Section 4: Problem 2 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Prove the following "laws of inequalities" for $\mathbb{R}$ , using axioms (1)-(6) along with the results of Exercise 1:

(a) $x>y$ and $w>z$ $\Rightarrow$ $x+w>y+z$ .

(b) $x>0$ and $y>0$ $\Rightarrow$ $x+y>0$ and $x\cdot y>0$ .

(d) $x>y$ $\Leftrightarrow$ $-x<-y$ .

(e) $x>y$ and $z<0$ $\Rightarrow$ $xz<yz$ .

(f) $x\neq0$ $\Rightarrow$ $x^{2}>0$ , where $x^{2}=x\cdot x$ .

(g) $-1<0<1$ .

(h) $xy>0$ $\Leftrightarrow$ $x$ and $y$ are both positive or both negative.

(i) $x>0$ $\Rightarrow$ $1/x>0$ .

(j) $x>y>0$ $\Rightarrow$ $1/x<1/y$ .

(k) $x<y$ $\Rightarrow$ $x<(x+y)/2<y$ .

(a) $x+w>x+z>y+z$ .

(b) Using (a), $x+y>0+0=0$ , and using Exercise 1(b), $x\cdot y>0\cdot y=0$ .

(d) Similar to (c) (from left to right add $(-x)+(-y)$ to both sides, from right to left add $x+y$ to both sides).

(e) Using (c) and Exercise 1(d), $z=-(-z)<0$ implies $-z>0$ , so, using Exercise 1(e), $x\cdot(-z)=-(xz)>y\cdot(-z)=-(yz)$ , and by (d), $xz<yz$ .

(f) If $x>0$ then (b) implies $x^{2}>0$ , and if $x<0$ then (e) and Exercise 1(b) imply $0\cdot x=0<x^{2}$ .

(g) $1\cdot1=1$ implies two things: a) from Exercise 1(b) it follows that $1\neq0$ , and b) from (f) it follows that $1^{2}=1>0$ . Now, using (c), $-1<0$ .

(h) If $x$ and $y$ are both positive, then, by (b), $xy>0$ . If they are both negative, then, by (e) and Exercise 1(b), $0\cdot y=0<yx$ . If at least one of them is $0$ , then Exercise 1(b) implies $xy=0$ . If $x>0$ and $y<0$ , then, by (e) and Exercise 1(b), $xy<0\cdot y=0$ . Similarly, if $x<0$ and $y>0$ , $xy<0$ .

(i) Using (g) and (h), $x\cdot(1/x)=1>0$ implies that $x$ and $1/x$ are both positive or both negative.

(j) Using transitivity, $x>0$ , using (i), $1/x>0$ and $1/y>0$ , and now we can just multiply both sides of $x>y$ by $1/x$ and then by $1/y$ .

(k) First, add $x$ or $y$ to both sides of $x<y$ to obtain $x+x<x+y$ and $x+y<y+y$ . Now, argue that for any $z$ , $(z+z)/2=(z\cdot1+z\cdot1)\cdot(1/2)$ $=z\cdot(1+1)\cdot(1/(1+1))$ $=z$ . Further, using (g) and (a), $1+1=2>0+0=0$ , and (i) implies $1/2>0$ . Hence, $(x+x)/2=x<(x+y)/2<(y+y)/2=y$ .

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Section 4: Problem 2 Solution

Section 4: Problem 2 Solution