Chapter 2. The Real Number System
7. Integers and Rationals
Definition 1.
All naturals in a field F, their additive inverses, and the zero element 0 are called the integral elements or integers (in F).
Below we denote by J the set of all integers in F and by N the set of all naturals, as before. In order to investigate J, we need a lemma.
Lemma. If m, n∈N in a field F, then m−n is an integer in F (m−n∈J).
Proof. We proceed by induction.1 Fixm∈N, and letP(n) meanm−n∈J. (i) P(1) is true. Indeed, m−1 = 0 or m−1 ∈ N by Example (B) in §5.
Thus m−1∈J, by definition. But this means thatP(n) holds forn= 1. (ii) P(k) =⇒ P(k+ 1). SupposeP(n) holds for some particularn=k. This means thatm−k ∈J; that is,m−k ∈N orm−k = 0 or −(m−k)∈N. We must show that this implies [m−(k+ 1)] ∈J, i.e., [(m−k)−1]∈J.
1If F is an ordered field, one can simply apply Example (D) in §5. Indeed, we have
m−n∈N, m−n= 0, or −(m−n)∈N accordingly as m > n, m =n, or m < n. Thus
§7. Integers and Rationals 75
Now, ifm−k ∈N, then (m−k)−1 = 0 or (m−k)−1∈N byExample (B)
in §5. Hence (m−k)−1∈J, as required. This settles the case m−k ∈N. Ifm−k = 0, then (m−k)−1 =−1∈J by definition.
Finally, if−(m−k)∈N, then−(m−k)+1∈N; that is,−[m−(k+1)] ∈N, and so, by definition, [m−(k+ 1)]∈J. But this means that P(k+ 1) is true. Thus, in all three cases, P(k + 1) results from P(k). This completes the induction, and so P(n) holds for every n∈ N, i.e., m−n∈ J for any m, n ∈
N.
Theorem 1. If x and y are integers in a field F, so are x+y and xy.2 Proof. As x, y∈J, we must consider the following possible cases.
(i) If x and y are both naturals, so are x+y and xy by Example (A) in §5. Thus they are integers, as claimed.
(ii) If x or y is 0, all is trivial (we leave this case to the reader).
(iii) If x and y are both additive inverses of naturals, then −x and −y are
naturals; hence so is their sum, (−x) + (−y) = −(x+y). This shows that x+y is the additive inverse of a natural element; so x+y ∈ J by definition. Similarly xy = (−x)(−y)∈N; hence certainly xy ∈J.
(iv) Suppose that one ofx and y (say x) is a natural element while the other (y) is not. Then either y = 0 or −y ∈N. The case y = 0 was taken care of in (ii). If, however, −y ∈ N, the lemma yields x−(−y) ∈ J; that is, x+y ∈ J, as claimed. Also, x(−y) ∈ N. Hence xy is an integer, being the additive inverse of the natural element x(−y) =−xy.
Thus, in all cases, x+y ∈J and xy ∈J. The theorem is proved.
We also have an induction rule for integers similar to that applying to natural elements.
Induction Law for Integers. A proposition P(n) holds for all integers n
greater than a fixed integer p in an ordered field if
(i′) P(n) holds for n=p+ 1, and
(ii′) whenever P(n) holds for all integers n such that p < n < m, then P(n) also holds for n=m (m∈J).
This is proved fromTheorem 2′ in§5 by substitutingx−p fornand noting
that x−p runs over all natural values when x takes on integral values greater than p. (Here we say that “induction starts with p+ 1.”)
Definition 2.
An element x of a field F is said to be rational iff x = p/q for some integral elements p and q, with q 6= 0.3
2So also isx−ysince it reduces to x+ (−y), wherexand−yare integers. 3In particular, the rationals inE1 are called rationalnumbers.
Theorem 2. The sum, the difference, and the product of two rationals x and y in a field F are rational. So also is x/y if y 6= 0.
Proof. Let x = p/q and y = r/s, where p, q, r, s are integers, with q and s different from 0. Then, as is easily seen (cf. Problem 3 in §4),
x±y= ps±qr qs , xy = pr qs, and x y = ps qr
(the latter provided that y and r, too, are different from zero). Thus x±y, xy, andx/y can be written as fractions with integral numerators and denomi- nators. (The fact that numerators and denominators are integers follows from Theorem 1. It is also easily seen that these denominators are not 0 since q, r, s6= 0.) By Definition 2, they are rational elements of F, as required.
It follows, in particular, that −x is rational whenever x is; similarly for x−1 = 1/x if x 6= 0. All integers (including 0 and 1) are rationals since an integer m can be written as m/1.
It is easy to verify that Axioms I to IX remain valid ifE1 is replaced by the set R of all rational elements of an ordered field F. This means that R is an ordered field. It is called the rational subfield of F.
Problems on Integers and Rationals
1. Prove in detail the induction law for integers, stated above.2. Show that the result of Problem 20 in §6, i.e., the division theorem, holds also with N′ replaced by J, the set of all integers.
3. Verify that the set J of all integers in an ordered field F satisfies Ax- ioms I–IX of §2except Axiom V(b). Thus J is not a field.
Structures satisfying Axioms I–IX, except possibly IV(b) and V(b), are called ordered commutative rings. In particular, J is such a ring.
4. Verify that the setRof all rationals in F is a field ifF is and an ordered field if F is.
5. Show that every rationalr in an ordered field F has a unique represen- tation r = m/n in lowest terms, i.e., such that n > 0 and |m| has the least possible value (along with |n|). Also prove that, in this case, m and nare relatively prime, i.e., have no common divisors >1.
[Hint: Ifr >0, letAbe the set of all naturalsmoccurring in various representations
r=m/n. Then applyTheorem 2of§5. The rest follows from the minimality ofm.] 6. Let A be a nonempty set of integers (A ⊂ J) in an ordered field F. Show that if all elements of A are greater than some integer p, then A has a least element.
[Hint: The differences x−p (x ∈ A) are naturals; so by Theorem 2 of §5, one of them is theleast; the corresponding xis the least inA.]
§7. Integers and Rationals 77
7. Let A be as in Problem 6. Show that if all elements of A are less than some integer p, then A has a largest element.
[Hint: Apply the result of Problem 6 to the set F of all additive inverses −x of elementsx∈A, noting that−x >−pfor allx∈A.]
8. From Problems 6 and 7 infer that in any ordered field, two nonzero integers m and n always have a least common multiple and a greatest common divisor.
[Hint: Show first that all common multiples (such asmn) are≥ |m|, while all common divisors are≤ |m|.]
9. Prove: Every integer n > 1 in an ordered field is the product of some finite sequence ofprimes, i.e., integers ≥2, each divisible only by 1 and itself.
[Hint: LetP(n) mean that ncan be so factored, and use induction. P(n) is trivial ifnis itself a prime (e.g.,n= 2).
Now suppose P(n) is true for alln less than somem. If m is not a prime, then m=n1n2for some integersn1andn2greater than 1 butless than m(why?); so by our assumption,n1 andn2 factor into primes, and the same follows form.]
Note: It can be shown that the factorization into primes is unique except for the order in which they occur.
10. Show that there are infinitely many primes.
[Hint: Seeking a contradiction, suppose all primes can be put in a finite sequence
p1, . . . , pn. Then show that
1 + n Y k=1
pk
is not divisible by any of the pk (use the division algorithm theorem; cf. Prob- lem 2). Infer from Problem 9 that 1 +Qnk=1pk is a prime different from all pk (k= 1,2, . . . , n).]
11. Show that every strictly decreasing sequence of positive integers is nec- essarily finite.
[Hint: Use Problem 6 orTheorem 2in§5.]