1549901369-Elements_of_Real_Analysis__Denlinger_

14 Chapter 1 • The Real Number System

Theorem 1.2.10 (Further Algebraic Properties of Inequalities) In any ordered
field F , the fallowing properties hold:

(a) 0 < x < y? 0 < y-^1 < x-^1.

(b) If x < y and u < v, then x + u < y + v.

( c) If 0 < x < y and 0 < u < v, then xu < yv and :'.. < '}!_.

v u

x+y

( d) If x < y , then x < -

2

• < y.

Proof. (a) First, the "=~»' part. Suppose 0 < x < y. By Corollary 1.2.9,
x-^1 > O and y-^1 > 0. Then (xy)-^1 = x-^1 y-^1 > 0. Multiplying both sides of
the inequality x < y by x-^1 y-^1 , we have
x(x-^1 y-^1 ) < y(x-^1 y-^1 ), i.e.,
(xx-^1 )y-^1 < (yy-^1 )x-^1 , i.e.,
y-1 < x-1.
The "~" part is Exercise 14.
(b) Exercise 15.
(c) Suppose 0 < x < y and 0 < u < v. Since u > 0, we may multiply both
sides of the inequality x < y by u and have xu < yu.
Since y > 0, we may multiply both sides of the inequality u < v by y and
have yu < yv. Combining t his inequality with that of the previous paragraph
and the transitive property, we have xu < yv.
The proof of the second claim is Exercise 16.
( d) Exercise 17. •

Theorem 1.2. 11 (The Large and Small of It)

(a) An ordered field has no largest element (and no smallest element).

(b) An ordered field has no smallest positive element (and no largest negative

element).

( c) In any ordered field, P (and consequently F itself) is an infinite set.

Proof. (a) Let F be an ordered field. Let x E F. Can x be the largest
element of F? Adding x to both sides of the inequality 1 > 0, we have x+l > x.
Thus, x cannot be the largest element of F. Hence, F cannot have a largest
element.
(b) Exercise 18.