Problem
Show that $(ab)x = a(bx)$ for all $x \in \mathbb{F}^n$ and $a,b \in \mathbb{F}$
Solution
Since multiplication of an element in $\mathbb{F}^n$ by a scalar is defined coordinate-wise, the result follows from the result shown in Problem 6 as such: $$(ab)x \stackrel{D1.10}{=} (ab) (x_1, x_2, \dots, x_n) \stackrel{D1.17}{=} ((ab)x_1, (ab)x_2, \dots, (ab)x_n)$$ $$\stackrel{P1A6}{=} (a(bx_1), a(bx_2), \dots, a(bx_n)) \stackrel{D1.17}{=} a(bx_1, bx_2, \dots, bx_n) \stackrel{D1.10 \land D1.17}{=} a(bx)$$ Proving the associativity of multiplication of an element in $\mathbb{F}^n$ by a scalar.