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Solution to Principles of Mathematical Analysis Chapter 1 Part A


Chapter 1 The Real and Complex Number Systems

Exercise 1

(By ghostofgarborg) Note that $\mathbb{Q}$ is closed under the arithmetic operations of addition, subtraction, multiplication and taking multiplicative inverses.

If $r+x$ were rational, so would $(r+x) – r = x$ be, a contradiction.

If $rx$ were rational, so would $\frac 1 r rx= x$ be, a contradiction.


Exercise 2

(By ghostofgarborg) We will assume without proof that $\mathbb{Z}$ has unique prime factorizations. (This follows from $\mathbb Z$ being a primary ideal domain, and is covered in an abstract algebra course.) This implies that $p$ is a prime iff $ p \mid ab \Rightarrow p \mid a $ or $p \mid b$.

Assume there are $m,n$ coprime such that $(\frac m n)^2 = 12$. This implies that $12n^2 = m^2$. Consequently, $3 \mid m^2$, and since $3$ is prime, $3 \mid m$. This means that there is a $p$ such that $m = 3p$, so that $12n^2 = 9p^2$, and consequently $4n^2 = 3p^2$. Using the primality of $3$, we can conclude that since $3 \nmid 4$, $3 \mid n^2$, and therefore $3 \mid n$. This contradicts the coprimality of $m$ and $n$. Therefore there cannot be any such $m,n$.

(By analambanomenos) If $r$ is a rational number whose square is 12, then $(r/2)^2=3$, so this is equivalent to showing there is no rational number whose square is 3. So suppose $(m/n)^2=3$ where $m,n$ are integers with no common factors. Then $m^2=3n^2$. Either $m=3p$, or $m=3p\pm 1$, and since $(3p\pm 1)^2=3(3p^2\pm 2p)+1$, $m$ must be a multiple of 3. But then we have $(3p)^2=3n^2$, or $3p^2=n^2$. Hence $n$ is also a multiple of 3, contradicting the assumption that $m$ and $n$ have no common factors.


Exercise 3

(By ghostofgarborg)

(a): Solution 1: $x \neq 0$ implies that $x$ has a multiplicative inverse. Multiply by $1/x$.

However, we can prove this without resorting to the existence of inverses, and get a solution that carries over to a larger class of rings, i.e. integral domains like $\mathbb Z$. Solution 2: Note that proposition 1.16b implies that $xy=0$ implies that $x=0$ or $y=0$. If $xy = xz$, then $x(y-z)=0$. If $ x \neq 0$, then by prop. 1.16b, $(y-z) = 0$. The claim follows.

(b): Follows from (a) by letting $z = 1$.

(c): Follows from (a) by letting $z = x^{-1}$.

(d): Follows by applying (a) to
\[ \left( \frac 1 x \right) \frac 1 {\left( \frac 1 x \right)} = \left( \frac 1 x \right) x.\]


Exercise 4

(By ghostofgarborg) Let $x \in E$. By definition of lower and upper bounds, $\alpha \leq x \leq \beta$.


Exercise 5

(By ghostofgarborg) Assume $\alpha$ is the greatest lower bound of $A$.

If $x \in (-A)$ then $-x \in A$, so $\alpha \leq -x$, and therefore $ – \alpha \geq x$. This implies that $-\alpha$ is an upper bound for $(-A)$.

If $ \beta < -\alpha$ then $-\beta > \alpha$, and there is an $x \in A$ such that $x < -\beta$. Then $-x \in -A$, and $-x > \beta$. This shows that $-\alpha$ is the least upper bound of $(-A)$, and we are done.


Exercise 6

(By analambanomenos)

(a): Since $np=qm$, $$((b^p)^{1/q})^{np} = ((b^p)^{1/q})^{qm}=b^{mp}.$$ Hence by Theorem 1.21, $$((b^p)^{1/q})^n=b^m.$$ A second application of Theorem 1.21 gives us $$(b^p)^{1/q}=(b^m)^{1/n}.$$ (b): Let $r=m/n$, $s=p/q$.
\begin{align*}
(b^rb^s)^{nq}=((b^m)^{1/n}(b^p)^{1/q})^{nq} &= b^{mq}b^{np} \\
(b^{r+s})^{nq}=((b^{mq+np})^{1/nq})^{nq} &= b^{mq}b^{np}
\end{align*}Hence by Theorem 1.21, $b^rb^s=b^{r+s}$.

(c): Note that the positive rational powers of $b$ are greater than 1 since the positive integral powers and roots of real numbers greater than 1 are also greater than 1. Hence if $t<r$,
$$b^t<b^tb^{r-t}=b^{t+r-t}=b^r.$$ Hence, since $b^r\in B(r)$, we have $b^r=\sup B(r)$.

(d): We need to show that $\sup B(x+y)=\sup B(x)\sup B(y)$. If $S$ and $T$ are sets of positive real numbers, it isn’t hard to show that $\sup ST = \sup S\sup T$, where $ST$ is the set of products of elements of $S$ with elements of $T$. Hence we want to show that $\sup B(x+y)=\sup(B(x)B(y))$.

Let $p,q$ be rational numbers such that $p\le x$ and $q\le y$. Then $b^pb^q=b^{p+q}\le\sup B(x+y)$, so $\sup(B(x)B(y))\le\sup B(x+y)$.

To get the reverse inequality, let $t$ be any rational number such that $t<x+y$, and let $\epsilon >0$ such that $t<x+y-\epsilon$. By Theorem 1.20(b), there are rational numbers $p\le x$, and $q\le y$ such that $x-\epsilon/2<p$ and $y-\epsilon/2<q$. Then $t<x+y-\epsilon<p+q$ so that $$b^t<b^{p+q}=b^pb^q\le\sup B(x)B(y).$$ Hence $\sup B(x+y)\le\sup(B(x)B(y))$.


Exercise 7

(By analambanomenos)

(a): Using the binomial expansion, for any positive integer $n$, $$b^n=(1+(b-1))^n=1^n+n1^{n-1}(b-1)+A$$ where $A$ is a non-negative sum of terms involving higher powers of $(b-1)$. Hence
$$b^n-1=n(b-1)+A\ge n(b-1).$$

(b): Replace $b$ in case (a) with $b^{1/n}$ to get $b-1\ge n(b^{1/n}-1)$.

(c): From case (b) we have
\begin{align*}
\frac{b-1}{t-1}\,(b^{1/n}-1) &< n(b^{1/n}-1)\le b-1 \\
b^{1/n}-1 &< t-1 \\
b^{1/n} &< t
\end{align*}(d): By case (c), if $n>(b-1)/(yb^{-w}-1)$, then $b^{1/n}<yb^{-w}$, or $b^{w+(1/n)}<y$.

(e): Applying case (c) to $t=y^{-1}b^w>1$, if $n>(b-1)/(y^{-1}b^w-1)$, then $b^{1/n}<y^{-1}b^w$, or $y<b^{w-(1/n)}$.

(f): If $b^x<y$, then from case (d) there is a sufficiently large integer $n$ such that $b^{x+(1/n)}<y$, that is, $x+(1/n)\in A$, contradicting $x=\sup A$. And if $b^x>y$, then from case (e) there is a sufficiently large integer $n$ such that $b^{x-(1/n)}>y$, so that $x-(1/n)$ is an upper bound of $A$, contradicting $x=\sup A$.

(g): Suppose there are real numbers $x_1<x_2$ such that $b^{x_1}=b^{x_2}$. Then $b^{x_1}b^{x_2-x_1}=b^{x_2}=b^{x_1}$, so that $b^{x_2-x_1}=1$. Using the definition of real powers given in exercise 6, this means there are positive integers $m,n$ such that $(b^m)^{1/n}\le 1$. However, since the positive integeral powers of numbers greater than 1 are also greater than 1, this is impossible.


Exercise 8

(By ghostofgarborg) By proposition 1.18d, an ordering $<$ that makes $\mathbb{C}$ an ordered field would have to satisfy $-1 = i^2 > 0$, contradicting $1 > 0$.


Exercise 9

(By ghostofgarborg) Let $z_i = a_i + b_i$.

Property 1.5(i): Assume $z_1 \neq z_2$. If $a_1 \neq a_2$, either $z_1 < z_2$ or $z_2 < z_1$. Otherwise $b_1 \neq b_2$, in which case $z_1 < z_2$ or $z_2 < z_1$.

Property 1.5(ii): Assume $z_1 < z_2$ and $z_2 < z_3$. Then $ a_1 \leq a_2 \leq a_3 $. If either of the inequalities is strict, $a_1 < a_3$, and $z_1 < z_3$. Otherwise, $b_1 \leq b_2 \leq b_3$ and all inequalities are strict, so that $z_1 < z_3$.

This proves that the order is well-defined.

The set does not have the least upper bound property. Consider the set $ \mathbb R i = \{ 0 + xi : i \in \mathbb R \}$. The set is bounded above by $1$. Assume for contradiction that $\alpha = a + bi$ is a least upper bound. It is clear that we must have $a \geq 0$. If $a > 0$, then $\frac 1 2 \alpha$ is another upper bound that is strictly smaller than $\alpha$, contradicting minimality. Therefore, we must have $a=0$. Then $\alpha + i \in \mathbb R i$ is greater than $\alpha$, contradicting $\alpha$ being an upper bound.


Exercise 10

(By analambanomenos}) We have
$$(a^2-b^2) = \frac{|w|+u}{2}-\frac{|w|-u}{2} = u$$ and $$2ab = (|w|+u)^{1/2}(|w|-u)^{1/2} = (|w|^2-u^2)^{1/2} = (v^2)^{1/2} = |v|.$$ Hence $$z^2=(a^2-b^2)+2abi=u+|v|i=w$$ if $v\ge 0$, and $$(\overline{z}^2)=(a^2-b^2)-2abi=u-|v|i=w$$ if $v\le 0$. Hence every nonzero $w$ has two square roots $\pm z$ or $\pm\overline{z}$. Of course, 0 has only one square root, itself.

Baby Rudin 数学分析原理完整第一章习题解答

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

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