Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to prove that the $\gcd(a,b,c)$ = $\gcd(\gcd(a,b),c)$.

I think it has something to do with $\gcd$'s being able to be represented by a linear combination (that is $\gcd(a,b) = ax + by > 0$, for some integer $x$ and $y$).

Any help is appreciated!

share|cite|improve this question

Here is one proof:

If $d$ divides each of $a,b,c$ then $d|\gcd(a,b)$ and $d|c$. Conversely if $d|\gcd(a,b)$ and $d|c$ then $d$ divides each of $a,b,c$. Hence any number which divides $a$, $b$ and $c$ must also divide both $\gcd(a,b)$, $c$, and the converse is also true.

Since their divisors are identical, the greatest common divisor will also be the same, and we have $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$.

share|cite|improve this answer
That's essentially the fact that set intersection is associative: $D(a,b,c)=D(a) \cap D(b) \cap D(c) = (D(a) \cap D(b)) \cap D(c)$ = $D(a,b) \cap D(c) = D(\gcd(a,b))\cap D(c) = D(\gcd(a,b),c)$. – lhf Sep 22 '11 at 11:18

$$\begin{align*} d|\gcd(a,b,c) & \Longleftrightarrow d|a,b,c\\ &\Longleftrightarrow d|a,b\text{ and }d|c\\ &\Longleftrightarrow d|\gcd(a,b),c\\ &\Longleftrightarrow d|\gcd(\gcd(a,b),c). \end{align*}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.