Can I Use Factor Sets to Solve GCD Problems?

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SUMMARY

The discussion focuses on proving that if gcd(a, b) = 1, then gcd(ac, b) = gcd(c, b). The user attempts to demonstrate this by defining d as gcd(ac, b) and g as gcd(c, b), aiming to show that g divides d and vice versa. The approach involves using linear combinations and factor sets, specifically the unique prime factor sets A, B, and C for a, b, and c respectively. The conclusion emphasizes that the factor set method is a reliable technique for establishing the relationship between the gcd values.

PREREQUISITES
  • Understanding of the greatest common divisor (gcd) concept
  • Familiarity with prime factorization and unique prime factor sets
  • Knowledge of linear combinations in number theory
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of gcd and its applications in number theory
  • Learn about unique prime factorization and its implications for gcd calculations
  • Explore linear combinations and their role in proving gcd relationships
  • Investigate advanced gcd algorithms and their computational efficiency
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Mathematics students, educators, and anyone interested in number theory, particularly those studying gcd properties and factorization techniques.

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Homework Statement


if 1 = gcd(a,b), show that gcd(ac,b) = gcd(c,b)

Homework Equations


None

The Attempt at a Solution


My attempt at a solution:

Let d = gcd(ac,b),
Let g = gcd(c,b),
I want to show that g|d and that d|g. I then went on to make a bunch of circular writing and get nowhere... I set up things like:

1 = ax + by
d = acw + bz --> 1 = (ac/d)w + (b/d)z
g = cn + bm --> 1 = (c/d)n + (b/d)m

Is my approach here a solid method? I can't think of any other way to show that gcd(ac,b) = gcd(c,b) besides assigning each of them a value and showing that they divide each other.
 
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That seems to be a reliable method.
I prefer to work with factor sets:
Let A be the unique prime factor set of a, B be the unique prime factor set of b, and C be the unique prime factor set of c.
##A\cap B =\{1\}##
##AC=A \cup C##
##AC\cap B= (A \cup C) \cap B = (A\cap B) \cup (C \cap B) = 1 \cup (C \cap B) ##
A similar argument should apply with repeated prime factors and combinations to find the gcd.
 

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