Ordered Fields Archives - Page 2 of 2

Prove the following consequences of the order axioms.

If $a < b$ and $c < 0$ , then $ac > bc$ .
If $a < b$ , then $-a > -b$ .
If $ab > 0$ , then $a$ and $b$ are both both positive or $a$ and $b$ are both negative.
If $a < c$ and $b < d$ , then $a + b < c + d$ .

Proof. First, by definition of $<$ we have
$a < b \implies b-a \in \mathbb{R}^+$

Then, by Axiom 8 (and since $c < 0$ implies $c \notin \mathbb{R}^+$ and implies $c \neq 0$ ), we have

$c < 0 \implies -c \in \mathbb{R}^+.$

Thus, by Axiom 7 $(b-a)(-c) \in \mathbb{R}^+$ , and then using the field properties (Section I.3.3) we have

$(b-a)(-c) = ac - bc$

Hence, $ac - bc \in \mathbb{R}^+$ , i.e., $ac - bc > 0. \qquad \blacksquare$
Proof. First, $a < b$ implies $b-a \in \mathbb{R}^+$ , and then we have,
$\begin{align*} (b-a) &= -(-(b-a)) & (\text{Theorem I.4})\\ &= -(-b+a) & (\text{Theorems I.3 and I.4})\\ &= -a - (-b) & (\text{I.3.3, Exercise #5}) \end{align*}$

Thus, $-a - (-b) \in \mathbb{R}^+$ ; hence, $-a > -b. \qquad \blacksquare$
Proof. First, we cannot have $a = 0$ or $b = 0$ since by Theorem I.11 this would mean $ab = 0$ ; hence, we could not have $ab > 0$ .
Assume then that $a \neq 0$ and $b \neq 0$ . By assumption $ab > 0$ . Now, if $a$ is positive then

$\begin{align*} ab > 0 &\implies -(ab) < 0 & (\text{Theorem I.23}) \\ &\implies a(-b) < 0 & (\text{Theorem I.12}) \\ &\implies -b < 0 & (\text{Axiom 7}) \\ &\implies b > 0 & (\text{Theorem I.23}) \end{align*}$

On the other hand, if $a < 0$ then

$\begin{align*} ab > 0 &\implies -(ab) < 0 & (\text{Theorem I.23})\\ &\implies (-a)b < 0 & (\text{Theorem I.12}) \\ &\implies b < 0 & (-a > 0 \text{ since } a > 0, \text{ so } b < 0 \text{ by Axiom 7}) \end{align*}$

Thus, $a < 0$ implies $b < 0$ as well. Hence, if $a$ is positive then so is $b$ , and if $a$ is negative then so is $b$ , i.e., either $a$ and $b$ are both positive or both negative $. \qquad \blacksquare$
Proof. Since $a < c$ we have $c-a > 0$ , and since $b < d$ we have $d-b > 0$ . Then,
$\begin{align*} (c-a) + (d-b) &> 0 & (\text{Axiom 7})\\ \implies (c+d) - (a+b) &>0 &(\text{Field properties and Theorem I.3.3, Ex. #5})\\ \end{align*}$

Hence, $a+b < c+d. \qquad \blacksquare$

Stumbling Robot

Tag: Ordered Fields

Sum of negatives is negative.

Prove some consequences of the order axioms