Exercise 1.2.33 ($(a,b) = (a,b, ax +by)$)

Proof. Let $d=a\wedge b$ and $\delta =a\wedge b\wedge (\mathit{ax}+\mathit{by})$.

• By the characterization of the gcd (Theorem 1.4),
  $\delta \mid a$ and $\delta \mid b$, thus $\delta \mid a\wedge b=d$.
• Since $d\mid a,d\mid b$, then $d\mid \mathit{ax}+\mathit{by}$, therefore $d\mid a\wedge b\wedge (\mathit{ax}+\mathit{by})=\delta$.

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

For $x=y=1$, we obtain $a\wedge b=a\wedge b\wedge (a+b).$ □