Exercise 1.2.36 ($(a,b,c) = ((a,b),c)$)

Proof. Let $d=a\wedge b\wedge c$ and $\delta =(a\wedge b)\wedge c$.

By the characterization of the gcd (Theorem 1.4), we obtain

• From $d\mid a,d\mid b$, we infer $d\mid a\wedge b$. Moreover $d\mid c$, therefore $d\mid (a\wedge b)\wedge c=\delta$.
• Since $\delta \mid a\wedge b$, $\delta \mid a$ and $\delta \mid b$. Moreover $\delta \mid c$, therefore $\delta \mid a\wedge b\wedge c=d$.

From $d\mid \delta$ and $\delta \mid d$, where $d\ge 0,\delta \ge 0$, we deduce $d=\delta$, so

(Alternatively, we can use

Since $a\wedge b\wedge c\ge 0$ and $(a\wedge b)\wedge c\ge 0$, we obtain $a\wedge b\wedge c=(a\wedge b)\wedge c$.) □