where, as indicated, $C=\sum_{k=0}^{n-1}\left(\begin{array}{c}n \ k\end{array}\right)|x|^{k}$. Let $\delta$ be small enough that the right side is less than $\left|x^{n}-a\right|$. For example, you could let $\delta=\min \left(\frac{\left|x^{n}-a\right|}{2 C}, 1\right)$. Then if $y \in(x-\delta, x+\delta)$
$$
\left|y^{n}-x^{n}\right| \leq C|x-y|<C \delta \leq\left|x^{n}-a\right|
$$
It follows that on the number line, $y^{n}$ must be between $a$ and $x^{n}$. Consequently, $y^{n}-a \neq 0$ and has the same sign as $x^{n}-a$. (Draw a picture.)
Theorem $2.11 .2$ Let $a>0$ and let $n>1$. Then there exists a unique $x>0$ such that $x^{n}=a$.
Proof: Let $S$ denote those numbers $y \geq 0$ such that $t^{n}-a<0$ for sll $t \in[0, y]$. One such number is 0 . If $a \geq 1$, then a short proof by induction shows $a^{n}>a$ and so, in this case, $S$ is bounded above by $a$. If $a<1$, then another short argument shows $(1 / a)^{n}>a$ and so $S$ is bounded above by $1 / a$. By completeness, there exists $x$, the least upper bound of $S$. Thus for all $y \leq x, y^{n}-a<0$ since if this is not so, then $x$ was not a least upper bound to $S$. If $x^{n}-a>0$, then by the lemma, $y^{n}-a>0$ on some interval $(x-\delta, x+\delta)$. Thus $x$ fails to be a the least upper bound because an upper bound is $x-\delta / 2$. If $x^{n}-a<0$, then by the lemma, $y^{n}-a<0$ on some interval $(x-\delta, x+\delta)$ and so $x$ is not even an upper bound because $S$ would then contain $[0, x+\delta)$. Hence the only other possibility is that $x^{n}-a=0$. That is, $x$ is an $n^{\text {th }}$ root of $a$.
This has shown that $a$ has a positive $n^{\text {th }}$ root. Could it have two? Suppose $x, z$ both work. If $z>x$, then by the binomial theorem,
$$
\begin{aligned}
z^{n} &=(x+z-x)^{n}=\sum_{k=0}^{n}\left(\begin{array}{c}
n \
k
\end{array}\right) x^{n-k}(z-x)^{k} \
&=x^{n}+\sum_{k=0}^{n-1}\left(\begin{array}{c}
n \
k
\end{array}\right) x^{n-k}(z-x)^{k}=a+\sum_{k=0}^{n-1}\left(\begin{array}{c}
n \
k
\end{array}\right) x^{n-k}(z-x)^{k}>a .
\end{aligned}
$$
Turning the argument around, it is also not possible that $z<x$. Thus the $n^{\text {th }}$ root is also unique.
From now on, we will use this fact that $n^{t h}$ roots exist whenever it is convenient to do so.
$2.12$ Exercises
Let $S=[2,5]$. Find sup $S$. Now let $S=[2,5)$. Find sup $S$. Is sup $S$ always a number in $S$ ? Give conditions under which $\sup S \in S$ and then give conditions under which inf $S \in S$.
Show that if $S \neq \emptyset$ and is bounded sbove (below) then sup $S$ (inf $S$ ) is unique. That is, there is only one least upper bound and only one greatest lower bound. If $S=\emptyset$ can you conclude that 7 is an upper bound? Can you conclude 7 is a lower bound? What sbout $13.5 ?$ What about any other number?
Let $S$ be a set which is bounded sbove and let $-S$ denote the set ${-x: x \in S}$. How are inf $(-S)$ and sup $(S)$ related? Hint: Draw some pictures on a number line. What about sup $(-S)$ and inf $S$ where $S$ is a set which is bounded below?

4. 在下面的 $0=2 ?$ Let $x=y=1$ 的论证中，哪个领域公理被滥用了。然后 $$ 0=x^{2}-y^{2}=(x-y)(x+y) $$ 所以$0=(x-y)(x+y)$。现在将两边除以$x-y$，得到$0=x+y=$$1+1=2$。 5. 给出三角不等式中等式成立的条件。 6. 设$k \leq n$ 其中$k$ 和$n$ 是自然数。 $P(n, k)$，每次取 $k$ 的 $n$ 个事物的排列，定义为形成 $k$ 个数字的有序列表的不同方式的数量，$\{1, 2，\cdots，n\}$。显示 $$ P(n, k)=n \cdot(n-1) \cdots(n-k+1)=\frac{n !}{(n-k) !} 。 $$ 7. 使用前面的问题，显示从 $n$ 个事物集合中选择 $k$ 个事物集合的方法数是 $\left(\begin{array}{l}n \\ k\end{array }\右）$。 8. 证明问题 7 中的二项式定理。提示：当你取 $(x+y)^{n}$ 时，注意结果将是 $\mathrm{s}$ 形式的项的总和，$a_{k} x^{nk} y^{ k}$ 并且您需要确定 $a_{k}$ 应该是什么。想象一下写成 $(x+y)^{n}=(x+y)(x+y) \cdots(x+y)$ ，其中乘积中有 $n$ 个因子。现在考虑当你乘法时会发生什么。每个因素对一个典型术语贡献一个 $x$ 或一个 $y$。 9. 通过归纳证明 $n0$ 并且 $x$ 是一个满足二次方程的实数， $$ 一个 x^{2}+b x+c=0 。 $$ 根据 $a$ 和 $b$ 以及涉及这些数字的表达式的平方根，找到 $x$ 的公式。提示：首先除以 $a$ 得到 $x^{2}+\frac{b}{a} x+\frac{c}{a}=0$。然后加减数量 $b^{2} / 4 a^{2}$。验证 $$ x^{2}+\frac{b}{a} x+\frac{b^{2}}{4 a^{2}}=\left(x+\frac{b}{2 a}\right)^ {2} $$