Home » Blog » Prove some consequences of the order axioms

Prove some consequences of the order axioms

Prove the following consequences of the order axioms.

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

  1. 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

  2. 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

  3. 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

  4. 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

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):