0

How do I prove that If $d\mid a, d\mid b$, and $d\mid c$, then $d\mid \gcd(a,b,c)$ without using gcd(a,b,c)=gcd(gcd(a,b),c)?

I understand that the Common Divisor Divides GCD is only defined for two variables.

Bill Cui
  • 11
  • 3
    Duplicate of Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$., which proves both associativity and the sought universal property. The proof obviously extends n-ary gcds. – Bill Dubuque Oct 22 '19 at 15:32
  • See also $d\mid a,b \iff d\mid\gcd(a,b) \ $ [GCD Universal Property] – Bill Dubuque Oct 22 '19 at 15:33
  • Re: your edit "without using...": the linked proof in the dupe doesn't use that (associativity) rather it uses Bezout. Here is a proof of Bezout for any number of arguments. – Bill Dubuque Oct 22 '19 at 15:39
  • The GCD of $n>0$ integers is defined to be their greatest common divisor - same as when $n = 2$. – Bill Dubuque Oct 22 '19 at 15:49
  • "I understand that the Common Divisor Divides GCD is only defined for two variables." Why do you say that? "common divisor divides GCD" is not a definition. It's a provable statement of fact. And you can prove it for any number of terms the exact same way. If $d< \gcd(a,b,c)$ is a common divisor $d\not \mid \gcd(a,b,c)$ then $\gcd(gcd(a,b,c), d)*\gcd(a,b,c)$ is a common divisor contradicting that $\gcd(a,b,c)$ is the greatest. The number of terms with a common divisor have nothing relevant to do with it. – fleablood Oct 22 '19 at 20:20
  • @fleablood Correct is: $,g,d\mid a,b,c\iff [g,d]\mid a,b,c,,$ and $,[g,d]=gd/(g,d),,$ not $,(g,d)g,,$ i.e. common divisors are closed under lcm (which increases for divisibly-incomparable args) – Bill Dubuque Oct 24 '19 at 03:52

0 Answers0