Let $G$ be a semisimple algebraic group and $K$ any field. Then the $K$-rank of $G$ is the maximal rank $r$ of a $K$-splitting torus $T \cong (K^*)^r $ contained as a subgroup in $G(K)$, the group of $K$-valued points of $G$.

In Alexander Lubotzky's 'Discrete Groups, Expanding Graphs and Invariant Measures' is clamed without proof on

I) page 27, 2nd line: that for every field $K$, the $K$-rank of $SL_n(K) $ is $n-1$

II) page 29, Example 3.2.4 (B): the $\mathbb{Q}_5$-rank of $ SO(n)(\mathbb{Q}_5)$ is $[\frac{n}{2}]$, since $x^2 \equiv -1$ has a solution in $\mathbb{Q}_5$

How to see these two statements? For I) we can obviously embed for every field $K$ the torus $(K^*)^{(n-1)} $ into $SL_n(K)$ via $(a_1,..., a_{n-1}) \mapsto \operatorname{diag}(a_1,..., a_{n-1}, \frac{1}{\prod_{i=1}^{n-1} a_i})$. How to show that it's not possible to embed in $SL_n(K)$ somehow a torus of bigger rank?

On II) I have no clue how to exploit that the congruence relation $x^2 \equiv -1$ has a solution in $\mathbb{Q}_5$, to show that the rank of $ SO(n)(\mathbb{Q}_5)$ is $[\frac{n}{2}]$.