Prove that $\operatorname {gcd}(\operatorname {gcd}(m,n), p) = \operatorname {gcd}(m,\operatorname {gcd}(n, p))$

Question

Prove that $$\operatorname {gcd}(\operatorname {gcd}(m,n), p) = \operatorname {gcd}(m,\operatorname {gcd}(n, p))$$

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How to begin ?

Answer 1

The simplest way to prove it is the following. Let $d_1=\operatorname {gcd}(\operatorname {gcd}(m,n), p)$ and $d_2=\operatorname {gcd}(m, \operatorname {gcd}(n,p))$. Consider $d_1$. Then we have: $d_1$ divides both $\operatorname {gcd}(m,n)$ and $p$, hence $d_1$ divides $m, n$. Further, $d_1$ divides $p$, then $d_1$ divides $\operatorname {gcd}(n, p)$. It follows that $d_1$ divides $(m, \operatorname {gcd}(n,p))$, so it divides $d_2=\operatorname {gcd}(m, \operatorname {gcd}(n,p))$. In particular, $d_1$ divides $d_2$. Now exchange the roles of $d_1, d_2$ and conclude that $d_2$ divides $d_1$, which implies $d_1=d_2$.

Answer 2

This is a special case of a much more general result, and is best proved at a higher level of generality.

Proposition. Let $P$ denote a poset. Let $a,b$ and $c$ denote elements of $P$, such every subset of $\{a,b,c\}$ has a join. Then: $$(a \vee b) \vee c = a \vee (b \vee c).$$

We then get the desired result in the special case where $P$ is the poset $(\mathbb{N},\mid^{op})$, in which case $\vee$ is just the $\mathrm{gcd}$. (If you don't know what the hell I'm talking about, I suggest getting yourself a good book on the matter. Order theory is both beautiful and useful. Davey and Priestley's book is a pretty good place to start.)

Proof.

Consider $k \in P$. Then the following are equivalent:

• $k \geq (a \vee b) \vee c$
• $k \geq (a \vee b)$ and $k \geq c$
• $k \geq a$ and $k \geq b$ and $k \geq c$
• $k \geq a$ and $k \geq (b \vee c)$
• $k \geq a \vee (b \vee c)$

Hence: $$\forall k \in P(k \geq (a \vee b) \vee c \iff k \geq a \vee (b \vee c))$$

Hence: $$(a \vee b) \vee c = a \vee (b \vee c)$$

Like every proof in order theory, the above is based on the "Yoneda lemma for posets":

Proposition. Let $P$ denote a partially ordered set. Let $a$ and $b$ denote elements of $P$. Then: $$a \geq b \iff \forall k \in P(k \geq a \implies k \geq b)$$

This implies

Corollary. Let $P$ denote a partially ordered set. Let $a$ and $b$ denote elements of $P$. Then: $$a = b \iff \forall k \in P(k \geq a \iff k \geq b)$$

Answer 3

Hint $\ d\mid (m,n),p \iff d\mid m,n,p\iff d\mid m,(n,p)\ $ so $\,(m,n),p\,$ and $\,m,(n,p)\,$ have the same set $\,S\,$ of common divisors, hence the same greatest common divisor $(= \max S)$.

Above we employed twice the gcd universal property $\, d\mid (a,b)\iff d\mid a,b$.