Exercise 3, § 3

Let $\mathfrak{b}$ be a Borel subalgebra of $\mathfrak{g}$, and let $l=$ rk($\mathfrak{g}$). Show that the minimum number of generators of the algebra $\mathfrak{b}$ is $l$ if $l\not= 1$, and 2 if $l= 1$.