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

  • Thread starter Thread starter MI5
  • Start date Start date
AI Thread Summary
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.
MI5
Messages
8
Reaction score
0
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.
 
Mathematics news on Phys.org
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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top