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 Sincex x x is rational, we see that one of the three cases must be true:x = 0 , x > 0 , x < 0 x=0,x>0,x<0 x=0,x>0,x<0.
Ifx = 0 x=0 x=0, letn = 0 n=0 n=0, we haven ≤ x < x + 1 n≤x Ifx > 0 x>0 x>0, letx = a / b x=a/b x=a/b, in whicha , b a,b a,b are positive integers. Thus by Proposition 2.3.9, we can have
a = b n + q , n , q ∈ N , q < b a=bn+q,\quad n,q∈\mathbf N,q This means
a b = n + q b \frac{a}{b}=n+\frac{q}{b} ba?=n+bq?
Asq < b q1 ? q b = b ? q b = ( b ? q ) ? 1 b > 0 ? 1 > q b ≥ 0 1-\frac{q}{b}=\frac{b-q}{b}=(b-q)?\frac{1}{b}>0 ?1>\frac{q}{b}≥0 1?bq?=bb?q?=(b?q)?b1?>0?1>bq?≥0
So
n ≤ a b = n + q b < n + 1 n≤\frac{a}{b}=n+\frac{q}{b} Ifx < 0 x<0 x<0, letx = ? a / b x=-a/b x=?a/b, in whicha , b a,b a,b are positive integers. Thus by Proposition 2.3.9, we can have
a = b m + q , n , q ∈ N , q < b a=bm+q,\quad n,q∈\mathbf N,q This means
? a = ? b m ? q ? x = ? a b = ? m ? q b = ? m ? 1 + ( 1 ? q b ) -a=-bm-q ?x=-\frac{a}{b}=-m-\frac{q}{b}=-m-1+(1-\frac{q}{b}) ?a=?bm?q?x=?ba?=?m?bq?=?m?1+(1?bq?)
As
0 < 1 ? q b ≤ 1 0<1-\frac{q}{b}≤1 0<1?bq?≤1
We can further divide cases
If0 < 1 ? q / b < 1 0<1-q/b<1 0<1?q/b<1, then letn = ? m ? 1 n=-m-1 n=?m?1, we have
n < x = n + ( 1 ? q b ) < n + 1 n If1 ? q / b = 1 1-q/b=1 1?q/b=1, thenq = 0 q=0 q=0, thus letn = ? m n=-m n=?m, we have
n ≤ x = ? m = n < n + 1 n≤x=-m=n We finished the proof of the existence ofn n n for everyx ∈ Q x∈\mathbf Q x∈Q. To show this n is unique, supposen 1 ≤ x < n 1 + 1 , n 2 ≤ x < n 2 + 1 n_1≤x( n 2 < n 1 + 1 ) ∧ ( n 1 < n 2 + 1 ) ? ( n 2 ≤ n 1 ) ∧ ( n 1 ≤ n 2 ) ? n 1 = n 2 (n_2 To see that there’sN > x , ? x ∈ Q N>x,? x∈\mathbf Q N>x,?x∈Q, we could find an n n such that
n ≤ x < n + 1 n≤x And then letN = n + 1 N=n+1 N=n+1.
Exercise 4.4.2 ( a ) Assume we find a infinite descent sequence{ a n } \{a_n \} {an?} inN \mathbf N N, then we use induction onk k k to show thata n ≥ k , ? k , n ∈ N a_n≥k,?k,n∈\mathbf N an?≥k,?k,n∈N:
First letk = 0 k=0 k=0, then as all thea n a_n an? are natural numbers, we havea n ≥ 0 , ? n ∈ N a_n≥0,?n∈\mathbf N an?≥0,?n∈N.
Now suppose the conclusion is true forK K K, considerK + 1 K+1 K+1, assume we can find aN N N such that
a N < K + 1 ? a N ≤ K a_N Then as{ a n } \{a_n\} {an?} is infinite descent, we can have
a N + 1 < a N ≤ K ? a N + 1 < K a_{N+1}aN+1? But the induction hypothesis showsa N + 1 ≥ K a_{N+1}≥K aN+1?≥K, thus we can’t find suchN N N, which means
a n ≥ K + 1 , ? n a_n≥K+1,\quad ?n an?≥K+1,?n
This completes the induction.
Now that we have showna n ≥ k , ? k , n ∈ N a_n≥k,?k,n∈\mathbf N an?≥k,?k,n∈N, we further show this is impossible:
As{ a n } \{a_n\} {an?} is inN \mathbf N N, we havea 1 ∈ N a_1∈\mathbf N a1?∈N, letk = a 1 k=a_1 k=a1?, we shall have
a n ≥ a 1 , ? k , n ∈ N a_n≥a_1,\quad ?k,n∈N an?≥a1?,?k,n∈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 } \{a_n \} {an?} inZ \mathbf Z Z as
a n = ? n , n ∈ N a_n=-n,\quad \mathbf n∈N an?=?n,n∈N
Or infinite descent sequence{ a n } \{a_n \} {an?} inQ \mathbf Q Q as
a n = 1 n , n ∈ N a_n=\frac{1}{n},\quad n∈\mathbf N an?=n1?,n∈N
Exercise 4.4.3 We find there’s 3 gaps marked why?
Gap 1: A natural number is even ifp = 2 k p=2k p=2k, odd ifp = 2 k + 1 p=2k+1 p=2k+1, in whichk ∈ N k∈\mathbf N k∈N, so every natural number is even or odd, but not both.
For? n ∈ N ?n∈\mathbf N ?n∈N, we can findm , q ∈ N m,q∈\mathbf N m,q∈N such that
n = 2 m + q , q < 2 n=2m+q,\quad q<2 n=2m+q,q<2
So ifq = 0 q=0 q=0, thenn n n is even, ifq = 1 q=1 q=1, thenn n n is odd.
If a number is both odd and even, then we may havem , n ∈ N m,n∈\mathbf N m,n∈N such that
2 m = 2 n + 1 ? m = n + 1 2 2m=2n+1 ?m=n+\frac{1}{2} 2m=2n+1?m=n+21?
This is absurd.
Gap 2:p p p is odd ? p 2 p^2 p2 is odd
We have
( pis odd ) ? ( p = 2 k + 1 , k ∈ N ) ? ( p 2 = 4 k 2 + 4 k + 1 = 2 ( 2 k 2 + 2 k ) + 1 ) ? ( p 2is odd ) \begin{aligned}(p\text{ is odd})&?(p=2k+1,k∈\mathbf N)\\&?(p^2=4k^2+4k+1=2(2k^2+2k)+1)\\&?(p^2\text{ is odd})\end{aligned} (p is odd)??(p=2k+1,k∈N)?(p2=4k2+4k+1=2(2k2+2k)+1)?(p2 is odd)?
Gap 3: For positive integersp , q p,q p,q we havep 2 = 2 q 2 ? q < p p^2=2q^2 ?q【陶哲轩实分析（上）4.4及习题-Analysis I 4.4】p2=2q2?q We letr = p ? q r=p-q r=p?q, then
p 2 = 2 q 2 ? p 2 ? q 2 = q 2 ? r ( p + q ) = q 2 p^2=2q^2 ?p^2-q^2=q^2 ?r(p+q)=q^2 p2=2q2?p2?q2=q2?r(p+q)=q2
Now sincep > 0 , q > 0 p>0,q>0 p>0,q>0, we haveq 2 > 0 q^2>0 q2>0, and
p + q > 0 + q > 0 ? 1 p + q > 0 ? q 2 p + q = r > 0 p+q>0+q>0 ? \frac{1}{p+q}>0 ? \frac{q^2}{p+q}=r>0 p+q>0+q>0?p+q1?>0?p+qq2?=r>0
This meansp > q p>q p>q by definition.