The Order Axioms — Integrated Mathematics

17.3 The Order Axioms

The second group of axioms orders the real numbers.

Positive numbers and the three axioms

These axioms allow us to establish “ordering.” This allows us to tell if one real number () is larger or smaller than another. We want a new axiom of “positiveness” and then define “less than” or “more than” in axiomatic-positiveness terms. Let us say that there is a subset of () called the set of positive numbers.

Definition 17.3.1 (Axiom 7). If and are in , so are and .

Definition 17.3.2 (Axiom 8). For every real , either or , but not both.

Definition 17.3.3 (Axiom 9). .

Inequalities

Definition 17.3.4. Less than, greater than, less than or equal to, greater than or equal to, respectively:


	is positive

	either or

if is positive, and if is negative. If , is nonnegative, and if , is nonpositive. “ and ” can be rewritten as . These axioms and definitions result in more theorems:

Theorem 17.3.5 (Theorem I.16 (Trichotomy law)). For arbitrary real numbers and , exactly one of the three relations , , holds.

Theorem 17.3.6 (Theorem I.17 (Transitive law)). If and , then .

Theorem 17.3.7 (Theorem I.18). If , then .

Theorem 17.3.8 (Theorem I.19). If and , then .

Theorem 17.3.9 (Theorem I.20). If , then .

Theorem 17.3.10 (Theorem I.21). .

Theorem 17.3.11 (Theorem I.22). If and , then .

Theorem 17.3.12 (Theorem I.23). If , then . In particular, if , then .

Theorem 17.3.13 (Theorem I.24). If , then both and are positive or both are negative.

Theorem 17.3.14 (Theorem I.25). If and , then .

(Scratch for Theorem I.25: .)

Theorems I.22–I.25 are proved in the exercises at the end of this page, along with some further results.

Exercises

Exercise 17.3.15. Prove Theorem I.22: if and , then .

Show solution

Apply Theorem I.19 and reverse the to become , and the to become .

Remark (Correction). Reversing the inequalities is not a step the axioms allow, so this restates the theorem. A proof: means , so Theorem I.19 gives , that is, by Theorem I.12. This says is positive, so .

Exercise 17.3.16. Prove Theorem I.23.

Show solution

Apply Theorem I.22.

Exercise 17.3.17. Prove Theorem I.24.

Show solution

By Axiom 8, we know that or , but not both. Set to ; then is true. Set to ; then is also true.

Remark (Correction). This does not reach the conclusion. A proof: neither nor is , since otherwise (Theorem I.6) and (Axiom 9). If and had opposite signs, say , then Theorem I.19 (multiplying by ) gives , contradicting by Theorem I.16. So and are both positive or both negative.

Exercise 17.3.18. Prove Theorem I.25.

Show solution

Apply Theorem I.18.

Exercise 17.3.19. Prove that has no solution .

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; there is no such number in that satisfies .

Remark (Correction). The sentence ‘there is no real number equal to ’ is what has to be proved, so the argument is circular. A proof: if then (Theorem I.20), and if then ; either way , so (Theorems I.18 and I.21), and therefore .

Exercise 17.3.20. Prove that the sum of two negative numbers is negative.

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Remark (Correction). Rewriting the sum does not show that it is negative. A proof: if and , then and are in , so is in by Axiom 7. Since , this says .

Exercise 17.3.21. Given , then , and if , then .

Show solution

By Theorem I.21 we understand that , thus: and , which is true. Then, by Axiom 8, we know that for every real , either or . Then if and , then , and by the method we see that must exist as a negative number not within but in . By Axiom 3: , which is true.

Remark (Correction). The argument assumes the conclusion at the start, and the ‘th root’ step is undefined, so nothing is proved. A proof: , since . If and , then Theorem I.22 applied to and gives , contradicting Theorem I.21; so . If , then , so by the first case (because ), and therefore .

Exercise 17.3.22. Given , then .

Show solution

By Theorem I.8 we know that if then . Then by Axiom 9, , so it must be that and and . Also by Theorem I.8, and are and , respectively. If , then as greater than . (How?)

Remark (Correction). Theorem I.8 says nothing about signs; the positivity of and is the previous exercise. The last step also has the inequality backwards: the claim is . A proof: and by the previous exercise, so by Axiom 7, and multiplying by (Theorem I.19) gives .

Exercise 17.3.23. Given and , then .

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By the transitive property, we know that if and , then . Furthermore, by Theorem I.17, we know that if and , then . Since means or , these transitive properties apply to .

Exercise 17.3.24. Given and , and , then .

Show solution

Let us assume that and and , but . Then must be . This results in . But this cannot be: since and , cannot be true. Thus we have a contradiction.