Exercises — Integrated Mathematics

I 3.5 Exercises

Exercise I 3.5.1 (1). Prove Theorem I.22: if and , then .

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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 I 3.5.2 (2). Prove Theorem I.23.

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Apply Theorem I.22.

Exercise I 3.5.3 (3). Prove Theorem I.24.

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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 I 3.5.4 (4). Prove Theorem I.25.

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Apply Theorem I.18.

Exercise I 3.5.5 (5). 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 I 3.5.6 (6). 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 I 3.5.7 (7). Given , then , and if , then .

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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 I 3.5.8 (8). Given , then .

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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 Exercise 7 above. The last step also has the inequality backwards: the claim is . A proof: and by Exercise 7, so by Axiom 7, and multiplying by (Theorem I.19) gives .

Exercise I 3.5.9 (9). 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 I 3.5.10 (10). Given and , and , then .

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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.