$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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'][1]
is claimed 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 $[\frac{n}{2}]$ (due to the text that's 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 conclude that the rank of 
 $ \SO(n)(\mathbb{Q}_5)$ is $[\frac{n}{2}]$.


  [1]: https://link.springer.com/book/10.1007/978-3-0346-0332-4