城南讲马堂

陶哲轩实分析4.4及习题-Analysis I 4.4

2019年4月10日2019年12月13日christangdt4条评论

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. Prove Proposition 4.4.1.
Solution: 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 $n\leq x0$ , let $x=a/b$ , in which $a,b$ are positive integers. Thus by Proposition 2.3.9, we can have

$a=bn+q,\quad n,q\in \mathbf N,q<b$

This means

$\dfrac{a}{b}=n+\dfrac{q}{b}$

As $q<b$ and $q\in \mathbf N$ , we have

$1-\dfrac{q}{b}=\dfrac{b-q}{b}=(b-q)\cdot\dfrac{1}{b}>0 \implies 1>\dfrac{q}{b}\geq 0$

$n\leq \dfrac{a}{b}=n+\dfrac{q}{b}<n+1$

If $x<0$ , let $x=-a/b$ , in which $a,b$ are positive integers. Thus by Proposition 2.3.9, we can have

$a=bm+q,\quad n,q\in \mathbf N,q<b$

This means

$-a=-bm-q \implies x=-\dfrac{a}{b}=-m-\dfrac{q}{b}=-m-1+(1-\dfrac{q}{b})$

$0<1-\dfrac{q}{b}\leq 1$

We can further divide cases

If $0<1-q/b<1$ , then let $n=-m-1$ , we have

$n<x=n+(1-\dfrac{q}{b})<n+1$

If $1-q/b=1$ , then $q=0$ , thus let $n=-m$ , we have

$n\leq x=-m=n<n+1$

We finished the proof of the existence of $n$ for every $x\in \mathbf Q$ . To show this n is unique, suppose $n_1\leq x<n_1+1,n_2\leq x<n_2+1$ , then we have

$(n_2<n_1+1)\wedge (n_1<n_2+1) \implies (n_2\leq n_1 )\wedge (n_1\leq n_2 ) \implies n_1=n_2$

To see that there’s $N>x,\forall x\in \mathbf Q$ , we could find a $n$ such that

$n\leq x<n+1$

And then let $N=n+1$ .

$\blacksquare$

Exercise 4.4.2. A definition: a sequence $a_0,a_1,a_2,\dots$ of numbers (naturalnumbers, integers, rationals, or reals) is said to be in infinite descent if we have $a_n>a_{n+1}$ for all natural numbers $n$ (i.e., $a_0>a_1>a_2>\dots$ ).
( a ) Prove the principle of infinite descent: that it is not possible to have a sequence of natural numbers which is in infinite descent.
( b ) Does the principle of infinite descent work if the sequence $a_1,a_2,a_3,\dots$ is allowed to take integer values instead of natural number values? What about if it is allowed to take positive rational values instead of natural numbers? Explain.

Solution:
( a ) Assume we find a infinite descent sequence $\{a_n \}$ in $\mathbf N$ , then we use induction on $k$ to show that $a_n\geq k,\forall k,n\in \mathbf N$ :
First let $k=0$ , then as all the $a_n$ are natural numbers, we have $a_n\geq 0,\forall n\in \mathbf N$ .
Now suppose the conclusion is true for $K$ , consider $K+1$ , assume we can find a $N$ such that

$a_N<K+1 \implies a_N\leq K$

Then as $\{a_n\}$ is infinite descent, we can have

$a_{N+1}<a_N\leq K \implies a_{N+1}<K$

But the induction hypothesis shows $a_{N+1}\geq K$ , thus we can’t find such $N$ , which means

$a_n\geq K+1,\quad \forall n$

This completes the induction.
Now that we have shown $a_n\geq k,\forall k,n\in \mathbf N$ , 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

$a_n\geq a_1,\quad \forall k,n\in \mathbf N$

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

$a_n=-n,\quad \mathbf n\in \mathbf N$

Or infinite descent sequence $\{a_n \}$ in $\mathbf Q$ as

$a_n=\dfrac{1}{n},\quad n\in \mathbf N$

$\blacksquare$

Exercise 4.4.3. Fill in the gaps marked (why?) in the proof of Proposition 4.4.4.
Solution: 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 $n=2m+q,\quad q<2$ .
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 $2m=2n+1 \implies m=n+\frac{1}{2}$ .
This is absurd.

Gap 2: $p$ is odd \implies $p^2$ is odd
We have

$\begin{aligned}(p\text{ is odd})&\implies (p=2k+1,k\in \mathbf N)\\&\implies (p^2=4k^2+4k+1=2(2k^2+2k)+1)\\&\implies (p^2\text{ is odd})\end{aligned}$

Gap 3: For positive integers $p,q$ we have $p^2=2q^2 \implies q<p$
We let $r=p-q$ , then

$p^2=2q^2 \implies p^2-q^2=q^2 \implies r(p+q)=q^2$

Now since $p>0,q>0$ , we have $q^2>0$ , and

$p+q>0+q>0 \implies \dfrac{1}{p+q}>0 \implies \dfrac{q^2}{p+q}=r>0$

This means $p>q$ by definition.

$\blacksquare$

陶哲轩实分析4.3及习题-Analysis I 4.3

2019年4月7日2019年12月13日christangdt2条评论

绝对值和指数值是再简单不过的概念，不过如果要精确定义（特别是指数）还是比较麻烦。这一节有个挺经典的处理，即引入ε-close的概念，其实很有利于初学者理解ε-δ语言。对于绝对值和指数的性质在这一节有比较完整的说明，证明都留作习题，做起来很痛苦但做完比较清晰了。

Exercise 4.3.1. Prove Proposition 4.3.3.
Solution:
( a ) By Lemma 4.2.7, we must have one of $x>0,x=0,x<0$ valid.
If $x$ is positive, then $x>0$ and $|x|=x>0$
If $x$ is negative, then $x=-y,y$ is positive, and $|x|=-x=-(-y)=y>0$
If $x=0$ then $|x|=0$ .
And the conclusion follows.

( b ) By Lemma 4.2.7, we have the trichotomy of rationals, also if one of $x,y=0$ , then the inequality reduces to $|x|\leq |x|$ or $|y|\leq |y|$ , which is always true. So we only consider cases which both $x$ and $y\neq 0$ .
If $x,y$ are positive, then $x+y$ is positive (can be proved by definition), thus
$|x+y|=x+y=|x|+|y|$
If $x,y$ are negative, then $x+y$ is negative (can be proved by definition), thus
$|x+y|=-(x+y)=-x+(-y)=|x|+|y|$
If one of $x,y$ is positive, the other negative, then without loss of generality assume $x>0,y<0$ , then $y=-p,p>0$ , and we have

$|x+y|=|x-p|=\begin{cases}0,\quad x=p\\x-p,\quad x>p\\p-x,\quad x<p\end{cases},\quad |x|+|y|=x+p$

We have

$|x|+|y|-(|x+y|)=\begin{cases}x+p,\quad x=p\\ 2p,\quad x>p\\2x,\quad x<p \end{cases}$

Thus $|x|+|y|-(|x+y|)$ is always positive, which means $|x|+|y|>|x+y|$ .

( c ) If $y\geq |x|$ , by ( a ) we have $y\geq |x|\geq 0$ , thus if $x\geq 0$ we have $0\leq x\leq y$ , if $x<0$ we have $y\geq -x$ or $0>x\geq -y$ , together we get $-y\leq x\leq y$ is true. On the other hand, we have

$\begin{aligned}(-y\leq x\leq y)&\implies (-y\leq x<0)\vee (x=0)\vee (0<x\leq y)\\&\implies (0<-x\leq y)\vee (x=0)\vee (0<x\leq y)\\&\implies (|x|\leq y)\vee (|x|=0)\vee (|x|\leq y)\end{aligned}$

Now since $-y\leq y$ we have $y\geq 0$ , thus

$(|x|\leq y)\vee (|x|=0)\vee (|x|\leq y)\implies |x|\leq y$

Let $y=|x|$ , then $|x|\geq |x|$ , thus $-|x|\leq x\leq |x|$ .

( d ) If $x,y$ are positive, then $xy$ is positive by definition. $|xy|=xy=|x||y|$
if $x,y$ are negative, then $xy$ is positive by definition. $|xy|=xy=(-x)(-y)=|x||y|$
if $x<0,y>0$ , then $|xy|=-xy=(-x)y=|x||y|$
if $x>0,y<0$ , then $|xy|=-xy=x(-y)=|x||y|$
let $y=-1$ , we have $|x\cdot -1|=|-x|=|x||-1|=|x|$

( e ) $d(x,y)=|x-y|\geq 0$ by (a), and

$d(x,y)=0\iff |x-y|=0\iff x-y=0\iff x=y$

( f ) by (d), $d(x,y)=|x-y|=|-(x-y)|=|y-x|=d(y,x)$
( g ) by (b), $d(x,z)=|x-z|=|x-y+y-z|\leq |x-y|+|y-z|=d(x,y)+d(y,z)$

$\blacksquare$

Exercise 4.3.2. Prove the remaining claims in Proposition 4.3.7.
Solution:
( a ) $(x=y)\implies (d(y,x)=0<\varepsilon ,\forall \varepsilon >0)$ . Conversely, assume $x\neq y$ , then $d(x,y)>0$ , and we have $d(x,y)>d(x,y)/2$ , contradicting the given condition.

( b ) We have

$(x\text { is }\varepsilon \text{-close to }y)\implies (d(y,x)\leq \varepsilon )\implies (d(x,y)\leq \varepsilon)\implies (y\text{ is }\varepsilon \text{-close to }x)$

( c ) $(d(y,x)\leq \varepsilon )\wedge (d(z,y)\leq \delta )\implies (d(z,x)\leq d(y,x)+d(z,y)\leq \varepsilon +\delta )$

( e ) $(d(x,y)\leq \varepsilon)\wedge (\varepsilon'>\varepsilon)\implies d(x,y)\leq \varepsilon'$

( f ) We have

$(y\leq w\leq z)\vee (z\leq w\leq y)\\ \implies (y-x\leq w-x\leq z-x)\vee (z-x\leq w-x\leq y-x)$

Now let $d=\max (|y-x|,|z-x|)$ , then $d\leq \varepsilon$ and in both case we have $-d\leq w-x\leq d$ . By Proposition 4.3.3 ( c ), we have $\varepsilon \geq d\geq |w-x|=d(w,x)$

( g ) $d(xz,yz)=|xz-yz|=|x-y||z|\leq \varepsilon |z|$ .

$\blacksquare$

Exercise 4.3.3. Prove Proposition 4.3.10.
Solution:
(a) Use induction.
Case I: we fix $m$ and induction on $n$ . $x^0 x^m=1\cdot x^m=x^m=x^(0+m)$ , and

$x^n x^m=x^{n+m}\implies x^{n+1} x^m=x^n xx^m=x^n x^m x=x^{n+m} x=x^{n+m+1}$

Case II: we fix $n$ and induction on $m$ . $(x^n )^0=1=x^0=x^(n\cdot 0)$ , and

$(x^n )^m=x^nm \implies (x^n )^{m+1}=(x^n )^m\times x^n=x^nm x^n=x^{nm+n}=x^{n(m+1)}$

Case III: induction on $n$ . $(xy)^0=1=1\cdot 1=x^0 y^0$ , and

$(xy)^n=x^n y^n \implies (xy)^{n+1}=(xy)^n\times xy=x^n y^n xy=x^n xy^n y=x^{n+1} y^{n+1}$

( b ) $x=0 \implies x^n=x\cdot x^{n-1}=0$ . On the other hand, if $x^n=0$ , we suppose $x\neq 0$ , then

$|x|>0\implies |x|^n>0\implies |x^n |>0\implies x^n\neq 0$

which is a contradiction.

( c ) $(x\geq y)\implies (x=y)\vee (x>y)$ . If $x=y\geq 0$ , then $x^n=y^n\geq 0$ . Assume $x>y\geq 0$ , then $x=y+z,z$ is a positive rational, we induction on $n>0$ :
If $n=1$ , then $x^n=x,y^n=y$ , so $x^n>y^n\geq 0$ . And

$(x^n>y^n\geq 0)\implies x^{n+1}=x^n x=x^n (y+z)>y^n (y+z)=y^{n+1}+y^n z>y^{n+1}\\y^{n+1}=y^n y\geq 0\cdot y=0$

( d ) $|x|^0=1=|1|=|x^0 |$ , and

$|x^n |=|x|^n \implies |x^{n+1} |=|x^n x|=|x^n ||x|=|x|^n |x|=|x|^{n+1}$

$\blacksquare$

Exercise 4.3.4. Prove Proposition 4.3.12
Solution:
If $m,n$ are natural numbers, then Proposition 4.3.10 guarantees each statement to be true. Now we only consider the cases when $n$ or $m$ is negative integer. In such cases, let $n=-p,m=-q,p,q\in \mathbf N$ .

( a ) First we prove $x^n x^m=x^{n+m}$ , we divide into three cases:
Case I: $n+m\geq 0$ , one of $n,m$ is negative, without loss of generality, assume $n<0$ . Then

$(x^p x^n x^m=(x^p/x^p ) x^m=x^m=x^{p+n+m}=x^p x^{n+m})\wedge (x^p\neq 0)\\ \implies x^n x^m=x^{n+m}$

Case II: $n+m<0$ , one of $n,m$ is negative, without loss of generality, assume $n<0$ . In this case $p>m$ , thus

$\begin{aligned}(x^m x^{p-m}=x^{m+p-m}=x^p )&\implies x^{-p} (x^m x^{p-m} ) x^{m-p}=x^{m-p}\\&\implies x^n x^m=\dfrac{1}{x^{p-m}} =\dfrac{1}{x^{-(n+m)} }=x^{n+m}\end{aligned}$

Case III: $n+m<0$ , both $n,m$ are negative

$x^n x^m=\dfrac {1}{x^p} \cdot \dfrac {1}{x^q} =\dfrac{1}{x^p x^q}=\dfrac{1}{x^{p+q}} =\dfrac {1}{x^{-(n+m)} }=x^{n+m}$

Next we prove $(x^n )^m=x^{nm}$ , we divide into three cases:

Case I: $n<0,m>0$ :

$\left((x^n )^m=\left(\dfrac{1}{x^p}\right )^m \right)\wedge \left(\left(\dfrac{1}{x^p}\right )^m (x^p )^m=(1)^m \right)\\ \implies (x^n )^m=(x^{pm} )^{-1}=\dfrac{1}{x^{pm}} =\dfrac{1}{x^{-nm}} =x^{nm}$

Case II: $n>0,m<0$ :

$(x^n )^m=(x^n )^{-q}=\dfrac{1}{(x^n )^q }=\dfrac{1}{x^{nq}} =\dfrac {1}{x^{n(-m)}} =\dfrac{1}{x^{-nm}} =x^{nm}$

Case III: $n<0,m<0$ : Use the result when $n<0,m>0,(x^n )^m=x^{nm}$ , we have

$(x^n )^m=\dfrac{1}{(x^n )^q} =\dfrac{1}{x^{nq}} =\dfrac {1}{x^{-nm}} =x^{nm}$

Last, we prove $(xy)^n=x^n y^n$ when $n<0$ :

$(xy)^n=(xy)^{-p}=\dfrac {1}{(xy)^p} =\dfrac{1}{x^p} \cdot \dfrac{1}{y^p} =x^n y^n$

( b ) If $n$ is positive then by Proposition 4.3.10 (b) $x^n\geq y^n\geq 0$ , and as $y\neq 0, y^n\neq 0$ . Now if $n$ is negative, $n=-p,p\in \mathbf N$ . we have $x\geq y>0\implies (x=y)\vee (x>y)$ , if $x=y$ , then $x^n=1/x^p =1/y^p =y^n$ , if $x>y>0$ , then by definition

$\begin{aligned}(\dfrac{1}{x}>0)\wedge (\dfrac{1}{y}>0)&\implies \dfrac{1}{xy}>0\\&\implies x\left(\dfrac{1}{xy}\right)>y\left(\dfrac{1}{xy}\right)>0\\&\implies \dfrac{1}{y}>\dfrac{1}{x}>0\\&\implies \left(\dfrac{1}{y}\right)^p>\left(\dfrac{1}{x}\right)^p>0\end{aligned}$

Since $\left(\frac{1}{y}\right)^p=(y^{-1} )^p=y^{-p}=y^n, \left(\frac{1}{x}\right)^p=x^n$ , the conclusion follows.

( c ) We suppose $x\neq y$ , then $(x>y)\vee (x0$ , then by Proposition 4.3.10( c ) we have $x^n\neq y^n$ , a contradiction. If $n<0$ , we can have

$\dfrac{1}{x}\neq \dfrac{1}{y} \implies \left(\dfrac{1}{x}\right)^p\neq \left(\dfrac{1}{y}\right)^p \implies x^n\neq y^n$

also a contradiction.

( d ) When $n<0$ , we have

$\begin{aligned}|x|^n=\dfrac{1}{|x|^p }=\dfrac {1}{|x^p |}&\implies |x|^n |x^p |=1\\&\implies |x|^n |x^p ||x^{-p} |=|x|^n |x^p x^{-p} |=|x^{-p} |=|x^n |\\&\implies |x|^n=|x^n |\end{aligned}$

$\blacksquare$

Exercise 4.3.5. Prove that $2^N\geq N$ for all positive integers $N$ .
Solution: Use induction on $N$ :
$N=1$ : $2\geq 1$ is true.
If $2^N\geq N$ , then

$2^(N+1)=2^N\times 2\geq N\times 2=N+N\geq N+1$

We use the fact that $2>0$ and $N\geq 1$ .

$\blacksquare$

陶哲轩实分析4.2及习题-Analysis I 4.2

2019年4月6日2019年12月13日christangdt2条评论

采用一种变异除法从整数定义有理数的做法，类似于从自然数到整数的过程，更好理解。这一节也是采用先讨论相等的性质，再讨论不等号连接的序的性质这种做法。结论也基本都是看起来显然的，但是这种显然的结论背后的基础是非常严格的。本节习题和4.1节一样是非常值得做的。

Exercise 4.2.1. Show that the definition of equality for the rational numbers is reflexive, symmetric and transitive.
Solution:
Reflexive: $ab=ab \implies a//b=a//b$
Symmetric: $a//b=c//d \implies ad=bc \implies cb=da \implies c//d=a//b$
Transitive:

$\begin{aligned}(a//b=c//d)\wedge (c//d=e//f)&\implies (ad=bc)\wedge (cf=de)\\&\implies (adcf=bcde)\\&\implies (afdc=bedc)\end{aligned}$

Now since $b,d,f\neq 0$ , we need to consider whether $c=0$ or not. If $c=0$ , then we can see that $a=0,e=0$ , thus $a//b=e//f$ is true. If $c\neq 0$ , then $cd\neq 0$ , thus by Corollary 4.1.9, $(afdc=bedc)\implies (af=be)\implies a//b=e//f$

$\blacksquare$

Exercise 4.2.2. Prove the remaining components of Lemma 4.2.3.
Solution: If $a//b=a' //b'$ , then $ab'=a'b$
Product:
As $acb' d=ab' cd=a' bcd=a'cbd$ , we have $(ac)//(bd)=(a'c)//(b'd)$ , thus

$(a//b)(c//d)=(ac)//(bd)=(a' c)//(b' d)=(a'//b')(c//d)$

The proof for $c//d$ can be done in a similar way.
Negation:
As $(-a) b'=-(ab' )=-(a' b)=(-a' )b$ , we have $(-a)//b=(-a' )//b'$ , thus

$-(a//b)=(-a)//b=(-a' )//b'=-(a'//b')$

$\blacksquare$

Exercise 4.2.3. Prove the remaining components of Proposition 4.2.4.
Solution: The second equality has been proved. Uniformly, we define $x=a//b,y=c//d,z=e//f$

$x+y=y+x$
$x+y=(ad+bc)//(bd)=(cb+da)//(db)=y+x$
$x+0=0+x=x$
Let $y=0$ in statement 1, the first equality is true. And
$x+0=(a//b)+(0//1)=(a\cdot 1+b\cdot 0)//b=a//b=x$
$x+(-x)=(-x)+x=0$
Let $y=-x$ in statement 1, the first equality is true. And
$x+(-x)=a//b+(-a)//b=(ab-ab)//b^2=0//b^2=0$
$xy=yx$
$xy=(a//b)(c//d)=(ac)//(bd)\\=(ca)//(db)=(c//d)(a//b)=yx$
$(xy)z=x(yz)$
It’s easy to verify both equal $(ace)//(bdf)$
$x1=1x=x$
Let $y=1$ in statement 4, the first equality is true. And
$x1=(a//b)(1//1)=(a\cdot 1)//(b\cdot 1)=a//b=x$
$x(y+z)=xy+xz$
$x(y+z)=(a//b)((cf+de)//df)=(acf+ade)//(bdf)\\=(acbf+abde)//(b^2 df)=(ac//bd)+(ae//bf)=xy+xz$
$(y+z)x=yx+zx$
A direct result from statement 4 and 7.
$xx^{-1}=x^{-1} x=1$
Let $y=x^{-1}$ in statement 4, the first equality is true. And
$xx^{-1}=(a//b)*(b//a)=ab//ab=1//1=1$

$\blacksquare$

Exercise 4.2.4. Prove Lemma 4.2.7.
Solution: Let $x\in \mathbf Q$ , then if $x=0$ we are done, suppose $x=ab\neq 0$ , then $a\neq 0,b\neq 0$ , we have by Lemma 4.1.11(6), exactly one of $(a>0),(a<0)$ and exactly one of $(b>0),(b<0)$ is true. Thus use Lemma 4.1.5, we have two positive integers m,n such that

$(a>0)\implies a=m,(a<0)\implies a=-m\\ (b>0)\implies b=n,(b<0)\implies b=-n$

So one and only one of the four statements below must be true:

$(a>0)\wedge (b>0)\implies x=m/n>0\\(a<0)\wedge (b>0)\implies x=(-m)n\implies x<0\\ (a>0)\wedge (b<0)\implies x=m/(-n)=(-m)/n\implies x<0\\(a<0)\wedge (b<0)\implies x=(-m)/(-n)=m/n\implies x>0$

This concludes the case.

$\blacksquare$

Exercise 4.2.5. Prove Proposition 4.2.9.
Solution:
( a ) Let $z=x-y\in \mathbf Q$ , then due to Lemma 4.2.7, exactly one of $z>0,z=0,z<0$ is true.

( b ) We have

$\begin{aligned}(x>y)&\iff (z=x-y\text{ is positive})\\&\iff (-z=-x+y=y-x\text{ is negative})\\&\iff (y<x)\end{aligned}$

( c ) $(x<y)\wedge (y<z)\implies (p=y-x\text{ is positive})\wedge (q=z-y\text{ is positive})$
Let $p=a/b,q=c/d,a,b,c,d$ are positive, then

$z-x=q+p=c/d+a/b=(cb+ad)/bd \text{ is positive } \implies x<z$

( d ) We have

$\begin{aligned}(x<y)&\implies (p=y-x\text{ is positive})\\&\implies p=(y+z)-(x+z)\text{ is positive}\\&\implies y+z>x+z\end{aligned}$

( e ) $(x<y)\implies (p=y-x\text{ is positive})$ . Let $p=a/b,a,b$ are positive, and $z=c/d,c,d$ are positive. Then $ac$ and $bd$ are positive since they are natural nonzero numbers, thus $pz=ac/bd\text{ is positive} \implies (y-x)z=yz-xz\text{ is positive} \implies yz>xz$ .

$\blacksquare$

Exercise 4.2.6. Show that if $x,y,z$ are rational numbers such that $x<y$ and $z$ is negative, then $xz>yz$ .
Solution: If $z$ is negative, $\exists q\in \mathbf Q,q$ is positive, s.t. $z=-q$ , let $q=cd,c,d$ are positive, and

$(x<y)\implies (p=y-x\text{ is positive})$

Let $p=a/b,a,b$ are positive, then

$yz-xz=(y-x)z=pz=p(-q)=-pq=(-ac)/bd$

which is a negative number, thus $yz<xz$ .

$\blacksquare$

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