 1 are both true?

Yes, for example, if a = 12, b = 3, and c = 4, then c|ab is true since 4 is a factor of 12 and 3, and gcd(b, c) = 1 is also true since the only common factor of 3 and 4 is 1.

Is the statement "If c|ab and gcd(b, c) = 1, then c|ac" always true?

Yes, the statement is always true. This is because if c is a factor of both ab and ac, then it must also be a factor of a (since c is a common factor of both ab and ac, it can be factored out of both terms leaving c multiplied by a). This is known as the transitive property of divisibility.

How does this statement relate to the concept of prime factorization?

This statement is closely related to the concept of prime factorization, which states that every positive integer can be expressed as a unique product of prime numbers. Since gcd(b, c) = 1 in this statement, it shows that c is a prime factor of a, which aligns with the concept of prime factorization.

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