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
