For Problem D-1:

(a) If $r \neq 0$, then since scalars are in a field we can multiply the equation $rv =$   0 on both sides by $r^{-1}$ on the left, obtaining $r^{-1}(rv) = r^{-1}$0. By laws (f) and (e) and the lhs (left-hand-side) is $(r^{-1}r)v = 1v = v$, while the rhs is 0 by (i). Therefore we get $v =$   0. In other words, if $r$ isn't the zero scalar then $v$ is the zero vector, so one of them is zero, as stated.

(b) The outline is

$$r(v-w) = rv - rw =$$   0$$\Rightarrow r(v-w) =$$   0$$\Rightarrow v - w =$$   0$$\Rightarrow v = w,$$

where $r$ is canceled using part (a) (not law (a)).

For a full proof we need to justify our use of familiar algebra for binary minus, since binary minus and laws for it are not in our definition of a vector space. Instead, binary minus is defined outside the definition of a vector space by $v - w = v + (-w)$. We do have laws such as $r(-w) = -(rw)$ because by laws (k) and (f) we have $r(-w) = r((-1)w) = (r(-1))w = ((-1)r)w = (-1)(rw)$.

The first equation in the outline above comes from $r(v-w) = r(v + (-w)) = rv + r(-w) = rv + (-(rw)) = rv -rw$. The last implication in the outline comes from adding $w$ to both sides: $v - w =$   0$\Rightarrow (v + (- w)) + w =$   0$+ w \Rightarrow v + (-w + w) = w \Rightarrow v = w$, where various defining laws have been used.

(c) In outline,

$$rv = sv \Rightarrow rv - sv =$$   0$$\Rightarrow (r-s)v =$$   0$$\Rightarrow r-s = 0 \Rightarrow r = s.$$

The justification of using familiar laws for binary minus is as in part (b).

Notice that in the original handout there was a misprint in law (h), which should say $(r+s)v = rv + sv$.

For Problem D-2:

We can define $-v$ to mean $(-1)v$. Then $v + (-v) = 1v + (-1)v$ by this definition and by (e), which equals $(1 + (-1))v$ by (h), which equals $0v$ by the fact that $1 + (-1) = 0$ in a field, which equals 0 by (j). We have shown $v + (-v) = \ldots =$   0, which is (d).