(A13) (Completeness Axiom26) Every nonempty subset S of F that has an upper bound has the least upper bound, which is an element of F.
To quickly review the terminology from Definition 1.2.14 and the discussion that follows it, if F is an ordered field, and S C F, then an element s E F is called an upper bound for S if and only if for every x E S, x s. Furthermore, t is the least upper bound of S. or supremum of S, denoted by sup S, if and only if t is an upper bound of S, and for every other upper bound r for S, we have t < r. Proof of the following theorem is the subject of Exercise 9.
THEOREM 1.7.7. Suppose that S is a nonempty subset of JR and k is an upper bound of S.
Then k is the least upper bound of S if and only i f for each e > 0 there exists s E S such that
k - e <s.
The set Q is not a complete ordered field. (See Exercise 14.) Intuitively, if we were to line up the rational numbers, we would see plenty of "holes" between them. Real numbers, on the other hand, would not have that property. Real numbers are a complete ordered field. R is the underlying number system throughout this textbook.
As mentioned before, from the axioms above we can prove all the properties of real num- bers. Some of these are presented next and some in the next section.
THEOREM 1.7.$. (Archimedean27 Order Property of). Let x be any real number. Then there exists a positive integer n* greater than x.
26Completeness axiom is sometimes referred to as Dedekind's axiom. Julius Wilhelm Richard Dedekind (1831--1916), a prominent German contributor to the theory of algebraic functions and number theory, is famous for the idea of "cuts" named after him. Dedekind also worked in the theory of irrational numbers.
27Archimedes (c.287-212 BC.), born to a Greek family in Sicily, is considered one of the world's great- est mathematicians. Even though algebra and a convenient number system had not been devel- oped, Archimedes devised methods for finding areas, volumes, and square roots. He created a discipline of hydrostatics and used it to find equilibrium positions for certain floating bodies.
Archimedes also formulated postulates for mechanics and calculated centers of gravity for solids.
When Archimedes was killed during the Roman invasion of Sicily, the practice of great mathemat- ics also disappeared for a very long period of time.
Proof We use proof by contradiction. Suppose that x is an upper bound of N. Since N and Jl satisfies the completeness axiom (A13), there exists m E )l such that m = sup N. Thus, m - 1 is not an upper bound of N. By Theorem 1.7.7 with a 1, there exists ni E N such that m - 1 <n1. Therefore, m < n l + 1, where n 1 + 1 E N. But this contradicts the assumption that m is the least upper bound of N. Hence, a natural number greater than x must exist.
See the proof of Theorem 2.4.8 for a different proof of Theorem 1.7.8.
THEOREM 1.7.9. The following are equivalent statements:
(a) The Archimedean order property of Jl.
(b) For any x, y E there exists n E N such that y < nx.
(c) For any x E 9l there exists n E N such that! <x.
n
(d) The set N is unbounded.
(e) For any x E Jl+, there exists n E N such that n - 1 x <n.
Proofs of Theorem 1.7.9 as well as the next theorem are left as Exercises 10 and 12.
THEOREM 1.7.10. Every open interval (a, b) contains both a rational number and an irrational number.
Thus, between any two irrational numbers, a rational number exists, and between any two rational numbers, an irrational number exists. In addition, we can conclude that every open interval (a, b) contains infinitely many rational and irrational numbers. Why? Now, if a set S c l satisfies the property that S fl I c for any interval I, then S is called dense in R.
Hence, both Q and JR \ Q are dense in R. Can you prove that? Some mathematicians say that S. a subset of JR, is dense if and only if between every two real numbers there exists an element of S. Is this definition equivalent to ours?
Real numbers can also be broken up into two disjoint sets, algebraic and transcendental, that is, not algebraic. An algebraic number is a real root of a polynomial equation
an_lx' 1 +... + alx + a0 = 0
with integer coefficients a, i = 0, 1, 2, ... , n, n E N. A familiar algebraic number is since it is a root of x2 -- 2 = 0. Examples of transcendental numbers are it ,28 e, e'T, and called the Hilbert29number. Cantor proved that the set of transcendental numbers is much larger than 28Johann Heinrich Lambert (1728-1777) was a Swiss-German analyst, number theorist, astronomer, physicist, and philosopher who wrote valuable books on geometry, cartography, and art. Lam- bert introduced hyperbolic functions, foreshadowed the discovery of non-Euclidean geometry, and proved that it is irrational. (See the definition of it in Section 1.1).
29David Hilbert (1862-1943)was born in Germany and is considered by many to be the greatest math- ematician of the twentieth century. Much of the mathematical research of the twentieth century has focused on problems stated by Hilbert. He contributed greatly to many areas of mathematics, but his work in geometry has been more influential than that of anyone since Euclid, who around 300 ac. provided the sole approach to geometry until the nineteenth-century development of non-Euclidean geometry.
c
2,
Sec. 1.7 * Ordered Field and a Real Number S stem 47 the set of algebraic numbers. Great difficulty arises in proving that a given number is tran- scendental. The first transcendental number was constructed by Liouville3Q in 1844. In 1873, Hermite31 proved that a is transcendental (see Part 2 of Section 8.8). In 1882, Lindemann32 proved that 'r is transcendental. Whether or not the numbers 'rand e + it are transcendental is still undetermined.
Exercises 1.7
1. Prove Theorem 1.7.2.
2. Suppose that F is a field and a, b, c E F. Prove that (a) (-a) (-b) = ab.
(b) if a + b = c + b, then a = c.
(c) -a - b = (a + b).
1 1
(d) if a 0, then - 0 and 1 = a.
a a
3. Suppose that F is an ordered field. Prove that (a) if a E F and a > 0, then -a <0.
(b) 0 <1.
(c) if a, b E F and ab > 0, then either both a, b > 0, or both a, b <0. Thus, saying that ab > 0 implies that both a and b are of the same sign.
(d) if a, b E F and ab <0, then either a > 0 and b <0, or a <0 and b > 0. Thus, saying that ab <0 implies that a and b are of opposite sign.
e if a b E F and 0< a<b, then 0< 1< 1.
4. Prove Theorem 1.7.4.
5. Prove that if r> 1 is a real number, then r2' - r and 1r2 < -. r
6. Prove Theorem 1.7.5.
7. If S b is a subset of real numbers that is bounded below, prove that inf S exists.
30Joseph Liouville (1809-1882), a Frenchman, is remembered for his work in differential equations, number theory, and complex analysis. Liouville, although unable to show that the numbers it and e are transcendental, constructed other transcendental numbers.
31 Charles Hermite (1822-1901), an influential French mathematician, taught a number of world- recognized mathematicians. His main areas of expertise were algebra, analysis, and number theory. Among the famous mathematicians Hermite taught was Henri Poincare (1854-1912) of France, a leading mathematician of his time. Poincare initiated both algebraic topology and the theory of analytic functions of several complex variables. He also made major contributions to algebraic geometry, number theory, optics, electricity, telegraphy, elasticity, thermodynamics, po- tential theory, the theory of relativity, celestial mechanics, the theory of light, electromagnetic waves, and so on. Poincare is considered to be the last of the great "Renaissance" mathematicians.
32 Ferdinand Lindemann (18521939), a German mathematician, did research on projective geometry and Abelian functions and developed a method of solving equations of any degree using transcendental functions. Lindemann also proved that it is impossible to square the circle using a ruler and compass.
b a
8. Prove that if a set A has a supremum, then sup A is unique.
9. Prove Theorem 1.7.7.
10. Prove Theorem 1.7.9.
11. Prove that the product of a nonzero rational number together with an irrational number is an irrational number.
12. Prove Theorem 1.7.10.
13. If possible, give an example of a nonempty bounded subset of Q that (a) has a least upper bound and a maximum in Q.
(b) has a least upper bound but no maximum in Q.
(c) does not have a least upper bound in Q.
14. Prove that Q is not a complete ordered field. (See Exercise 6.) 15. Prove that there exists a real number x such that x2 = 2.
16. Prove that 3 2 + is an algebraic number.
1.8* Some Properties of Real Numbers
In this section we review only the properties of the real numbers that will be most needed in the rest of this textbook. Many of them will be proved. We start with roots and factorization of polynomials.
If f (x) = axe + bx + c, with a, b, and c real constants with a 0, then the roots of f are given by the quadratic formula
-b t b2 - 4ac
x= 2a
provided that b2 - 4ac ? 0. (See Exercise 22 for a proof.) The expression b2 - 4ac is called a discriminant. Thus, if b2 - 4ac ? 0, then f (x) can be factored, that is, written as a product of polynomials of lower degree, as
-b + b2 - 4ac -b - b2 - 4ac .f(x)=a x- 2a x-
If f (x) is a difference of two squares, then it can easily be factored. Recall that for real numbers A and B , A2 -- B 2 = (A - B) (A + B). Replacing A and B with their nth powers, we obtain A2n - Ben = (An - B) (A'2 + B'2) for n E N. Also, recall the formulas
A3 - B3 =(A -B)(AZ +AB + BZ) and
A3+B3 =(A-i-B)(A2-AB+B2),
2a 2a
Sec. 1.8 * Some of Real Numbers 49 which can be used to factor polynomials of higher order. These expressions, as well as the ones given in Example 1.3.4, can be generalized to yield
xn -- an - (x - a)(xn-' + axn-2+ a2xn-3 +... +a"-1) ` (x - a) xn-1-kak, n E N,
k=0
x" + a" = (x + a) (xn- ' - ax"-2 + an-' ) for odd natural number n.
Can you prove these formulas? Compare the first expression above to Exercise 2(e) in Sec- tion 1.3. If in the first expression x = 1 and a = r, we obtain 1 - r" _ (1 - r) k=o rk,which is equivalent to
,l - It
1 - r
>Jrk= 1- if r
k=o
See Example 1.3.4. When possible, some more complicated polynomials can be factored using the following theorem.
THEOREM 1.8.1. (Rational Root Theorem) If a rational number expressed in lowest terms as is a root of a polynomial
q
n -
f(x) =a,1x +a,,_lx,t1 +...+ajx+ao,
thenp divides a0 and q divides
Hence, the rational root theorem says that the only candidates for the rational solutions of
2x4-7x3+7x2-7x+5=0
are the numbers p wherepdivides 5 andqdivides 2. Thus, ±, ±52 2q ±!,and ±1 are the only possibilities for rational solutions of this equation. Note that some or all of these might not, in fact, be solutions.
The proof of Theorem 1.8.1 is in Exercise 1. Observe that any polynomial can be written as a product of two nontrivial expressions. For example,
x2+1= x+1- 2x x+1+ 2x .
However, this is not a factorization of x2 + 1. Why?
Example 1.8.2. Prove that + is irrational. (See Exercise 4(d) from Section 1.4.) Proof Let x = +. Then x - = and squaring gives
which can be written as 2J 2x = x2 - 1. With the radical isolated again, we square and reduce to obtain
x4 - lOx2+1 =0.
Because of the construction of this equation, we know that the equation has 4 + as one of its solutions. But in view of Theorem 1.8.1, the only choices for rational solutions of this equation are x = 1 and x = -1, and neither of them satisfy this equation; f (x) = x4 - l Ox2 has no rational roots. Hence, + must be irrational.
- _ ^ ` n-1
J
-
+
2
In Example 1.4.3 we proved that is irrational by a lengthy contradictory argument.
With the help of the rational root theorem, we can draw the desired conclusion quite readily, following the steps in Example 1.8.2. We observe simply that x2 - 2 = 0 has as a solution.
But according to Theorem 1.8.1, the only possible rational solutions of this equation are 1, -1, 2, or -2. Upon direct substitution into the equation, none of these values satisfy it. Therefore,
cannot be a rational number. Hence, it is irrational.
Recall from earlier studies that Descartes' rule of signs33 provides information regarding the number of roots that a polynomial, p, possesses. In addition, the number of sign variations between p (x) and p (-x) also leads to information regarding the number of positive and nega- tive roots of p. Furthermore, the following particular case34 of Gerschgorin's circle theorem35 yields a convenient bound on the roots of p.
THEOREM 1.8.3. Consider a polynomial p (x) = anx' + an_ 1xn -1 + , , + a 1 x + a0. If r I an I + I an -1 l - +I a° I
is a root ofp and M = then rII M.
lanI
A proof of this result is Exercise 10. The generalized version of Theorem 1.8.3, the Ger- schgorin's circle theorem, can be found in linear algebra, numerical analysis, and/or complex analysis texts when studying "eigenvalues"
We close this section with the presentation of properties for inequalities and absolute values.
We will use these results extensively throughout the text.