Problem Prove that $-(-v) = v$ for every $v \in V$ Solution By direct computation: $$-(-v) \stackrel{T1.31}{=} (-1)(-v) \stackrel{T1.31}{=} (-1)\left((-1)v\right) \stackrel{P1A13}{=} \left((-1)(-1)\right)v = 1v \stackrel{D1.19}{=} v$$ Proving that the additive inverse of an additive inverse of $v$ is $v$ itself.

Problem Show that $(a + b)x = ax + bx$ for all $a,b \in \mathbb{F}$ and all $x \in \mathbb{F}^n$ Solution Multiplication of elements in $\mathbb{F}^n$ by scalars is defined coordinate-wise and from problem 9 we know that multiplication of elements in $\mathbb{F}$ is distributive. Therefore: $$(a + b)x \stackrel{D1.10}{=} (a + b)(x_1, x_2, \dots, x_n) $$ $$\stackrel{D1.17}{=} ((a + b)x_1, (a+b)x_2, \dots, (a+b)x_n)$$ $$\stackrel{E1.4 \land P1A9}{=} (ax_1 + bx_1, ax_2 + bx_2, \dots, ax_n + bx_n)$$ $$ \stackrel{D1....

Problem Show that $\lambda (x + y) = \lambda x + \lambda y$ for all $\lambda \in \mathbb{F}$ and all $x, y \in \mathbb{F}^n$ Solution Multiplication of an element in $\mathbb{F}^n$ by a scalar is defined coordinate-wise and by problem 9 we know that multiplication in $\mathbb{F}$ is distributive, therefore: $$\lambda (x + y) \stackrel{D1.10 \land D1.12}{=} \lambda (x_1 + y_1, x_2 + y_2, \dots, x_n + y_n)$$ $$\stackrel{D1.17}{=} (\lambda (x_1 + y_1), \lambda (x_2 + y_2), \dots, \lambda (x_n + y_n))$$ $$\stackrel{P1A9}{=} (\lambda x_1 + \lambda y_1, \lambda x_2 + \lambda y_2, \dots, \lambda x_n + \lambda y_n)$$ $$\stackrel{D1....

Problem Show that $1x = x$ for all $x \in \mathbb{F}^n$ Solution Multiplication by scalar is done coordinatewise, and since $1a = a$ for all $a \in \mathbb{F}$ we must have that: $$1x \stackrel{D1.10}{=} 1(x_1, x_2, \dots, x_n) \stackrel{D1.17}{=} (1x_1, 1x_2, \dots, 1x_n) = (x_1, x_2, \dots, x_n) \stackrel{D1.10}{=} x$$ Proving that $1$ is a multiplicative identity for $\mathbb{F}^n$.

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....