The discriminants of the quadratic fields $K_{2}=\mathbb{Q}(\sqrt{2}),\ K_{3}=\mathbb{Q}(\sqrt{3})$ and $K_{{13}}=\mathbb{Q}(\sqrt{13})$ are $D_{{K_{2}}}=8,\ D_{{K_{3}}}=12$ and $D_{{K_{{13}}}}=13$ respectively. For all three $n=2=r_{1}$ and $r_{2}=0$. Therefore, the Minkowski’s constants are:

so in the three cases:

Now, suppose that $C$ is an arbitrary class in $\operatorname{Cl}(K_{i})$. By the theorem, there exists an ideal $\mathfrak{A}$, representative of $C$, such that:

and therefore ${\bf N}(\mathfrak{A})=1$. Since the only ideal of norm one is the trivial ideal $\mathcal{O}_{{K_{i}}}$, which is principal, the class $C$ is also the trivial class in $\operatorname{Cl}(K_{i})$. Hence there is only one class in the class group, and the class number is one for the three fields $K_{2},\ K_{3}$ and $K_{{13}}$.

Let $K=\mathbb{Q}(\sqrt{17})$. The discriminant is $D_{K}=17$ and the Minkowski’s bound reads:

Suppose that $C$ is an arbitrary class in $\operatorname{Cl}(K)$. By the theorem, there exists an ideal $\mathfrak{A}$, representative of $C$, such that:

and therefore ${\bf N}(\mathfrak{A})=1$ or $2$. However,

so the ideal $2\mathcal{O}_{K}$ is split in $K$ and the prime ideals

are the only ones of norm $2$. Since they are principal, the class $C$ is the trivial class, and the class group $\operatorname{Cl}(K)$ is trivial. Hence, the class number of $\mathbb{Q}(\sqrt{17})$ is one.