Chapter 2. The Real Number System
10. Some Applications of the Completeness Axiom
From everyday experience, one knows that even a large distance y can be measured by a small yardstick x; one only has to mark x off sufficiently many times. This fact was noticed by ancient Greeks; it goes back to the Greek geometer and scientist Archimedes. Mathematically, it means that, given a positive numberx(no matter how small) and another numbery(no matter how large), there always is a natural number n such thatnx > y. This fact, known as the Archimedean property, holds not only for real numbers (i.e., in E1) but
also in many other ordered fields. All such fields are called Archimedean fields to distinguish them from other fields in which this property fails. In particular, we shall now prove that everycomplete field (such asE1) is Archimedean. That is, we have following.
Theorem 1 (Archimedean property). If x and y are elements of a complete ordered field F (e.g., E1) and if x > 0, then there always is a natural n ∈ F such that nx > y.
We shall prove this theorem by showing that the opposite assertion is im- possible since it leads to a contradiction; it will then follow that our theorem must be true.
Thus, given a fixed element x > 0, assume (seeking a contradiction) that there is no natural n with nx > y. Then, for all natural n, we have nx≤y.
This means thaty is an upper bound of the set of all products nx (n= 1, 2, 3, . . .);
call this set M. Clearly, M is nonvoid and bounded above (by y); so, by the assumed completeness of F, M has a supremum, say, q = supM. As q is an upper bound of M, we have (by the definition of M) that nx ≤ q for each natural element n. But if n is a natural element, so is n+ 1. Thus, replacing n by n+ 1, we get (n+ 1)x≤q, whence
nx≤q−x, n= 1, 2, 3, . . . .
In other words, q−x (which is less than q since x > 0) is another upper bound of all nx, i.e., of the set M. But this is impossible because q = supM is by definition the least upper bound of M; so no smaller element, such as q−x, can be its upper bound. This contradiction shows that the negation of our theorem must be false. The theorem is proved.
Note 1. The theorem also holds, with the same proof, for “natural multi- ples” nx=x+x+· · ·+x as defined in §6 (see the note after Definition 3).
Note 2. Theorem 1 shows that no complete ordered field, such as E1 can contain so-called “infinitely small” elements, supposedly 6= 0 but such that all their integral multiples are less than 1. (However, such elements do exist in non-Archimedean fields; and recent research, due to A. Robinson, made use of them in what is now generally called “Nonstandard Analysis”.)
Corollary 1. Given any element y in an Archimedean field F, there always are naturals m, n ∈N such that −m < y < n.
Proof. Given any y ∈ F, use the Archimedean property (with x= 1) to find a natural n ∈ F such that n·1 > y, i.e., n > y. Similarly there is another natural m such that m >−y, i.e., −m < y < n.
Corollary 2. In any Archimedean field, the set N of all naturals has no upper bound, and the set J of all integers has neither upper nor lower bounds. (The negative integers are not bounded below.)
For, by Corollary 1, no element y ∈ F can be an upper bound of N (being exceeded byn∈N), nor can it be a lower bound of the negative integers (since
§10. Some Applications of the Completeness Axiom 87
it exceeds some −m, m∈N).
Although our next theorem is valid in all Archimedean fields (see Problem 2 below), a simpler proof (avoiding the use of Theorem 2of §5) can be given for
complete fields, such as E1. This is our purpose here.
Theorem 2. In an Archimedean field F, every nonvoid right-bounded set of integers has a maximum, and every nonvoid left-bounded set of integers has a minimum.
Proof forcomplete fields. LetM be a nonvoid right-bounded set of integers in a complete field F. By completeness, M has a supremum, call it q. The theorem will be proved if we show thatq ∈M (for, an upper bound that belongs to the set is its maximum). To prove it, we assume the opposite, q /∈M, and seek a contradiction.
Consider the elementq−1. Asq−1< q,Corollary 1of§9 shows thatq−1 is exceeded by some element x ∈M. Since q /∈M, q cannot equal x. Therefore, as q is an upper bound of M, we have x < q, so that q−1 < x < q. Now, as x < q, Corollary 1 of §9 yields another element y ∈ M such that x < y < q, and so
q−1< x < y < q.
But this is impossible becausex and y are integers (being elements of M), and no two distinct integers can lie between q−1 and q (indeed, this would imply 0< y−x <1, with y−xa positive integer, contrary to what was shown in Example (C) of §5).
This contradiction shows that q must belong to M, and hence q = maxM, proving the first clause of the theorem. The second clause is proved quite similarly. We leave it to the reader.
We now use Theorem 2 to obtain two further results.
Corollary 3. Given any element x of an Archimedean field F, there always is a unique integer n∈F such that
n≤x < n+ 1.
(This integer is called the integral part of x, denoted [x].)
Proof. By Corollary 1, there are integers ≤ x. Clearly, the set of all such integers (call it M) is bounded above by x. Hence, by Theorem 2, M has a maximum; call it n. Thus, n is the greatest integer ≤x. It follows that n+ 1
cannot be ≤x, and so n+ 1 > x≥n. Thus n has the desired property. This property, in turn, implies that n = maxM. Hence n is unique, as maxM
is.
As we saw in §4, any ordered field isdense:
If a < b in F, there is x∈F such that a < x < b.
We shall now show that, inArchimedean fields, x can be chosenrational, even if a, b are not. We call this the density of rationals.
Theorem 3 (Density of rationals). Given any elements a and b (a < b) in an Archimedean field F, there always is a rational r ∈F such that a < r < b. (Briefly: The rationals are dense in any Archimedean field.)
Proof. Let p = [a] (the integral part of a); so p ∈ J, p ≤ a. The idea of the proof is to start with p, and then to mark off a small “yardstick” 1
n < b−a several (say m) times until p+ m
n lands inside the interval (a, b) (see Figure 11). p a r b Figure 11 p+m n 1 n
More precisely, as F is Archimedean, there aren, m ∈N, with n(b−a)>1 and m1
n
> a−p.1
Among all such m, fix the least one (it exists by Theorem 2). Then a−p < m n but (m−1) n ≤a−p, 2 so that p+ m n ≤a+ 1 n. Hence a < p+ m n ≤a+ 1 n < a+ (b−a) for 1 n < b−a, by construction . Setting r=p+ m n, we find that a < r < a+b−a =b.
1Here we apply the Archimedean propertytwice: first to find n, we takex= (b−a) and
y= 1; then (having fixedn) we findm, takingx= 1
n, y=a−p.
§10. Some Applications of the Completeness Axiom 89
Moreover,risrational, being the sum of two rationals, pand m
n. (The number p is even an integer, namely the integral part of a.) Thus r is the desired rational, with a < r < b.
Note 3. Having found one rationalr1,a < r1 < b, we can apply Theorem 3
to find another rationalr2, withr1 < r2 < b, then a third rationalr3, withr2 <
r3 < b, and so on, ad infinitum. Continuing, we obtaininfinitely many rationals
between a and b. Thus any interval (a, b), witha < b, in an Archimedean field (such as E1) contains infinitely many rationals.
Problems on Complete and Archimedean Fields
1. Prove the second part of Theorem 2.
2. Prove Theorem 2 for Archimedean fields.
[Hint: IfM6=∅is left-bounded (right-bounded), its elements are greater (less) than someinteger (why?); so one can use the results of Problems 6and7of§7.]
3. From Theorem 2, prove the induction law of§7 for integers in E1.
[Hint: LetAbe the set of those integersn∈E1that satisfyP(n) and are> p. Show
(as inTheorem 2′
of§5) thatAcontainsall integers > p.]
∗4. In Problem 11 of §9, show that if the intervals [a
n, bn] also satisfy (for a fixed d >0) bn−an ≤ d n, n= 1,2, . . . , then ∞ \ n=1
[an, bn] contains only one point, p,
and this p is both supan and infbn. Also show that, if F is only Archi-
medean, the same result follows, provided that ∞
\
n=1
[an, bn]6=∅.
[Hint: Seeking a contradiction, supposeT∞
n=1[an, bn] contains two pointsp, qwith
p−q = r > 0, say. Then, using the Archimedean property, show that there is an
n∈N such that
r > d
n ≥bn−an,
so thatpandqcannot beboth in [an, bn], let alone inT∞n=1[an, bn].]
∗5. Prove that if the principle of nested intervals (cf.Problem 11of§9) holds
in some Archimedean field F, then F is complete.
[Outline: IfM has an upper boundb, prove that supM exists as follows. Fix anya∈M and let
d=b−a, c= 1 2(a+b);
socbisects [a, b].
If there is ana1∈M witha1> c, replace [a, b] by the interval [a1, b]⊆[a, b],
noting that
b−a1< b−c= d 2. If, however, all elements ofM are ≤c, replace [a, b] by
[a, c]⊆[a, b].
In both cases, the new smaller interval (call it [a1, b1]) is such that
[a1, b1]⊆[a, b], b1−a1≤ d
2, a1∈M andb1 is an upper bound ofM. Now let c1 = 12(a1+b1), and repeat this process for [a1, b1] to obtain a new interval
[a2, b2]⊆[a1, b1]; b2−a2≤ d
4; b2an upper bound ofM , a2∈M.
Continuing this process indefinitely, obtain a contracting sequence of intervals [an, bn], with bn −an ≤ d/2n (cf. §6, Problem 8), such that an ∈ M and bn is
an upper bound of M for each n. Then obtain p as in Problem 4 and show that
p= supM, as required.]
6. Prove that an ordered fieldF is Archimedean iff, for any x, y∈F with x >0, there is a natural number n∈E1, with nx > y.
[Hint: UseProblem 11′
of §6.]