It may appear that the postulates for an ordered field are an adequate descrip-tion of the real number system. But this is far from true; our postulates, so far, allow some strange possibilities indeed. (See Chapter 32). We shall not discuss these, but merely state postulates which rule them out.
Throughout this section, F is an ordered field.
The easiest way to see the meaning of the following postulate is to think of it geometrically. Suppose we have given two linear segments, like this:
M
•
Figure 1.4
The case of interest is the one in which the first segment is "very long" and the second is "very short." It is reasonable to suppose that if you take enough copies of the second segment, and lay them end to end, you get a segment longer than the first one. And this should be true no matter how long the first may be, and no matter how short the second may be. If the lengths of the seg-ments are the real numbers M and 8, as indicated in the figure, and n copies of the second segment are enough, then we have ns > M.
(This is the idea that Archimedes had in mind when he said that, if you gave him a long enough lever and a fulcrum to rest it on, he could move the world. Let s be the weight of Archimedes, and let M be the weight of the world.
Then Archimedes wanted a lever long enough to give him a mechanical advan-tage of n to 1, where ns > M.)
The algebraic form of this statement follows.
A. The Archimedean Postulate. Let M ands be any two positive numbers.
Then there is a positive integer n such that ns > M .
An ordered field which satisfies this condition is called Archimedean. Hence-forth we shall assume that the real number system forms an Archimedean or-dered field.
32 The Algebra of the Real Numbers
Note that if a certain integer n gives us ns > M, then any larger integer has the same property. Therefore the postulate might equally well have gone on to say that we have ns > M for every integer n greater than or equal to a certain n0.
Even our latest postulate, however, is still not enough for the purposes either of algebra or of geometry. The easiest way to see this is to observe that the field al of rational numbers satisfies all our postulates so far, and in CD the number 2 has no square root. We need to know that our number field is com-plete in such a sense as to permit the ordinary processes of algebra. For a long time to come, it will be sufficient for us to know that every positive number has a square root.
If a > 0, then x is a square root of a if x2 = a. Obviously, if x is a square root of a, then so also is —x. Therefore, if a number has one square root, it must have two. On the other hand, no number a has two different positive square roots x, and x2; if this were so, we would have
2 = = 2 x2i - x 22 = 0,
xi - a — x 2, (x1 — x2) (x, + x2) = 0.
Here x, — x2 0 0, because x, 0 x2, and x, + x2 > 0, because x1 > 0 and x2 > 0.
Therefore the product (x, — x2) (x, + x2) cannot be = 0.
An ordered field is called Euclidean if it satisfies the following condition.
C-1. The Euclidean Completeness Postulate. Every positive number has a posi-tive square root.
We call this the Euclidean postulate because of the part that it will play in geometry. Eventually, this postulate will ensure that circles will intersect lines, and intersect each other, in the ways that we would expect.
It follows, as shown above, that every a > 0 has exactly one positive square root. This is denoted by V. The other square root of a is —Va. We agree that
= 0.
This terminology may be confusing. Consider the following statements.
(1) x is a square root of a.
(2) x = Va.
The second of these statements is not merely a shorthand transcription of the first. Statement (1) means merely that x 2 = a. Statement (2) means not only that x2 = a but also that x -. 0. Strictly speaking, it is never correct to speak of
"the square root of a," except when a = 0, because every a 0 has either two square roots or none at all (in the real number system). One way to avoid this confusion is to pronounce the symbol Vi as "root a," thus warning people that you are pronouncing a formula.
Much later, we shall need another completeness postulate, to guarantee, for example, the existence of 7r. We shall postpone this discussion until we need it.
The following trivial looking observations turn out to be surprisingly useful.
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THEOREM 1. For every real number a there is an integer n > a and an integer m < a.1.8 The Archimedean Postulate; Euclidean Completeness 33
PROOF. In getting n, we can assume a > 0. In the Archimedean postulate, take M = a, e = 1. This gives an n such that n • 1 > a, as desired. To get m < a, we merely take n > —a, and let m = —n. ❑
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THEOREM 2. Between any two real numbers, there is at least one ratio-nal number.(Obviously there are more.)
PROOF. Given x < y. If there is a rational number r, with x +n <r<y + n, then there is a rational number r' = r — n between x and y. We may therefore suppose that
1 < x < y .
0 1
Figure 1.5
Let e = y — x. By the Archimedean postulate, we have ps > 1
for some integer p. Thus
1 < E.
p
The rational numbers with denominator p now divide the whole number line into segments of length 1/p, like this:
x y
I I I I I I
II I
II
—2 —1 0 1 2 . k-1 k k+1 k+2
P P P P P P P P
Figure 1.6
If k/p is the first one of them that lies to the right of x, then k/p ought to be between x and y, because
1 < 8 = y — x . To be more precise, let
K= tn —>x . n
34 The Algebra of the Real Numbers
By the well-ordering principle, K has a least element k. Thus
but
Therefore
— k 1
x-F—<x+ e
P p
x + (y — x)
= y .
Therefore
x< —k
P
<y,which was to be proved. ❑
In using the Archimedean postulate to prove this trivial looking theorem, we are not making any sort of joke; the postulate is needed. There are ordered fields that are Euclidean but not Archimedean. (See Chapter 32.) In such fields, Theorems 1 and 2 do not hold true; in them, some numbers x and y are greater than every integer and hence greater than every rational number. For many of the purposes of geometry we need the Archimedean postulate to rule out such phenomena.