chapter1.tex

\section{The Real and Complex Number Systems}

Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood to be real.

\begin{questions}
  \question If $r$ is rational ($r\neq0$) and $x$ is irrational, prove that $r+x$ and $rx$ are irrational.
  \begin{solution}
    If $r+x$ were rational, then so too would be $(-r)+(r+x)=x$. If $rx$ were rational, then since $r\neq0$, $(1/r)\cdot(rx)=x$ would also be rational.

    As $x$ is irrational, it follows that $r+x$ and $rx$ are irrational too.
  \end{solution}

  \question Prove that there is no rational number whose square is 12.
  \begin{solution}
    Suppose there was a rational $r$ satisfying $r^2=12$. Then we may write $r=m/n$ where $m$ and $n$ are integers that are not both multiples of 3. The integers $m,n$ satisfy
    \[ m^2 = 12n^2.. \]
    Hence $m^2$ is a multiple of 3. Since 3 is prime it follows that $m$ is a multiple of 3. Similarly we may deduce that $m$ is a multiple of 2. Hence $m$ is a multiple of 6, so we may write $m=6k$ where $k$ is an integer. This yields
    \[ 36k^2 = 12n^2 \implies 3k^2 = n^2. \]
    Then $n^2$ is a multiple of 3, so that $n$ is a multiple of 3. This contradicts our choice of $m$ and $n$. Hence can be no rational number whose square is 12.
  \end{solution}

  \question Prove Proposition 1.15: The axioms for multiplication imply the following statements.
  \begin{parts}
    \part If $x\neq0$ and $xy=xz$ then $y=z$.

    \part If $x\neq0$ and $xy=x$ then $y=1$.

    \part If $x\neq0$ and $xy=1$ then $y=1/x$.

    \part If $x\neq0$ then $1/(1/x)=x$.
  \end{parts}
  \begin{solution}
    If $x\neq0$ and $xy=xz$, the axioms for multiplication give
    \begin{align*}
      y = 1\cdot y &= \left( \frac{1}{x}\cdot x \right)y = \frac{1}{x}\cdot(xy) \\
                   &= \frac{1}{x}\cdot(xz) = \left(\frac{1}{x}\cdot x\right)z = 1\cdot z = z.
    \end{align*}
    This proves (a). Take $z=1$ in (a) to obtain (b). Take $z=1/x$ in (a) to obtain (c). Since $(1/x)\cdot x=1$, (c) (with $1/x$ in place of $x$, and $x$ in place of $y$) gives (d).
  \end{solution}

  \question Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound of $E$. Prove that $\alpha\leq\beta$.
  \begin{solution}
    Let $x\in E$, then $\alpha \leq x$, since $\alpha$ is a lower bound of $E$, and $x\leq \beta$, since $\beta$ is an upper bound of $E$. If $\alpha=x$ then from $x\leq\beta$ we see $\alpha\leq\beta$. Similarly if $x=\beta$ then from $\alpha\leq x$ we see $\alpha\leq\beta$.

    If neither $\alpha=x$ nor $x=\beta$ hold, then $\alpha<x$ and $x<\beta$, in which case it follows from the definition of an ordered set that $\alpha<\beta$. Hence in all cases, $\alpha\leq\beta$.
  \end{solution}

  \question Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x\in A$. Prove that
  \[ \inf A = -\sup(-A). \]
  \begin{solution}
    Let $x\in A$, then $-x\in-A$, hence
    \[ -x \leq \sup(-A) \implies -\sup(-A) \leq x. \]
    Therefore $-\sup(-A)$ is a lower bound of $A$. Let $\alpha>-\sup(-A)$, then $-\alpha<\sup(-A)$. Hence $-\alpha$ is not an upper bound of $-A$. All elements of $-A$ have the form $-x$ for some $x\in A$, so there exists some $x\in A$ for which
    \[ -\alpha < -x \implies x < \alpha. \]
    Therefore $\alpha$ is not a lower bound of $A$. It follows that
    \[ \inf A = -\sup(-A). \]
  \end{solution}

  \question Fix $b>1$.
  \begin{parts}
    \part If $m$, $n$, $p$, $q$ are integers, $n>0$, $q>0$, and $r=m/n=p/q$, prove that
    \[ (b^m)^{1/n} = (b^p)^{1/q}. \]
    Hence it makes sense to define $b^r=(b^m)^{1/n}$.
    \begin{solution}
      Let $y=(b^m)^{1/n}$ be the unique positive $n$th root of $b^m$. Then $y^n=b^m$, hence $y^{nq}=b^{mq}$. Since $m/n=p/q$, $mq=np$, therefore $b^{mq}=b^{np}$. It follows that $y^q$ and $b^p$ are both positive $n$th roots of the common number $y^{nq}=b^{np}$. As positive $n$th roots are unique, $y^q=b^p$, and so $y=(b^p)^{1/q}$.
    \end{solution}

    \part Prove that $b^{r+s}=b^rb^s$ is $r$ and $s$ are rational.
    \begin{solution}
      Write $r$ and $s$ with a common positive denominator $d>0$, say $r=m/d$ and $s=n/d$, where $m$, $n$, and $d$ are all integers. Then
      \begin{align*}
        (b^rb^s)^d &= ((b^m)^{1/d}\cdot(b^n)^{1/d})^d \\
                   &= ((b^m)^{1/d})^d\cdot((b^{n})^{1/d})^d = b^mb^n = b^{m+n}.
      \end{align*}
      Therefore $b^rb^s$ is the unique positive $d$th root of $b^{m+n}$, hence
      \[ b^rb^s = (b^{m+n})^{1/d} = b^{(m+n)/d} = b^{r+s}. \]
    \end{solution}

    \part If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t\leq x$. Prove that
    \[ b^r = \sup B(r) \]
    when $r$ is rational. Hence it makes sense to define
    \[ b^x = \sup B(x) \]
    for every real $x$.
    \begin{solution}
      It suffices to show that $b^t\leq b^r$ for every rational $t\leq r$. Then $b^r$ will be an upper bound of $B(r)$, while also being an element of it, so it will follow that it is the least upper bound of $B(r)$.

      Let $r,t$ be rationals with $t\leq r$. Write them with a common positive denominator $d>0$, say $r=m/d$ and $t=n/d$, where $m$, $n$, and $d$ are all integers. Then $m\geq n$, so $m-n\geq0$. An easy induction will verify that $b^{m-n}\geq1$ for any nonnegative integer value of $m-n$. Therefore
      \[ b^m=b^n\cdot b^{m-n}\geq b^n. \]
      Suppose for a contradiction that $b^t>b^r$, i.e. $(b^n)^{1/d}>(b^m)^{1/d}$. It follows by induction that
      \[ ((b^n)^{1/d})^d>((b^m)^{1/d})^d \implies b^n > b^m, \]
      which is a contradiction. Hence $b^t\leq b^r$, and so the result follows.
    \end{solution}

    \part Prove that $b^{x+y}=b^xb^y$ for all real $x$ and $y$.
    \begin{solution}
      Define
      \[ B(x,y)=\{b^rb^s \mid r,s\in\Q, r\leq x, s\leq y\}. \]
      We claim that $b^xb^y=\sup B(x,y)$. Let $r,s\in\Q$ with $r\leq x$ and $s\leq y$, then certainly $0<b^r\leq b^x$ and $0<b^s\leq b^y$. Therefore $b^rb^s\leq b^xb^y$, so $b^xb^y$ is certainly an upper bound of $B(x,y)$. Now notice that for all $r,s\in\Q$ with $r\leq x$ and $s\leq y$, we have
      \[ b^rb^s \leq \sup B(x,y) \implies b^r \leq \frac{1}{b^s} \sup B(x,y). \]
      Hence for every $s\in\Q$ with $s\leq y$, $(1/b^s)\sup B(x,y)$ is an upper bound of $B(x)$. Therefore
      \[ b^x \leq \frac{1}{b^s}\sup B(x,y) \implies b^s \leq \frac{1}{b^x}\sup B(x,y). \]
      Hence $(1/b^x)\sup B(x,y)$ is an upper bound of $B(y)$, so
      \[ b^y \leq \frac{1}{b^x}\sup B(x,y) \implies b^xb^y \leq \sup B(x,y). \]
      Since $b^xb^y$ is an upper bound of $B(x,y)$, we must have $b^xb^y=\sup B(x,y)$.

      Now if $r,s\in\Q$, $r\leq x$, and $s\leq y$, then certainly $r+s\in\Q$ and $r+s\leq x+y$. As $b^rb^s=b^{r+s}$ for all rational $r,s$, it follows that
      \[ B(x,y) \subset B(x+y) \implies b^xb^y \leq b^{x+y}. \]

      Now let us define
      \[ B'(x+y) = \{ b^t \mid t\in\Q, t<x+y\}. \]
      We claim that $b^{x+y}=\sup B'(x+y)$. Clearly $B'(x+y)\subset B(x+y)$, so $b^{x+y}$ is an upper bound of $B'(x+y)$. Let $z<b^{x+y}$. Then $z$ is not an upper bound of $B(x+y)$, so there exists a $t\in\Q$ with $t\leq x+y$ and $z<b^t$. By the archimedian property of $\R$ there is a positive integer $n$ such that
      \[ n(b^t-z) > z(b-1) \implies n > \frac{b-1}{b^tz^{-1} - 1}. \]

      Note that \textbf{7}.(c) can be completed without using \textbf{6}.(d). By \textbf{7}.(c) we have
      \[ b^{1/n} < b^tz^{-1} \implies z < b^tb^{-1/n} = b^{t-(1/n)}. \]
      Now $t-(1/n)<x+y$, hence $b^{t-(1/n)}\in B'(x+y)$. Therefore $z$ is not an upper bound of $B'(x+y)$. It follows that $b^{x+y}$ is the least upper bound of $B'(x+y)$.

      Let $t\in\Q$, $t<x+y$. Then due to the density of $\Q$ in $\R$, there exist $r,s\in\Q$ such that
      \[ x - \frac{1}{2}(x+y-t) < r < x, \qquad y - \frac{1}{2}(x+y-t) < s < y.  \]

      Then $r<x$, $s<y$ and $t<r+s$. Hence $b^t<b^{r+s}=b^rb^s\in B(x,y)$. Hence $b^xb^y$ is an upper bound of $B'(x+y)$, and therefore $b^xb^y\geq b^{x+y}$. It follows that $b^{x+y}=b^xb^y$ for all real $x,y$.
    \end{solution}
  \end{parts}

  \question Fix $b>1$, $y>0$, and prove that there is a unique real $x$ such that $b^x=y$, by completing the following outline. (This is called the logarithm of $y$ to the base $b$.)
  \begin{parts}
    \part For any positive integer $n$, $b^n-1\geq n(b-1)$.
    \begin{solution}
      The result is clear when $n=1$. Suppose the result holds for some positive integer $n$, then since $b^n>1$ we have
      \begin{align*}
        b^{n+1}-1  &= b^n(b-1) + (b^n-1) \\
                   &> (b-1)+n(b-1) \\
                   &= (n+1)(b-1).
      \end{align*}
      By induction it follows that $b^n-1\geq n(b-1)$ for all positive integers $n$.
    \end{solution}

    \part Hence $b-1\geq n(b^{1/n}-1)$.
    \begin{solution}
      Apply part (a) with $b^{1/n}$ in place of $b$.
    \end{solution}

    \part If $t>1$ and $n>(b-1)/(t-1)$, then $b^{1/n}<t$.
    \begin{solution}
      Since $t-1>0$, multiplying both sides of $n>(b-1)/(t-1)$ by $t-1$, and applying (b) yields
      \[ n(t-1) > (b-1) \geq n(b^{1/n}-1) \implies t-1 > b^{1/n} - 1. \]
      Hence $b^{1/n}<t$.
    \end{solution}

    \part If $w$ is such that $b^w<y$, then $b^{w+(1/n)}<y$ for sufficiently large $n$; to see this, apply part (c) with $t=y\cdot b^{-w}$.
    \begin{solution}
      By the archimedian property of $\R$, there exists a positive integer $N$ such that
      \[ N(y-b^w) > b^w(b-1) \implies N > \frac{b-1}{y\cdot b^{-w} - 1}. \]
      Then for all $n\geq N$, if we let $t=y\cdot b^{-w}$ then $n>(b-1)/(t-1)$, so by (c)
      \[ b^{1/n} < y\cdot b^{-w} \implies b^{w+(1/n)} = b^{1/n}b^w < y. \]
    \end{solution}

    \part If $b^w>y$, then $b^{w-(1/n)}>y$ for sufficiently large $n$.
    \begin{solution}
      We have $b^{-w}<y^{-1}$, so  from the previous part, $b^{-w+(1/n)}<y^{-1}$ for sufficiently large $n$. Hence $b^{w-(1/n)}>y$ for sufficiently large $n$.
    \end{solution}

    \part Let $A$ be the set of all $w$ such that $b^w<y$, and show that $x=\sup A$ satisfies $b^x=y$.
    \begin{solution}
      Suppose $b^x<y$. By (d) there is a positive $n$ such that $b^{x+(1/n)}<y$, so that $x+(1/n)\in A$, which contradicts $x$ being an upper bound of $A$.

      Suppose $b^x>y$. By (e) there is a positive $n$ with $b^{x-(1/n)}>y$, so that $b^{x-(1/n)}>b^w$ for all $w\in A$. Therefore $b^w$ is not an upper bound of $B(x-(1/n))$, for all $w\in A$. So for all $w\in A$ there is a $t\in \Q$ with $t\leq x-(1/n)$ such that
      \[ b^t > b^w \implies t > w. \]

      By the density of $\Q$ in $\R$ it follows that
      \[ x-(1/n)=\sup\{t\in\Q \mid t\leq x-(1/n)\}. \]
      Therefore $x-(1/n)\geq w$ for all $w\in A$, hence $x-(1/n)$ is an upper bound of $A$. This contradicts $x$ being the least upper bound of $A$. The only possibility which remains is $b^x=y$.
    \end{solution}

    \part Prove that this $x$ is unique.
    \begin{solution}
      If $x>z$, then there exists a $t\in\Q$ with $0<t<x-z$, so that $b^t<b^{x-z}$. Since $t>0$ then $b^t>b^0=1$, hence
      \[ b^x = b^{x-z}b^z > b^tb^z > b^z. \]
      Similarly if $x<z$ then $b^x<b^z$. So if $b^x=y=b^z$ we must have $x=z$.
    \end{solution}
  \end{parts}

  \question Prove that no order can be defined in the complex field that turns it into an ordered field. \emph{Hint:} $-1$ is a square.
  \begin{solution}
    In an ordered field, if $x\neq0$, then $x^2>0$. For, if $x>0$ then $x^2>0$, and if $x<0$ then $-x>0$, meaning $x^2=(-x)^2>0$. Therefore $1=1^2>0$, and hence $-1<0$. However if we could define an order on the complex field which turns it into an ordered field, then
    \[ -1 = i^2 > 0, \]
    which contradicts $-1<0$. Hence there is no way of defining such an order.
  \end{solution}

  \question Suppose $z=a+bi$, $w=c+di$. Define $z<w$ if $a<c$, and also if $a=c$ but $b<d$. Prove that this turns the set of all complex numbers into an ordered set. (This type of order relation is called a \emph{dictionary order}, or \emph{lexicographic order}, for obvious reasons.) Does this ordered set have the least-upper-bound property?
  \begin{solution}
    Let $z=a+bi$, $w=c+di$. Precisely one of $a<c$, $a=c$ and $c<a$ holds. In the first case, $z<w$, $z\neq w$ and $w\not<z$. In the last case, $w<z$, $z\neq w$ and $z\not>w$.

    If $a=c$, then precisely one of $b<d$, $b=d$ and $d<b$ holds. In the first case $z<w$, $z<w$ and $w\not<z$. In the second case $z=w$, $z\not<w$ and $w\not<z$. In the last case $w<z$, $z\neq w$ and $z\not<w$. In all cases, precisely one of $z<w$, $z=w$, $w<z$ holds.

    Now suppose $y=u+iv$ and that $z<w$ and $w<y$. If $a<c$, then since $w<y$ we have $c\leq u$. Hence $a<u$ and so $z<y$. The only other possibility is when $a=c$. If $c<u$ then $a<u$ and so $z<y$.

    Otherwise $c=u$, and so $a=u$. Since $z<w$ and $w<y$ we must have $b<d$ and $d<v$, hence $b<v$. Therefore $z<y$ in this case also. Therefore if $z<w$ and $w<y$, then $z<y$. Hence this order turns the set of all complex numbers into an ordered set.

    This ordered set does not have the least-upper-bound property. For example, define $E=\{0+vi \mid v\in\R\}$. Then $E$ is nonempty and bounded above by $1$ (as $0+vi<1+0i$ for every $v\in\R$).

    Suppose that $a+bi$ were a least-upper-bound of $E$, then certainly $0\leq a$. If $0<a$, then $0+vi<a/2+0i$ for all $v\in\R$, and $a/2+0i<a+bi$, which contradicts $a+bi$ being the least upper bound. Hence $E$ has no least-upper-bound.
  \end{solution}

  \question Suppose $z=a+bi$, $w=u+iv$, and
  \[ a = \left( \frac{\abs{w} + u}{2} \right)^{1/2}, \qquad b = \left( \frac{\abs{w} - u}{2} \right)^{1/2}. \]
  Prove that $z^2=w$ if $v\geq0$ and that $(\conj{z})^2$ if $v\leq0$. Conclude that every complex number (with one exception!) has two complex square roots.
  \begin{solution}
    Note that $x^{1/2}y^{1/2}$ is the unique positive square root of $xy$, hence $x^{1/2}y^{1/2}=(xy)^{1/2}$. We then have the following calculation
    \begin{align*}
      z^2 &= a^2-b^2 + 2abi \\
          &= \frac{\abs{w}+u}{2} - \frac{\abs{w}-u}{2} + 2i\left( \frac{\abs{w}+u}{2} \right)^{1/2}\left( \frac{\abs{w}-u}{2} \right)^{1/2} \\
          &= u + i(\abs{w}^2-u^2)^{1/2} \\
          &= u + i(u^2+v^2-u^2)^{1/2} \\
          &= u + i(v^2)^{1/2}.
    \end{align*}
    If $v\geq0$, then $v$ is the unique positive square root of $v^2$, hence $(v^2)^{1/2}=v$, and so in this case $z^2=w$. A similar calculation gives $(\conj{z})^2=u-i(v^2)^{1/2}$, and if $v\leq0$ then $-v$ is the unique positive square root of $v^2$, hence $(v^2)^{1/2}=-v$, and so in this case $(\conj{z})^2=w$.

    So we have demonstrated that for every complex number $w$, there exists a complex number $z$ for which $z^2=w$. It is clear that $z=0$ iff $w=0$, so provided $w\neq0$ we actually have two complex square roots, namely $z$ and $-z$. Moreover, let $x$ be a complex number with $x^2=w$. Then we have
    \[ x^2 - z^2 = 0 \implies (x-z)(x+z) = 0. \]
    So it follows that either $x=z$ or $x=-z$. Hence every complex number has exactly two complex square roots, with the exception of 0 which has exactly one.
  \end{solution}

  \question If $z$ is a complex number, prove that there exists an $r\geq0$ and a complex number $w$ with $\abs{w}=1$ such that $z=rw$. Are $w$ and $r$ always uniquely determined by $z$?
  \begin{solution}
    If $z\neq0$, then $\abs{z}\neq0$. In this case, define $r=\abs{z}$ and $w=z/\abs{z}$. Then it is clear that $z=rw$, and $\abs{w}=\abs{z}/\abs{z}=1$. If $z=0$ then define $r=0$ and $w=1$, then $\abs{w}=1$ and $z=0\cdot1=rw$.

    The number $r$ is uniquely determined by $z$, in fact we must have $r=\abs{z}$. For, $\abs{z}=\abs{rw}=\abs{r}$, and since $r\geq0$ it follows that $r=\abs{z}$. However $w$ is not uniquely determined by $z$. For example $0=0\cdot1=0\cdot(-1)$, and $\abs{1}=\abs{-1}=1$.
  \end{solution}

  \question If $z_1,\ldots,z_n$ are complex, prove that
  \[ \abs{z_1+z_2+\cdots+z_n} \leq \abs{z_1}+\abs{z_2}+\cdots+\abs{z_n}. \]
  \begin{solution}
    We prove this by induction on the number of variables. Note that we already have the triangle inequality for triangle numbers. The inequality is clear when $n=1$. Suppose that it holds for some $n\geq1$, then
    \begin{align*}
      \abs{z_1+z_2\cdots+z_n+z_{n+1}} &\leq \abs{z_1+z_2+\cdots+z_n} + \abs{z_{n+1}} \tag{Triangle inequality} \\
                                      &\leq \abs{z_1} + \abs{z_2} + \cdots + \abs{z_n} + \abs{z_{n+1}} \tag{Inductive hypothesis}
    \end{align*}
    The result follows by induction.
  \end{solution}

  \question If $x,y$ are complex, prove that
  \[ \abs{\abs{x} - \abs{y}} \leq \abs{x-y}. \]
  \begin{solution}
    We have
    \[ \abs{x} \leq \abs{x-y} + \abs{y} \implies \abs{x} - \abs{y} \leq \abs{x-y}. \]
    Since $\abs{x-y}=\abs{y-x}$, by symmetry $\abs{y}-\abs{x}\leq\abs{x-y}$ also. Because $\abs{\abs{x}-\abs{y}}$ is one of $\abs{x}-\abs{y}$, $\abs{y}-\abs{x}$, it follows that
    \[ \abs{\abs{x}-\abs{y}} \leq \abs{x-y}. \]
  \end{solution}

  \question If $z$ is a complex number such that $\abs{z}=1$, that is, such that $z\conj{z}=1$, compute
  \[ \abs{1+z}^2 + \abs{1-z}^2. \]
  \begin{solution}
    Let $z=a+bi$, then $a^2+b^2=1$, so
    \begin{align*}
      \abs{1+z}^2 + \abs{1-z}^2 &= (1+a)^2+b^2 + (1-a)^2+b^2 \\
                                &= 2 + 2(a^2+b^2) \\
                                &= 4.
    \end{align*}
  \end{solution}

  \question Under what conditions does equality hold in the Schwarz inequality?
  \begin{solution}
    The Schwarz inequality states that if $a_1,a_2\ldots,a_n$ and $b_1,b_2\ldots,b_n$ are complex numbers, then
    \[ \abs*{\sum_{j=1}^n a_j\conj{b_j}}^2 \leq \sum_{j=1}^n \abs{a_j}^2 \sum_{j=1}^n \abs{b_j}^2. \]

    We claim that equality holds precisely when either $b_1=b_2=\cdots=b_n=0$, or there exists a $\lambda\in\C$ such that $a_1=\lambda b_1$, $a_2=\lambda b_2$ \ldots, $a_n=\lambda b_n$. Clearly equality holds in the first case, so suppose not all of $b_1,b_2,\ldots,b_n$ are zero. The given proof of the inequality defines
    \[ A = \sum_{j=1}^n \abs{a_j}^2, \ B = \sum_{j=1}^n \abs{b_j}^2, \ C = \sum_{j=1}^n a_j\conj{b_j}, \]
    and proceeds to show that
    \[ \sum_{i=1}^n \abs{Ba_i-Cb_i}^2 = B(AB-\abs{C}^2). \]
    Since $B>0$ it follows that $\abs{C}^2\leq AB$, which is precisely the inequality. Now equality holds if and only if $Ba_i=Cb_i$ for each $i=1,\ldots,n$. Since $B>0$ we can define $\lambda = C/B$, and then $a_i=\lambda b_i$ for each $i=1,\ldots,n$. Conversely, suppose there is a $\lambda\in\C$ such that $a_i=\lambda b_i$ for each $i=1,\ldots,n$. Then
    \[ Ba_i = a_i\sum_{j=1}^n \abs{\lambda a_j}^2 = \conj{\lambda}\lambda a_i \sum_{j=1}^n a_j\conj{a_j} = b_i\sum_{j=1}^n a_j(\conj{\lambda a_j}) = Cb_i, \]
    and so equality holds. This proves the claim.
  \end{solution}

  \question Suppose $k\geq3$, $\vec{x}, \vec{y}\in\R^k$, $\abs{\vec{x}-\vec{y}}=d>0$, and $r>0$. Prove:
  \begin{parts}
    \part If $2r>d$, there are infinitely many $\vec{z}\in\R^k$ such that
    \[ \abs{\vec{z} - \vec{x}} = \abs{\vec{z} - \vec{y}} = r. \]
    \begin{solution}
      For any $\vec{z}\in\R^k$, define $\vec{w}=\vec{z}-\frac{1}{2}(\vec{x}+\vec{y})$. Then $\vec{z}$ satisfies $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{y}}=r$ if and only if $\vec{w}$ satisfies the following
      \begin{equation}
      \vec{w}\cdot(\vec{x}-\vec{y}) = 0, \qquad \abs{\vec{w}}^2 = r^2 - \frac{d^2}{4}.\label{eq:ch1ex16a}
    \end{equation}

      For, we have the equality
      \[ \abs{\vec{z}-\vec{x}}^2 = \abs*{\vec{w} + \frac{1}{2}(\vec{x}-\vec{y})}^2 = \abs{\vec{w}}^2 + \vec{w}\cdot(\vec{x}-\vec{y}) + \frac{1}{4}\abs{\vec{x}-\vec{y}}^2. \]
      Similarly $\abs{\vec{z}-\vec{y}}^2 = \abs{\vec{w}}^2 - \vec{w}\cdot(\vec{x}-\vec{y}) + \frac{1}{4}\abs{\vec{x}-\vec{y}}^2$. Hence if $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{y}}=r$, then $\vec{w}\cdot(\vec{x}-\vec{y})=0$, and $\abs{\vec{w}}^2=\abs{\vec{z}-\vec{x}}^2-\frac{1}{4}\abs{\vec{x}-\vec{y}}^2=r^2-\frac{d^2}{4}$. Conversely if $\vec{w}$ satisfies (\ref{eq:ch1ex16a}) then
      \[ \abs{\vec{z}-\vec{x}}^2 = \abs{\vec{z}-\vec{y}}^2 = \abs{\vec{w}}^2 + \frac{1}{4}\abs{\vec{x}-\vec{y}}^2 = r^2 - \frac{d^2}{4} + \frac{d^2}{4} = r^2, \]
      hence $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{x}}=r$, so the equivalence follows. Hence there is a one-to-one and onto correspondence between $\vec{z}\in\R^k$ satisfying $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{y}}=r$, and $\vec{w}\in\R^k$ satisfying (\ref{eq:ch1ex16a}).

      Let $\vec{x}=(x_1,x_2,\ldots,x_k)$, $\vec{y}=(y_1,y_2,\ldots,y_k)$ and $\vec{w}=(w_1,w_2,\ldots,w_k)$. Since $\vec{x}\neq\vec{y}$ we may assume without loss of generality that $x_1-y_1\neq0$. As $k\geq3$ if we let $w_2=\lambda$, $w_3=1$, and $w_i=0$ if $i>3$, then there is a unique $w_1\in\R$ such that $\vec{w}$ satisfies the first equation of (\ref{eq:ch1ex16a}), namely
      \[ w_1 = \frac{\lambda(x_2-y_2) + (x_3-y_3)}{y_1-x_1}. \]
      For each $\lambda\in\R$ this produces a nonzero solution $\vec{w}$ to the first equation. For each of these $\vec{w}$, the vector $t\vec{w}$ satisfies both equations in (\ref{eq:ch1ex16a}), where $t=(r^2-d^2/4)/\abs{\vec{w}}$, since $r^2-d^2/4>0$. So for each $\lambda\in\R$ we have a vector $t\vec{w}$ satisfying (\ref{eq:ch1ex16a}). Moreover these solutions are all distinct, since for a given $\lambda$ the induced solution has ratio $\lambda$ between its second and third coordinates.

      Therefore there are infinitely many $\vec{w}$ which satisfy (\ref{eq:ch1ex16a}) when $k\geq3$. Hence there are infinitely many $\vec{z}$ which satisfy $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{y}}=r$ if $2r>d$.
    \end{solution}

    \part If $2r=d$, there is exactly one such $\vec{z}$.
    \begin{solution}
      If $2r=d$, then $r^2-\frac{d^2}{4}=0$, so there is precisely one $\vec{w}$ satisfying (\ref{eq:ch1ex16a}), namely $\vec{w}=0$. Hence there is one such $\vec{z}$, namely $\vec{z}=\frac{1}{2}(\vec{x}+\vec{y})$.

      \textbf{Alternate Solution:} Suppose that $\vec{z}$ satisfies $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{y}}=r$. Then the Schwarz inequality gives
      \begin{align*}
        d^2 = \abs{\vec{x}-\vec{y}}^2 &= \abs{(\vec{x}-\vec{z}) + (\vec{z}-\vec{y})}^2 \\
                                      &= \abs{\vec{x}-\vec{z}}^2 + 2(\vec{x}-\vec{z})\cdot(\vec{z}-\vec{y}) + \abs{\vec{z}-\vec{y}}^2 \\
                                      &\leq \abs{\vec{x}-\vec{z}}^2 + 2\abs{\vec{x}-\vec{z}}\abs{\vec{z}-\vec{y}} + \abs{\vec{z}-\vec{y}}^2 \\
                                      &= (2r)^2.
      \end{align*}
      Equality must hold in the above application of the Schwarz inequality. By Exercise \textbf{15}, since $r>0$ we conclude there is a $\lambda\in\R$ with $\vec{x}-\vec{z}=\lambda(\vec{z}-\vec{y})$. Then
    \[ r = \abs{\lambda(\vec{z}-\vec{y})} = \abs{\vec{z}-\vec{y}} \implies \abs{\lambda}=1. \]
    If $\lambda=-1$, then $\vec{x}=\vec{y}$, contradicting $d>0$. Hence $\lambda=1$, and so $\vec{z}=(\vec{x}+\vec{y})/2$. So if $2r=d$, there is precisely one such $\vec{z}$, which answers (b).
    \end{solution}

    \part If $2r<d$, there is no such $\vec{z}$.
    \begin{solution}
      If $2r<d$, then $r^2-\frac{d^2}{4}<0$, so there are no $\vec{w}$ satisfying (\ref{eq:ch1ex16a}). Hence there are no such $\vec{z}$.

      \textbf{Alternate Solution:} Suppose that $\vec{z}$ satisfies $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{y}}=r$. Then the Schwarz inequality gives
    \begin{align*}
      d^2 = \abs{\vec{x}-\vec{y}}^2 &= \abs{(\vec{x}-\vec{z}) + (\vec{z}-\vec{y})}^2 \\
                                      &= \abs{\vec{x}-\vec{z}}^2 + 2(\vec{x}-\vec{z})\cdot(\vec{z}-\vec{y}) + \abs{\vec{z}-\vec{y}}^2 \\
                                      &\leq \abs{\vec{x}-\vec{z}}^2 + 2\abs{\vec{x}-\vec{z}}\abs{\vec{z}-\vec{y}} + \abs{\vec{z}-\vec{y}}^2 \\
                                      &= (2r)^2.
    \end{align*}
    Hence if there are any $\vec{z}$ satisfying the equation, then $d\leq2r$, so there are no solutions in this case.
    \end{solution}
  \end{parts}
  How must these statements be modified if $k$ is 2 or 1?
  \begin{solution}
    If $k=2$ and $2r>d$ then we claim there are precisely two $\vec{z}\in\R^2$ satisfying $\abs{\vec{z}-\vec{x}}=\abs{\vec{z}-\vec{y}}=r$. Write $\vec{x}=(x_1,x_2)$, $\vec{y}=(y_1,y_2)$, $\vec{w}=(w_1,w_2)$. Then $\vec{w}\cdot(\vec{x}-\vec{y})=0$ precisely when
    \[ w_1(x_1-y_1)+w_2(x_2-y_2)=0. \]
    Hence any $\vec{w}$ satisfying the first equation of (\ref{eq:ch1ex16a}) must be a multiple of the vector $\vec{w}_0=(x_2-y_2,y_1-x_1)\neq\vec{0}$. Now $\abs{\lambda\vec{w}_0}=r^2-d^2/4$ if and only if $\lambda=\pm(r^2-d^2/4)/\abs{\vec{w}_0}$, and since $r^2-d^2/4>0$, this shows there are precisely two vectors satsifying (\ref{eq:ch1ex16a}), hence precisely two such $\vec{z}$.

    If $k=1$ and $2r>d$ then there are no $z\in\R^1$ satisfying $\abs{z-x}=\abs{z-y}=r$, since $\abs{z-x}=\abs{z-y}=r$ implies that $\abs{x-y}=0$ or $\abs{x-y}=2r$. Neither of these are equal to $d$ in this case.

    The statements in (b) and (c) need not be modified if $k$ is 2 or 1.
  \end{solution}

  \question Prove that
  \[ \abs{\vec{x} + \vec{y}}^2 + \abs{\vec{x} - \vec{y}}^2 = 2\abs{\vec{x}}^2 + 2\abs{\vec{y}}^2 \]
  if $\vec{x}\in\R^k$ and $\vec{y}\in\R^k$. Interpret this geometrically, as a statement about parallelograms.
  \begin{solution}
    \begin{align*}
      \abs{\vec{x}+\vec{y}}^2 + \abs{\vec{x}-\vec{y}}^2 &= (\vec{x}+\vec{y})\cdot(\vec{x}+\vec{y}) + (\vec{x}-\vec{y})\cdot(\vec{x}-\vec{y}) \\
                                                            &= (\vec{x}\cdot\vec{x} + 2\vec{x}\cdot\vec{y} + \vec{y}\cdot\vec{y}) + (\vec{x}\cdot\vec{x} - 2\vec{x}\cdot\vec{y} + \vec{y}\cdot\vec{y}) \\
                                                            &= 2\vec{x}\cdot\vec{x} + 2\vec{y}\cdot\vec{y} \\
                                                            &= 2\abs{\vec{x}}^2 + 2\abs{\vec{y}}^2.
    \end{align*}
    Consider the parallelogram spanned by vectors $\vec{x},\vec{y}\in\R^k$. Then $\vec{x}+\vec{y}$ and $\vec{x}-\vec{y}$ represent the diagonals of this parallelogram. So geometrically this says that the sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides.
  \end{solution}

  \question If $k\geq2$ and $\vec{x}\in\R^k$, prove that there exists $\vec{y}\in\R^k$ such that $\vec{y}\neq\vec{0}$ but $\vec{x}\cdot\vec{y}=0$. Is this also true if $k=1$?
  \begin{solution}
    Suppose that $\vec{x}=(x_1,x_2,\ldots,x_k)$. If any one of the $x_j$ is equal to 0, then we may define $\vec{y}=(y_1,y_2,\ldots,y_k)$, where $y_j=1$ and $y_i=0$ whenever $i\neq j$. As $k\geq2$, $\vec{y}\neq\vec{0}$, and
    \[ \vec{x}\cdot\vec{y} = \sum_{i=1}^k x_iy_i = x_jy_j + \sum_{i\neq j} x_iy_i = 0\cdot1 + \sum_{i\neq j} x_i\cdot0 = 0. \]
    Otherwise all of $x_1,x_2,\ldots,x_k$ are nonzero. Since $k\geq 2$, in particular $x_1$ and $x_2$ are nonzero. Hence if we define $\vec{y}=(y_1,y_2,\ldots,y_k)$ by $y_1=x_2$, $y_2=-x_1$ and $y_i=0$ if $i>2$, then $\vec{y}\neq\vec{0}$. Moreover
    \[ \vec{x}\cdot\vec{y} = \sum_{i=1}^k x_iy_i = x_1x_2-x_2x_1 + \sum_{i=3}^k x_i\cdot0 = 0. \]
    So if $k\geq2$, for every $\vec{x}\in\R^k$ there exists $\vec{y}\in\R^k$ such that $\vec{y}\neq\vec{0}$ but $\vec{x}\cdot\vec{y}=0$. However if $k=1$ then we can take $x=1$, then whenever $x\cdot y=0$ we must have $y=0$, therefore the same result is not also true in this case.
  \end{solution}

  \question Suppose $\vec{a}\in\R^k$, $\vec{b}\in\R^k$. Find $\vec{c}\in\R^k$ and $r>0$ such that
  \[ \abs{\vec{x}-\vec{a}} = 2\abs{\vec{x}-\vec{b}} \]
  if and only if $\abs{\vec{x}-\vec{c}}=r$.
  (\emph{Solution:} $3\vec{c}=4\vec{b}-\vec{a}$, $3r=2\abs{\vec{b}-\vec{a}}$.)
  \begin{solution}
    We have the following equality
    \begin{align*}
      4\abs{\vec{x}-\vec{b}}^2 - \abs{\vec{x}-\vec{a}}^2 &= 4\abs{\vec{x}}^2 - 8\vec{x}\cdot\vec{b} + 4\abs{\vec{b}}^2 - \abs{\vec{x}}^2 + 2\vec{x}\cdot\vec{a} - \abs{\vec{a}}^2 \\
                                                             &= 3\abs{\vec{x}}^2 - 2\vec{x}\cdot(4\vec{b} - \vec{a}) + 4\abs{\vec{b}}^2 - \abs{\vec{a}}^2 \\
                                                             &= 3\abs*{\vec{x}-\frac{1}{3}(4\vec{b}-\vec{a})}^2 - \frac{1}{3}\abs{4\vec{b}-\vec{a}}^2 + 4\abs{\vec{b}}^2 - \abs{\vec{a}}^2 \\
                                                             &= 3\abs*{\vec{x}-\frac{1}{3}(4\vec{b}-\vec{a})}^2 - \frac{16}{3}\abs{\vec{b}}^2 + \frac{8}{3}\vec{b}\cdot\vec{a} - \frac{1}{3}\abs{\vec{a}}^2 + 4\abs{\vec{b}}^2 - \abs{\vec{a}}^2 \\
                                                             &= 3\abs*{\vec{x}-\frac{1}{3}(4\vec{b}-\vec{a})}^2 - \frac{4}{3}\abs{\vec{b}}^2 + \frac{8}{3}\vec{b}\cdot\vec{a} - \frac{4}{3}\abs{\vec{a}}^2 \\
                                                             &= 3\abs*{\vec{x}-\frac{1}{3}(4\vec{b}-\vec{a})}^2 - \frac{4}{3}\abs{\vec{b}-\vec{a}}^2.
    \end{align*}
    Hence $\abs{\vec{x}-\vec{a}}=2\abs{\vec{x}-\vec{b}}$ if and only if $\abs{\vec{x}-\frac{1}{3}(4\vec{b}-\vec{a})}=\frac{2}{3}\abs{\vec{b}-\vec{a}}$. So if we set
    \[ \vec{c}=\frac{1}{3}(4\vec{b}-\vec{a}),\quad r=\frac{2}{3}\abs{\vec{b}-\vec{a}}, \]
    then $\abs{\vec{x}-\vec{a}}=2\abs{\vec{x}-\vec{b}}$ if and only if $\abs{\vec{x}-\vec{c}}=r$.
  \end{solution}

  \question With reference to the Appendix, suppose that property (III) were omitted from the definition of a cut. Keep the same definitions of order and addition. Show that the resulting ordered set has the least-upper-bound property, that addition satisfies the axioms (A1) to (A4) (with a slightly different zero-element!) but that (A5) fails.
  \begin{solution}
    We will use $\R^\#$ to denote the collection of modified cuts (subsets of $\Q$ satisfying property (I) and (II)). The same proof that $\R$ is an ordered set shows that $\R^\#$ is an ordered set, as property (III) of cuts is never invoked in the former.

    The supremum of a bounded subset of $\R$ is the union of its elements. Only properties (I) and (II) of cuts were required to show that the union satisfied properties (I) and (II), and property (III) was also not required to prove that the union was a supremum. Hence the same construction produces a least-upper-bound of a bounded subset of $\R^\#$, so it has the least-upper-bound property also.

    To prove that the sum of two cuts in $\R$ satisfies properties (I) and (II) only requires properties (I) and (II) of the two cuts, hence (A1) holds in $\R^\#$. Property (III) of cuts is not required to show that (A2) and (A3) hold in $\R$, hence they hold in $\R^\#$.

    Define $0^\#$ to be the set of all nonpositive rationals (note this is different from the additive identity $0^*$ of $\R$ in the appendix). Then we claim that (A4) holds in $\R^\#$ with $0^\#$ as the additive identity. First we must show that $0^\#$ satisfies (I) and (II).

    Clearly $0\in0^\#$ and $1\notin0^\#$, so that $0^\#$ satisfies (I). If $p\in0^\#$, $q\in\Q$ and $q<p$, then $q<0$ so that $q\in0^\#$. Hence $0^\#$ satisfies (II), and so $0^\#\in\R^\#$. Next we seek to prove that $\alpha=\alpha+0^\#$ for all $\alpha\in\R^\#$.

    Let $\alpha\in\R^\#$, then if $r\in\alpha$ and $s\in0^\#$ then $r+s\leq r$, hence $r+s\in\alpha$. So $\alpha+0^\#\subset\alpha$. Conversely $r=r+0\in\alpha+0^\#$, and so $\alpha\subset\alpha+0^\#$. Hence $\alpha=\alpha+0^\#$, and so (A4) holds in $\R^\#$.

    Let $0^*$ be the set of all negative rationals, then $0^*\in\R\subset\R^\#$. We show that $0^*$ has no additive inverse in $\R^\#$. Suppose $0^*+\alpha=0^\#$ for some $\alpha\in\R^\#$. Then $r+s=0$ for some $r\in0^*$, $s\in\alpha$. However then $-r/2=r/2+s\in0^*+\alpha$, but since $-r/2>0$ we have $-r/2\notin0^\#$, which is a contradiction. Therefore (A5) fails to hold in $\R^\#$.
  \end{solution}
\end{questions}

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