Exercises — Integrated Mathematics

I 3.3 Exercises

Exercise I 3.3.1 (1). Prove Theorem I.5: .

Show solution

By Axiom 3 we know that , and by Axiom 5 we know that . Once again by Axiom 3 we distribute to and to , resulting in , which by Axiom 5 is simply . ( was previously shown by Theorem I.3.)

Remark (Correction). The step is Theorem I.12, which is proved later, so it is unavailable here. A proof from earlier results: by Axiom 3 and Theorem I.2, so is the number with , and that number is by Theorem I.2.

Exercise I 3.3.2 (2). Prove Theorem I.6: .

Show solution

By Axiom 4, , thus , and via elementary algebraic manipulation, . Thus multiples of zero will always result in zero.

Remark (Correction). The argument shows and does not show . A proof: by Axioms 1, 3 and 4, so by Theorem I.1; and by Axiom 1.

Exercise I 3.3.3 (3). Prove Theorem I.7: if and , then .

Show solution

By Axiom 3 we can factor: . By Theorem I.6 we know that . Thus either or in order to have . We know , thus must be equal to . By Axiom 5 we know that there is a real number such that , and by Theorem I.3 we know that . Thus .

Remark (Correction). The step ‘since , either or ’ is Theorem I.11, the converse of Theorem I.6, and Exercise 7 below proves Theorem I.11 using the present result, so the argument is circular. A direct proof: since , Axiom 6 gives with , and then .

Exercise I 3.3.4 (4). Prove Theorem I.8.

This does not seem fully right, double check/review.

Show solution

Given , there is such that . By Axiom 6 we know that, given , there is a such that . For any , in accordance with Axiom 6, the only possibility for ’s reciprocal such that is . Now multiplying by the reciprocal we get .

Remark (Correction). The argument assumes that the quotient exists, which is part of what has to be proved. A proof: by Axiom 6 there is with ; put , so that and a solution exists. If also , then and by Theorem I.7, so the solution is unique.

Exercise I 3.3.5 (5). Prove Theorem I.9.

Show solution

Proven via the proof of Theorem I.8.

Exercise I 3.3.6 (6). Prove Theorem I.10: if , then .

Show solution

By Theorems I.8–I.9 we know that . Then by Axiom 6 and Theorem I.8, , and is a version of , and by Axiom 4, , thus . Thus .

Remark (Correction). The step (flipping the fraction) is neither an axiom nor an earlier theorem. A proof: , since otherwise (Theorem I.6), contradicting ; and , so is a solution of . By the uniqueness in Theorem I.8, .

Exercise I 3.3.7 (7). Prove Theorem I.11: if , then or .

Show solution

From the proof of Theorem I.7 we learned that and thus either or . Let us now define ; then , and either or . Substitute the variable for : then if , either or (or both).

Remark (Correction). This is circular: the step ‘either or ’ in the proof of Theorem I.7 is the theorem being proved here. A proof: if , then by Theorem I.6, so or .

Exercise I 3.3.8 (8). Prove Theorem I.12: and .

Show solution

By Axiom 3 we can extend to , then to , then to . Furthermore, by Theorem I.4, , and by Axiom 3, . Thus and by Axiom 4, .

Remark (Correction). The identities and are instances of what is being proved, so they cannot be used. A proof: by Axiom 3 and Theorem I.6, so by Theorem I.2. Then , using this identity twice (with by Axiom 1) and Theorem I.4.

Exercise I 3.3.9 (9). Prove Theorem I.13: if and .

Show solution

By Axiom 4, Axiom 6 and Theorem I.8, we know that and . Thus Now by Axiom 3, and . Since , we can combine and to .

Exercise I 3.3.10 (10). Prove Theorem I.14: if and .

Show solution

By Axiom 3 and Theorem I.13 we distribute across , resulting in .

Remark (Correction). There is no sum to distribute over here, and Exercise 9 used this theorem (for ), so citing Theorem I.13 is circular. A proof: by Theorem I.9, , and because . So the product is .

Exercise I 3.3.11 (11). Prove Theorem I.15: if , and .

Show solution

By Axiom 6 we know that there is a such that . For , , as this is the reciprocal (Theorems I.8 and I.9). Via Axiom 3 we can distribute, while noting by Axiom 4 that does not change the value (). Now via distribution, The denominator is of the form , thus it becomes , which, as noted in the proof of Theorem I.10, is simply .

Exercise I 3.3.12 (12). Prove that .

Show solution

Given Axiom 4 and Theorem I.6, we know that . Furthermore, via Theorem I.3, we know that . Thus we may rewrite as . Knowing Theorem I.6, we get , which is true.

Remark (Correction). The identity follows from Theorem I.12 (put there), and Theorem I.3 says something different, namely . A more direct proof: by Axiom 4, so is the number with , and that number is by Theorem I.2. Hence .

Exercise I 3.3.13 (13). Prove that .

Show solution

Given Theorem I.8, we know that . Then, by Axiom 4 and Theorem I.10, .

Exercise I 3.3.14 (14). Prove that has no reciprocal.

Show solution

By Theorem I.8 we know that , but this cannot be, since is only defined when , by definition. Thus we form a contradiction.

Remark (Correction). The remark that is only defined when restates the claim. A proof: for every , (Theorem I.6), and (Apostol’s Axiom 4 provides two distinct numbers and ), so no satisfies .

Exercise I 3.3.15 (15). Prove that .

Show solution

By Axiom 3:

Exercise I 3.3.16 (16). Prove that .

Show solution

By Axiom 3, , and by Theorem I.4 we know that ; thus .

Exercise I 3.3.17 (17). Prove that .

Show solution

By Axiom 2 we rearrange to .

Exercise I 3.3.18 (18). Prove that if and , then .

Show solution

By Theorem I.8 we know that is and is ; then by Axiom 3 and Theorem I.14 we get .

Exercise I 3.3.19 (19). Prove that if .

Show solution

By Axiom 3 and Theorems I.8 and I.14 we find that , and can be distributed to or .

Exercise I 3.3.20 (20). Prove that if and .

Show solution

By Theorems I.3 and I.13, becoming is shown by Axiom 1.