陶哲轩实分析（上）4.4及习题-Analysis I 4.4

这一节的目的是说明rational numbers并不是稠密的，一方面在任何两个有理数之间都能插入一个有理数，另一方面实数轴上有些地方，有理数并不能达到。

本节的习题不多，但总体上这一章为下一章讲实数打好了基础。下一章的一些实数性质其实是有理数以（形式）极限形式作拓展后的结果。

Exercise 4.4.1

Since $x$ is rational, we see that one of the three cases must be true: $x=0,x>0,x<0$.
If $x=0$, let $n=0$, we have $n\le x<x+1$.
If $x>0$, let $x=a/b$, in which $a,b$ are positive integers. Thus by Proposition 2.3.9, we can have
$$a=bn+q,n,q\in \mathbf{N},q<b$$
This means
$$\frac{a}{b}=n+\frac{q}{b}$$
As $q<b$ and $q\in \mathbf{N}$, we have
$$1-\frac{q}{b}=\frac{b-q}{b}=(b-q)\cdot \frac{1}{b}>0\Rightarrow 1>\frac{q}{b}\ge 0$$
So
$$n\le \frac{a}{b}=n+\frac{q}{b}<n+1$$
If $x<0$, let $x=-a/b$, in which $a,b$ are positive integers. Thus by Proposition 2.3.9, we can have
$$a=bm+q,n,q\in \mathbf{N},q<b$$
This means
$$-a=-bm-q\Rightarrow x=-\frac{a}{b}=-m-\frac{q}{b}=-m-1+(1-\frac{q}{b})$$
As
$$0<1-\frac{q}{b}\le 1$$
We can further divide cases
If $0<1-q/b<1$, then let $n=-m-1$, we have
$$n<x=n+(1-\frac{q}{b})<n+1$$
If $1-q/b=1$, then $q=0$, thus let $n=-m$, we have
$$n\le x=-m=n<n+1$$
We finished the proof of the existence of $n$ for every $x\in \mathbf{Q}$. To show this n is unique, suppose $n_1\le x<n_1+1,n_2\le x<n_2+1$, then we have
$$(n_2<n_1+1)\wedge (n_1<n_2+1)\Rightarrow (n_2\le n_1)\wedge (n_1\le n_2)\Rightarrow n_1=n_2$$
To see that there’s $N>x,\forall x\in \mathbf{Q}$, we could find a $n$ such that
$$n\le x<n+1$$
And then let $N=n+1$.

Exercise 4.4.2

( a ) Assume we find a infinite descent sequence $\{a_n\}$ in $\mathbf{N}$, then we use induction on $k$ to show that $a_n\ge k,\forall k,n\in \mathbf{N}$:
First let $k=0$, then as all the $a_n$ are natural numbers, we have $a_n\ge 0,\forall n\in \mathbf{N}$.
Now suppose the conclusion is true for $K$, consider $K+1$, assume we can find a $N$ such that
$$a_N<K+1\Rightarrow a_N\le K$$
Then as $\{a_n\}$ is infinite descent, we can have
$$a_{N+1}<a_N\le K\Rightarrow a_{N+1}<K$$
But the induction hypothesis shows $a_{N+1}\ge K$, thus we can’t find such $N$, which means
$$a_n\ge K+1,\forall n$$
This completes the induction.
Now that we have shown $a_n\ge k,\forall k,n\in \mathbf{N}$, we further show this is impossible:
As $\{a_n\}$ is in $\mathbf{N}$, we have $a_1\in \mathbf{N}$, let $k=a_1$, we shall have
$$a_n\ge a_1,\forall k,n\in N$$
This contradicts the infinite descent condition.
( b ) the principle won’t work for integers or rationals. We can choose infinite descent sequence $\{a_n\}$ in $\mathbf{Z}$ as
$$a_n=-n,\mathbf{n}\in N$$
Or infinite descent sequence $\{a_n\}$ in $\mathbf{Q}$ as
$$a_n=\frac{1}{n},n\in \mathbf{N}$$

Exercise 4.4.3

We find there’s 3 gaps marked why?
Gap 1: A natural number is even if $p=2k$, odd if $p=2k+1$, in which $k\in \mathbf{N}$, so every natural number is even or odd, but not both.
For $\forall n\in \mathbf{N}$, we can find $m,q\in \mathbf{N}$ such that
$$n=2m+q,q<2$$
So if $q=0$, then $n$ is even, if $q=1$, then $n$ is odd.
If a number is both odd and even, then we may have $m,n\in \mathbf{N}$ such that
$$2m=2n+1\Rightarrow m=n+\frac{1}{2}$$
This is absurd.

Gap 2: $p$ is odd ⇒$p^2$ is odd
We have
$$\begin{array}{rl}(p\text{ is odd})& \Rightarrow (p=2k+1,k\in \mathbf{N})\\ & \Rightarrow (p^2=4k^2+4k+1=2(2k^2+2k)+1)\\ & \Rightarrow (p^2\text{ is odd})\end{array}$$

Gap 3: For positive integers $p,q$ we have p 2 = 2 q 2 ⇒ q < p p^2=2q^2 ⇒q

p2=2q2⇒q<p
We let $r=p-q$, then
$$p^2=2q^2\Rightarrow p^2-q^2=q^2\Rightarrow r(p+q)=q^2$$
Now since $p>0,q>0$, we have $q^2>0$, and
$$p+q>0+q>0\Rightarrow \frac{1}{p+q}>0\Rightarrow \frac{q^2}{p+q}=r>0$$
This means $p>q$ by definition.