Exercise 4.2

Main Problem.

(a)

Proof. We have

$$\begin{array}{lllllll}\hfill x+w& >y+w& \hfill \text{(by (6) since }x>y\text{)}& & \hfill & & \hfill \\ \hfill & =w+y& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & >z+y& \hfill \text{(by (6) since }w>z\text{)}& & \hfill & & \hfill \\ \hfill & =y+z& \hfill \text{(by (2))}& & \hfill & & \hfill \end{array}$$

as desired. □

(b)

Proof. First, we have

$$\begin{array}{lllllll}\hfill x+y& >0+y& \hfill \text{(by (6) since }x>0\text{)}& & \hfill & & \hfill \\ \hfill & =y+0& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =y& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & >0.& \hfill & & \hfill & \end{array}$$

Also

$$\begin{array}{lllllll}\hfill x\cdot y& >0\cdot y& \hfill \text{(by (6) since }x>0\text{ and }y>0\text{)}& & \hfill & & \hfill \\ \hfill & =0& \hfill \text{(by Exercise 4.1 part (b))}& & \hfill & & \hfill \end{array}$$

as desired. □

(c)

Proof. $\text{()}$ Suppose that $x>0$. Then we have

$$\begin{array}{lllllll}\hfill -x& =-x+0& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =0+(-x)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & <x+(-x)& \hfill \text{(by (6) since }0<x\text{)}& & \hfill & & \hfill \\ \hfill & =0.& \hfill \text{(by (4))}& & \hfill & & \hfill \end{array}$$

$\text{()}$ Suppose now that $-x<0$. Then we have

$$\begin{array}{lllllll}\hfill x& =x+0& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =0+x& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & >-x+x& \hfill \text{(by (6) since }0>-x\text{)}& & \hfill & & \hfill \\ \hfill & =x+(-x)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =0& \hfill \text{(by (4))}& & \hfill & & \hfill \end{array}$$

as desired. □

(d)

Proof. $\text{()}$ Suppose that $x>y$. Then we have

$$\begin{array}{lllllll}\hfill -y& =-y+0& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =-y+(x+(-x))& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =(x+(-x))+(-y)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =x+(-x+(-y))& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & >y+(-x+(-y))& \hfill \text{(by (6) since }x>y\text{)}& & \hfill & & \hfill \\ \hfill & =y+(-y+(-x))& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =(y+(-y))+(-x)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =0+(-x)& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =-x+0& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =-x.& \hfill \text{(by (3))}& & \hfill & & \hfill \end{array}$$

$\text{()}$ Now suppose that $-x<-y$. Then we have

$$\begin{array}{lllllll}\hfill x& =x+0& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =x+(y+(-y))& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =(y+(-y))+x& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =(-y+y)+x& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =-y+(y+x)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & >-x+(y+x)& \hfill \text{(by (6) since }-y>-x\text{)}& & \hfill & & \hfill \\ \hfill & =-x+(x+y)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =(-x+x)+y& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =(x+(-x))+y& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =0+y& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =y+0& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =y& \hfill \text{(by (3))}& & \hfill & & \hfill \end{array}$$

as desired. □

(e)

Proof. First, by Exercise 4.1 part (d), we have $-(-z)=z<0$ so that $-z>0$ by part (c). Then, since $x>y$, it follows from (6) that

$$\begin{array}{llllll}\hfill x(-z)& >y(-z)& \hfill & & \hfill & \\ \hfill -(\mathit{xz})& >-(\mathit{yz})& \hfill \text{(by Exercise 4.1 part (e) applied to both sides)}& & \hfill & & \hfill \\ \hfill \mathit{xz}& <\mathit{yz}& \hfill \text{(by part (d))}& & \hfill & & \hfill \end{array}$$

as desired. □

(f)

Proof. Since $x\ne 0$ we either have that $x>0$ or $x<0$ since the $<$ relation is an order (in particular a linear order since this is part of the definition of order in this text). If $x>0$ then we have $x^2=x\cdot x>0\cdot x=0$ by (6) (since $x>0$) and Exercise 4.1 part (b). If $x<0$ then we have $0=0\cdot x<x\cdot x=x^2$ by part (e) (since $0>x$) and Exercise 4.1 part (b). Together these show the desired result. □

(g)

Proof. By (4) we know that $1\ne 0$ so that $1^2>0$ by part (f). However, we have $1^2=1\cdot 1=1$ by (3). Hence $1=1^2>0$. It then follows from part (c) that $-1<0$ so that we have $-1<0<1$ as desired. □

(h)

Proof. $\text{()}$ Suppose that $\mathit{xy}>0$. It cannot be that $x=0$, for then we would have $0=0\cdot y=\mathit{xy}>0$ by Exercise 4.1 part (b), which is impossible by the definition of an order. Hence we have $x\ne 0$, and an analogous argument shows that $y\ne 0$ as well. We then have the following:

Case: $x>0$. Suppose that $y<0$. Then, by part (e) and Exercise 4.1 part (b), we have $\mathit{xy}<0\cdot y=0$ since $x>0$ and $y<0$, which contradicts our initial supposition. Thus, since we know that $y\ne 0$, it has to be that $y>0$ as well.

Case: $x<0$. Suppose that $y>0$. Then, by (6) and Exercise 4.1 part (b), we have $0=0\cdot y>\mathit{xy}$ since $0>x$ and $y>0$, which again contradicts the initial supposition. So it must be that $y<0$ also since $y\ne 0$.

Therefore in every case either both $x$ and $y$ are positive or they are both negative. Since $x\ne 0$, these cases are exhaustive so that this shows the result.

$\text{()}$ Suppose that either $x>0,y>0$ or $x<0,y<0$. In the case where both $x>0$ and $y>0$ we clearly have $\mathit{xy}>0\cdot y=0$ by (6) and Exercise 4.1 part (b). In the other case in which $x<0$ and $y<0$ we have $0=0\cdot y<\mathit{xy}$ by part (e) and Exercise 4.1 part (b) since $0>x$ and $y<0$. Hence $\mathit{xy}>0$ in both cases. □

(i)

Proof. First, it cannot be that $1∕x=0$ because then we would have $1=x(1∕x)=x\cdot 0=0\cdot x=0$ by (4), (2), and Exercise 4.1 part (b). This is clearly a contradiction since we know that $1\ne 0$ by (3). Hence $1∕x\ne 0$. Now suppose that $1∕x<0$ so that $1=x(1∕x)<0\cdot (1∕x)=0$ by part (e) since $x>0$ and $1∕x<0$, and we have also used Exercise 4.1 part (b). This is also a contradiction since it was proved in part (g) that $1>0$. Hence the only remaining possibility is that $1∕x>0$ as desired. □

(j)

Proof. First, since the order is transitive, we have $x,y>0$. It then follows from part (i) that $1∕x,1∕y>0$. Then $(1∕x)(1∕y)>0$ by part (h). We then have

$$\begin{array}{lllllll}\hfill \frac{1}{x}& =\frac{1}{x}\cdot 1& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =\frac{1}{x}\left(y\cdot \frac{1}{y}\right)& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =\left(\frac{1}{x}\cdot y\right)\frac{1}{y}& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =\left(y\cdot \frac{1}{x}\right)\frac{1}{y}& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =y\left(\frac{1}{x}\cdot \frac{1}{y}\right)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & <x\left(\frac{1}{x}\cdot \frac{1}{y}\right)& \hfill \text{(by (6) since }y<x\text{ and }(1∕x)(1∕y)>0\text{)}& & \hfill & & \hfill \\ \hfill & =\left(x\cdot \frac{1}{x}\right)\frac{1}{y}& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =1\cdot \frac{1}{y}& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =\frac{1}{y}\cdot 1& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =\frac{1}{y}& \hfill \text{(by (3))}& & \hfill & & \hfill \end{array}$$

as desired. □

(k)

Proof. First, we know by part (g) that $1>0$ so that

$$\begin{array}{lllllll}\hfill 2& =1+1& \hfill \text{(by the definition of 2)}& & \hfill & & \hfill \\ \hfill & >0+1& \hfill \text{(by (6) since }1>0\text{ )}& & \hfill & & \hfill \\ \hfill & =1+0& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =1& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & >0.& \hfill \text{(by part (g))}& & \hfill & & \hfill \end{array}$$

To summarize, $0<1<2$. It then follows from part (i) that $1∕2>0$. We then have

$$\begin{array}{llllll}\hfill x& <y& \hfill & & \hfill & \\ \hfill x+x& <x+y& \hfill \text{(by (6))}& & \hfill & & \hfill \\ \hfill 2x& <x+y& \hfill \text{(by Lemma }\text{1<!-- tex4ht:ref: lem:intreal:xpx -->}\text{ )}& & \hfill & & \hfill \\ \hfill (2x)\frac{1}{2}& <(x+y)\frac{1}{2}& \hfill \text{(by (6) since }1∕2>0\text{)}& & \hfill & & \hfill \\ \hfill (x\cdot 2)\frac{1}{2}& <\frac{x+y}{2}& \hfill \text{(by (2) and the definition of division)}& & \hfill & & \hfill \\ \hfill x\left(2\cdot \frac{1}{2}\right)& <\frac{x+y}{2}& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill x\cdot 1& <\frac{x+y}{2}& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill x& <\frac{x+y}{2}.& \hfill \text{(by (3))}& & \hfill & & \hfill \end{array}$$

Similarly, we have

$$\begin{array}{llllll}\hfill x& <y& \hfill & & \hfill & \\ \hfill x+y& <y+y& \hfill \text{(by (6))}& & \hfill & & \hfill \\ \hfill x+y& <2y& \hfill \text{(by Lemma }\text{1<!-- tex4ht:ref: lem:intreal:xpx -->}\text{ )}& & \hfill & & \hfill \\ \hfill (x+y)\frac{1}{2}& <(2y)\frac{1}{2}& \hfill \text{(by (6) since }1∕2>0\text{)}& & \hfill & & \hfill \\ \hfill \frac{x+y}{2}& <(y\cdot 2)\frac{1}{2}& \hfill \text{(by the definition of division and (2))}& & \hfill & & \hfill \\ \hfill \frac{x+y}{2}& <y\left(2\cdot \frac{1}{2}\right)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill \frac{x+y}{2}& <y\cdot 1& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill \frac{x+y}{2}& <y.& \hfill \text{(by (3))}& & \hfill & & \hfill \end{array}$$

This shows that $x<(x+y)∕2<y$ as desired. □