(a) Prove that Axiom 10 can be deduced from the other axioms.
(b) Prove that Axiom 10 cannot be deduced from the other axioms if Axiom 6 is replaced by Axiom 6': For every $x$ in $V$ there is an element $y$ in $V$ such that $x + y = O$.
Refer to Theorems 15.2 and 15.3 of Section 15.4 (Volume 1):
$\text{Theorem 15.2. $\quad$ Uniqueness of Negative Elements.}\quad$ In any linear space every element has exactly one negative. That is, for every $x,$ there is one and only one $y$ such that $x + y = O.$
$\text{Theorem 15.3. $\quad$}$ In any given linear space, let $x$ and $y$ denote arbitrary elements and let $a$ and $b$ denote arbitrary scalars. Then we have the following properties:
(a) $0x = O$
(b) $aO = O$
(c) $(-a)x = -(ax) = a(-x)$
(d) If $ax = O,$ then either $a = 0$ or $x = O.$
(e) If $ax = ay$ and $a \neq 0,$ then $x = y.$
(f) If $ax = bx$ and $x \neq O,$ then $a = b.$
(g) $-(x + y) = (-x) + (-y) = -x - y.$
(h) $x + x = 2x,$ $x + x + x = 3x,$ and in general, $\sum_{i = 1}^n x = nx.$
(a) We know from Theorem 15.2 that there is exactly one $y$ such that $x + y = O,$ and we know from Axiom 6 that $y = (-1)x,$ giving us $x + (-1)x = O.$ If we multiply both sides by $(-1),$ we get $(-1)x + 1x = O.$ In other words, we have $$x + (-1)x = 1x + (-1)x.$$Which implies that $1x = x.\,\blacksquare$
(b) Let $x$ and $y$ be elements in $V$ such that $x + y = O,$ and let $a$ be a real scalar. Setting $a = 1$ and using the distributive law for addition in $V,$ we find that $1(x + y) = 1x + 1y.$ Since $V$ has closure under multiplication by real numbers, we know that $1x$ is an element in $V.$ However, it is not necessary that $1x + 1y \neq O,$ thus we cannot deduce that $1x = x$ for all $x$ in $V$ since in this case it would mean that $1O \neq O.\,\blacksquare$
