Exercise 1.7

Proof. Suppose $d\ne 0$ (if $d=0$, then $a=b=0$, and $a^{\prime },b^{\prime }$ are any numbers in $\mathbb{Z}$ and the result may be false, so we must suppose $d\ne 0$).

As $d=\mathit{am}+\mathit{bn},m,n\in \mathbb{Z}$, $d=d(a^{\prime }m+b^{\prime }n)$, so $1=a^{\prime }m+b^{\prime }n$, which proves $a^{\prime }\wedge b^{\prime }=1$.

Conclusion : if $d=a\wedge b\ne 0$, and $a=d{a}^{\prime },b=d{b}^{\prime }$, then $a^{\prime }\wedge b^{\prime }=1$. □