1549901369-Elements_of_Real_Analysis__Denlinger_

1.2 The Order Properties 13

Theorem 1.2.8 (Algebraic Properties of Inequalities) For any ordered field F,

the following properties hold Vx, y, z E F:

(a) If x < y, and y < z, then x < z. (Transitive property)

(b) x < y iffx+z < y+z; similarly, x < y iffx-z < y-z. (That is, the

same element of F can be added to, or subtracted from, both sides of an

inequality.)

( c) If z > 0, then x < y ==? xz < yz. (That is, if both sides of an inequality
are multiplied by the same positive element, the inequality is preserved.)
( d) If z < 0, then x < y ==? xz > yz. (That is, if both sides of an inequality
are multiplied by the same negative element, the inequality reverses.)
(e) Ifx,y > 0, then x < y ¢=> x^2 < y2.
Proof. (a) Suppose x < y and y < z. By definition, this means y - x E P
and z -y E P. Then
z - x = z + ( -y + y) - x
= (z - y) + (y - x) E P by Axiom (01).
That is, x < z.
(b) Exercise 10.
(c) Suppose z > 0 and x < y. By definition, this means z E P and y-x E P.
Then, by Axiom (02), (y - x)z E P. That is, yz - xz E P. By definition, this
means xz < yz.
( d) Exercise 11.
(e) Suppose x,y > 0.
First, the "==?" part. Suppose x < y. Since x > 0, we may multiply both
sides of this inequality by x, and have

x^2 < xy.

Also, we may multiply both sides of the inequ ality x < y by y and have

xy < y2.

Now, applying the transitive property [Part (a) above],

x2 < y2.

The "¢=" part is Exercise 12. •

Corollary 1.2.9 (a) If x > 0, then x-^1 > 0. [Also, if x < 0, x -^1 < O.]

(b) If both sides of an inequality are divided by the same positive element,

the inequality is preserved.

(c) If both sides of an inequality are divided by the same negative element,

the inequality is reversed.

Proof. Exercise 13. •