S. D. V. Stephens

notes / Integrated Mathematics / chapter 17: The Real Numbers

17.2 The Field Axioms

The first group of axioms, the field axioms, governs addition and multiplication.

The six axioms

Definition 17.2.1 (Axiom 1 (Commutative laws)). and .

Definition 17.2.2 (Axiom 2 (Associative laws)). and .

Definition 17.2.3 (Axiom 3 (Distributive law)). .

Definition 17.2.4 (Axiom 4 (Existence of identity elements)). and are in , and and .

Definition 17.2.5 (Axiom 5 (Existence of negatives)). For every real number there is a real number such that .

Definition 17.2.6 (Axiom 6 (Existence of reciprocals)). For every real number there is a real number such that .

Consequences of the field axioms

From these axioms we can derive all the laws of elementary algebra and create a series of theorems.

Theorem 17.2.7 (Theorem I.1 (Cancellation law for addition)). If , then . (In particular, this shows that the number of Axiom 4 is unique.)

Theorem 17.2.8 (Theorem I.2 (Possibility of subtraction)). Given and , there is exactly one such that . This is denoted by . In particular, is written simply and is called the negative of .

Theorem 17.2.9 (Theorem I.3). .

Theorem 17.2.10 (Theorem I.4). .

Theorem 17.2.11 (Theorem I.5). .

Theorem 17.2.12 (Theorem I.6). .

Theorem 17.2.13 (Theorem I.7 (Cancellation law for multiplication)). If and , then . (In particular, this shows that the number of Axiom 4 is unique.)

Theorem 17.2.14 (Theorem I.8 (Possibility of division)). Given and with , there is exactly one such that . This is denoted by or and is called the quotient of and . In particular, is also written and is called the reciprocal of .

Theorem 17.2.15 (Theorem I.9). If , then .

Theorem 17.2.16 (Theorem I.10). If , then .

Theorem 17.2.17 (Theorem I.11). If , then or .

Theorem 17.2.18 (Theorem I.12). and .

Theorem 17.2.19 (Theorem I.13). if and .

Theorem 17.2.20 (Theorem I.14). if and .

Theorem 17.2.21 (Theorem I.15). if , , and .

Theorems I.5–I.15 are proved in the exercises at the end of this page, along with some further results.

Exercises

Exercise 17.2.22. 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 17.2.23. 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 17.2.24. 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 the exercise below that proves Theorem I.11 uses the present result, so the argument is circular. A direct proof: since , Axiom 6 gives with , and then .

Exercise 17.2.25. 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 17.2.26. Prove Theorem I.9.

Show solution

Proven via the proof of Theorem I.8.

Exercise 17.2.27. 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 17.2.28. 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 17.2.29. 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 17.2.30. 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 17.2.31. 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 the proof of Theorem I.13 in the previous exercise used this theorem (for ), so citing Theorem I.13 is circular. A proof: by Theorem I.9, , and because . So the product is .

Exercise 17.2.32. 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 17.2.33. 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 17.2.34. Prove that .

Show solution

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

Exercise 17.2.35. 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 (the full statement of Axiom 4 provides two distinct numbers and ), so no satisfies .

Exercise 17.2.36. Prove that .

Show solution

By Axiom 3:

Exercise 17.2.37. Prove that .

Show solution

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

Exercise 17.2.38. Prove that .

Show solution

By Axiom 2 we rearrange to .

Exercise 17.2.39. 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 17.2.40. 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 17.2.41. Prove that if and .

Show solution

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