Mathematical Immaturity

1.5 Exercises

• Prove parts (d) through (h) of Theorem 1.3.
• Recall Theorem 1.3. (Section 1.4) $\text{Theorem 1.3. $\quad$}$ 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, $\sum_{i = 1}^n x = nx.$
• $Proof.$ (d) Let $ax = O.$ Then we have $x + ax = (1 + a)x = 1x.$ If $a = 0,$ then the proof is complete. If $a \neq 0,$ then we can multiply both sides of the equation $ax = O$ by $c = 1/a$ to give us \begin{align*} x &= cO \end{align*}But we know from (b) that for any scalar $c,$ $cO = O,$ hence $x = O.$ (e) We know from the associative law of multiplication that for any scalars $a$ and $b,$ we have \begin{align*} a(bx) &= (ab)x \end{align*}Now, if $a \neq 0,$ set $b = 1/a.$ Then, if $ax = ay,$ we have \begin{align*} b(ax) &= b(ay) \\ (ba)x &= (ba)y \\ (a/a)x &= (a/a)y \\ 1x &= 1y \end{align*}And by the existence of identity, we conclude that if $ax = ay,$ $x = y.$ (f) Suppose $ax = bx.$ Then, if we add $(-b)x$ to both sides and apply the distributive law for addition of numbers, we find that \begin{align*} ax + (-bx) &= (a - b)x = O \end{align*}But from the proof of (d), we know that if $cx = O,$ then either $c = 0$ or $x = O.$ But since we know $x \neq O,$ then $c = (a - b)$ must be zero. Or in other words, $a = b.$ (g) From (c), we know that $-(ax) = (-a)x$ for some scalar $a$ and element $x$ in $V.$ Setting $a = -1,$ we can see that $-(x + y) = (-1)(x + y).$ But by the distributive law for addition in $V,$ we know $$(-1)(x + y) = (-1)x + (-1)y$$Applying the result of (c) once more, we can see that $(-1)x + (-1)y = 1(-x) + 1(-y).$ Using the existence of identity, the right-hand side becomes $-x - y$ giving us $$-(x + y) = (-x) + (-y) = -x - y$$ (h) This follows as a result of inductively applying the distributive law for addition of numbers. Let $n \geq 1$ be an integer. Then, using the existence of identity, we can write the sum $\sum_{i = 1}^nx$ as \begin{align*} \sum_{i = 1}^n x &= 1x + \sum_{i = 1}^{n - 1}1x \\ \end{align*} If $n \geq 2,$ we have \begin{align*} \sum_{i = 1}^n x &= 1x + 1x + \sum_{i = 1}^{n - 2}1x \\ &= (1 + 1)x + \sum_{i = 1}^{n - 2}1x \\ &= 2x + \sum_{i = 1}^{n - 2}1x \end{align*}Moreover, for each integer $k \leq n,$ we find that \begin{align*} \sum_{i = 1}^n x &= kx + \sum_{i = 1}^{n - k}1x \end{align*}Setting $k = n,$ we find that the rightmost sum vanishes, leaving us with \begin{align*} \sum_{i = 1}^n x &= nx \end{align*} This completes the proof.