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## Solution to Principles of Mathematical Analysis Chapter 1 Part B

### Chapter 1 The Real and Complex Number Systems

#### Exercise 11

(By ghostofgarborg) There is a solution, and it is unique whenever $z \neq 0$. Assume $z = rw$, $r \geq 0$ and $|w|^2 = w \bar w = 1$. Then$r = \sqrt{r^2} = \sqrt{(rw)(\overline{rw})} = \sqrt{z \bar z} = |z|.$If $r = 0$, we can take any $w$ with $|w| = 1$, e.g. $w = e^{i \theta}$. Otherwise $w = \frac{z}{r} = \frac{z}{|z|}.$It is easy to check that the uniquely determined $r$ and $w$ have the desired properties.

#### Exercise 12

(By ghostofgarborg) Note that by the triangle inequality, $|z_1 + z_2| \leq |z_1| + |z_2|$. Assume the statement holds for $n-1$. Then
$|z_1 + \cdots + z_{n-1} + z_n| \leq |z_1 + \cdots + z_{n-1}| + |z_n| \leq |z_1| + \cdots + |z_n|,$which establishes the claim by induction.

#### Exercise 13

(By ghostofgarborg) By the triangle inequality
$|x| = | (x – y) + y| \leq |x-y| + |y|$so that$|x| – |y| \leq |x-y|.$Interchanging the roles of $x$ and $y$ in the above, we also have$|y| – |x| \leq |x-y|$so that
$||x| – |y|| \leq |x-y|.$

#### Exercise 14

(By ghostofgarborg) We have\begin{align*}
|1+z|^2 + |1-z|^2 &= (1+z)(\overline{1+z}) + (1-z)(\overline{1-z}) \\
&= (1 + z)(1 + \bar z) + (1 – z)(1 – \bar z) \\
&= (1 + z + \bar z + z \bar z) + (1 -z – \bar z + z \bar z) \\
&= 2 + 2z \bar z \\
&= 4.
\end{align*}

#### Exercise 15

(By ghostofgarborg) We observe that $(AB-|C|^2) > 0$ iff $S = \sum |Ba_j – Cb_j|^2 > 0$.

If $S=0$, then $a_j = – \frac C B b_j = zb_j$. Assume $a_j = zb_j$. Then $C = zB$, so that $S = \sum |Bzb_j – zBb_j|^2 = 0$. I.e., equality holds iff there is a $z \in \mathbb C$ such that $a_j = zb_j$.

#### Exercise 16

(By analambanomenos)

(a): To simplify the calculations, we may assume without losing generality that $\mathbf{x}=(0,\ldots,0)$ and $\mathbf{y}=(d,0,\ldots,0)$. (Translate $\mathbb R^k$ by $-\mathbf x$, then rotate it to place $\mathbf y$ on the positive x-axis. Both transformations preserve distances and angles.)

For $0\le\theta<1$ let $$\mathbf{z}_\theta=(d/2,\cos\theta\sqrt{4r^2-d^2}/4,\sin\theta\sqrt{4r^2-d^2}/2,0,\ldots,0).$$ Then $$|\mathbf{z}_\theta-\mathbf{x}|^2=|\mathbf{z}_\theta-\mathbf{y}|^2= \frac{d^2}{4}+(\cos^2\theta+\sin^2\theta)\frac{4r^2-d^2}{4}=r^2.$$For $k=2$, the hyperplane of equidistant points is the line $(d/2,x_2)$, so there are two points with distance $r>d/2$ from $\mathbf x$ and $\mathbf y$, $(d/2,\pm\sqrt{4r^2-d^2}/2)$. For $k=1$, the hyperplane reduces to the single point $z=d/2$, so there are no points with distance $r>d/2$ from $\mathbf x$ and $\mathbf y$.

(b): If $d=2r$, then $|\mathbf{z}-\mathbf{x}|+|\mathbf{z}-\mathbf{x}|=|\mathbf{x}-\mathbf{y}|$, that is, we have equality in Theorem 1.37(f). Looking at the proof, this only happens if $$(\mathbf{z}-\mathbf{x})\cdot(\mathbf{z}-\mathbf{y})=|\mathbf{z}-\mathbf{x}| |\mathbf{z}-\mathbf{y}|.$$ By the solution to exercise 15, this only happens if $\mathbf{z}-\mathbf{x}=t(\mathbf{z}-\mathbf{y})$ for some real number $t$. Since $|\mathbf{z}-\mathbf{x}|=|\mathbf{z}-\mathbf{y}|$, $t$ must be $\pm 1$. If $t=1$, then $\mathbf{x}=\mathbf{y}$, so $t=-1$ and $\mathbf z$ is the midpoint $(\mathbf{x}-\mathbf{y})/2$.

(c): $|\mathbf{z}-\mathbf{x}|+|\mathbf{z}-\mathbf{y}|=2r<|\mathbf{x}-\mathbf{y}|$ contradicts Theorem 1.37(f).

#### Exercise 17

(By Matt frito Lundy)

This solution will use some linear algebra. In particular, for any inner product $\langle \cdot {,} \cdot \rangle$ the norm is $$| \cdot | = \sqrt{\langle \cdot {,} \cdot \rangle}$$ or
$$| \cdot |^2 = \langle \cdot {,} \cdot \rangle.$$ In this real vector space $R^k$, for any $x,y,z \in R^k$ we also have the following properties: $$\langle x + y , z \rangle = \langle x , z \rangle + \langle y , z\rangle$$ $$\langle x , y \rangle = \langle y , x \rangle.$$ Then
\begin{align*}
|x + y|^2 + |x – y|^2 &= \langle x , x \rangle + 2\langle x , y \rangle + \langle y , y \rangle
+ \langle x , x \rangle – 2\langle x , y \rangle +\langle y , y \rangle \\
&= 2\langle x , x \rangle + 2\langle y , y \rangle \\
&= 2|x|^2 + 2|y|^2
\end{align*}Geometrically, this says that for any parallelogram in Euclidian Space, the sum of the lengths of the diagonals is equal to the perimeter.

#### Exercise 18

(By Matt frito Lundy) This solution uses some linear algebra. If $x = 0$ then any non-zero $y$ will suffice, so assume that $x \neq 0$. Also choose any $z \in R^k$ such that $x$ and $z$ are linearly indepedent (this is possible because $k \geq 2$). Then let $$y = z – \frac{\langle z, x \rangle}{\langle x , x \rangle} x.$$ Notice that $y \neq 0$ because $x$ and $z$ are linearly independent and $y$ is a linear combination of $x$ and $z$ with not all coefficients equal to zero. And so we have
\begin{align*}
\langle x ,y \rangle &= \langle x , z – \frac{\langle z, x \rangle}{\langle x , x \rangle} x \rangle\\
&= \langle x , z \rangle -\frac{\langle z ,x \rangle \langle x , x \rangle}{\langle x, x \rangle} \\
&= \langle x , z \rangle – \langle x , z \rangle \\
&= 0
\end{align*} There is trouble when $k = 1$ because any two vectors in $R^1$ are linearly dependent.

#### Exercise 19

(By analambanomenos)\begin{align*}
4|\mathbf{x}-\mathbf{b}|^2 &= |\mathbf{x}-\mathbf{a}|^2 \\
4|\mathbf{x}|^2-8\mathbf{x}\cdot\mathbf{b}+4|\mathbf{b}|^2 &= |\mathbf{x}|^2-2\mathbf{x}\cdot\mathbf{a}+|\mathbf{a}|^2 \\
|\mathbf{x}|^2-2\mathbf{x}\cdot\bigl((1/3)(4\mathbf{b}-\mathbf{a})\bigr) &= (1/3)|\mathbf{a}|^2-(4/3)|\mathbf{b}|^2 \\
|\mathbf{x}|^2-2\mathbf{x}\cdot\bigl((1/3)(4\mathbf{b}-\mathbf{a})\bigr) + \big|(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/3)|\mathbf{a}|^2 – (4/3)|\mathbf{b}|^2 + \big|(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 \\
\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/9)(3|\mathbf{a}|^2-12\mathbf{b}^2+|\mathbf{a}|^2-8\mathbf{a}\cdot\mathbf{b}+16|\mathbf{b}|^2) \\
\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/9)(4|\mathbf{a}|^2 – 8\mathbf{a}\cdot\mathbf{b} + 4|\mathbf{b}|^2) \\
\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (4/9)\big|\mathbf{a}-\mathbf{b}\big|^2
\end{align*}
This describes a sphere with center $(1/3)(4\mathbf{b}-\mathbf{a})$ and radius $(2/3)|\mathbf{a}-\mathbf{b}|$.

#### Exercise 20

(By Jack Gallagher)Proof is omitted whenever the proof in the book does not use property (III).

In this case, a cut is defined as any inhabited set $\alpha \subset Q$ such that

1. $\alpha \neq Q$
2. If $p \in \alpha$, $q \in Q$, and $q < p$, then $q \in \alpha$

I’ll adopt the same convention as the book, labelling rationals with Roman letters and cuts with Greek. The proof that $R’$ has the least-upper-bound property is the same as that given in the book.

The proofs of axioms (A1-3) are again the same. For (A4), we define
$$0_R = \left\{ n\ |\ n \in Q,\ n \leq 0 \right\}.$$(A4) For any $a \in \alpha$ and $s \in 0_r$, we have either that $$a + s = a \text{ or } a + s < a.$$ In either case we have that $a + s \in \alpha$, either by equality or by (II). Thus $\alpha + 0_{R’} \subset \alpha$.

Similarly, if $b \in \alpha + 0_{R’}$, we have that $b = a + s$ for some $a \in \alpha$ and $s \in 0_{R’}$, and the same analysis applies.

Finally, to negate (A5), we shall first show that any addition with an open cut will produce another open cut.

Consider two cuts $\alpha \in R$, $\beta \in R’$. For any $r \in \alpha + \beta$, we have that $r = a + b$ for some $a \in \alpha$, $b \in \beta$. But, because $\alpha$ is open, we can pick some $a’ \in \alpha$ such that $a’ > a$, and therefore we have $a’ + b > a + b \in \alpha + \beta$.

(A5) fails! Suppose we had a valid negation operation. Then we would have
$$0^* – 0^* = 0_{R’}$$ But this leads to a contradiction, as with the given definition of addition we cannot add any number to the open $0^*$ to produce the closed $0_{R’}$.

Baby Rudin 数学分析原理完整第一章习题解答