Chapter 2. Real Numbers. 1. Rational Numbers

Chapter 2. Real Numbers 1. Rational Numbers A commutative ring is called a field if its nonzero elements form a group under multiplication. Let (F, +, ) be a filed with 0 as its additive identity element and 1 as its multiplicative identity element. We always assume that 1 0. For b F, its additive inverse is denoted by b. The subtraction in F is defined as follows: a b := a + ( b), a, b F. For b F \ {0}, its multiplicative inverse is denoted by b 1. The division in F is defined as follows: a/b := a b := ab 1, a F, b F \ {0}. For n IN 0 and a F, na is defined recursively: na := 0 for n = 0 and (n + 1)a := na + a. Recall that ( Z, +, ) is a commutative ring with 1 as its (multiplicative) identity element. But any integer m Z \ {0, 1, 1} has no multiplicative inverse in Z. Thus, it is desirable to extend the ring ( Z, +, ) to a field. Let us introduce a relation in the set Z IN. For two elements (m, n) and (p, q) in Z IN, we define (m, n) (p, q) if mq = np. We claim that it is an equivalence relation. Clearly, this relation is reflexive and symmetric. Suppose (m, n) (p, q) and (p, q) (r, s). Then mq = np and ps = rq. Hence, mqs = nps = nrq. Since q 0, it follows that ms = nr, that is, (m, n) (r, s). This shows that the relation is transitive. For (m, n) Z IN, we use m n or m/n to denote the equivalence class of (m, n): m := {(r, s) Z IN : rn = sm}. n Then m/n = m /n if and only if mn = nm. Let Q be the set of equivalence classes. An element of Q is called a rational number. The addition in Q is defined by the rule m n + p q mq + np :=. nq The addition is well defined. Indeed, if m/n = m /n and p/q = p /q, then mn = m n and pq = p q. It follows that (mq + np)(n q ) = (m q + n p )(nq). It is easily seen that the addition is associative and commutative. Moreover, 0/1 is the identity element for addition. Each element m/n Q has an addictive inverse: m n + m n = m n + m n = 0 1. 1

Thus, ( Q, +) is an abelian group. The multiplication in Q is defined by the rule m n p q := mp nq. If m/n = m /n and p/q = p /q, then mn = m n and pq = p q. It follows that (mp)(n q ) = (m p )(nq). So the multiplication is well defined. It is easily seen that the multiplication is associative and commutative. Moreover, the multiplication is distributive with respect to addition: m ( p n q + r ) = m s n p q + m n Clearly, 1/1 is the identity element for multiplication. If m/n 0/1 and p/q 0/1, then (m/n)(p/q) 0/1. Furthermore, if m/n Q and m > 0, then n/m is its multiplicative inverse; if m/n Q and m < 0, then ( n)/( m) is its multiplicative inverse. We conclude that ( Q, +, ) is a field. Let ϕ be the mapping from Z to Q given by ϕ(m) = m/1, m Z. Then ϕ is injective. We have ϕ(m + n) = ϕ(m) + ϕ(n) and ϕ(mn) = ϕ(m)ϕ(n) for all m, n Z. Thus, the ring ( Z, +, ) is embedded into the field ( Q, +, ) via the mapping ϕ. We identify ϕ( Z) with Z and write m for m/1. In particular, 0 = 0/1 is the additive identity element, and 1 = 1/1 is the multiplicative identity element. For two elements m/n and p/q in Q, we write m/n p/q if mq np. is a linear ordering in Q. r s. Then Indeed, it is obvious that is reflexive and antisymmetric. Suppose m/n p/q and p/q r/s, where q, s IN. Then mq np and ps qr; hence mqs nps nqr. Since mqs nqr and q > 0, we obtain ms nr, i.e., m/n r/s. This shows that is transitive. Moreover, either mq np or np mq. So is a linear ordering. Furthermore, if a, b Q and a b, then a + c b + c for all c Q, and ac bc for all c 0. A field (F, +, ) with a linear ordering is called an ordered field if the following two conditions are satisfied for all x, y, z F : (a) x y implies x + z y + z; (b) if z 0, then x y implies xz yz. Thus, ( Q, +, ) with its natural ordering is an ordered field. Let F be an ordered field. An element a of F is said to be positive if a > 0, and it is said to be non-negative if a 0. Suppose x, y F. Then x < y implies x + z < y + z for all z F and xz < yz for all z > 0. Moreover, if x 0, then x 2 = x x > 0. Indeed, this is true if x > 0. If x < 0, then 0 = ( x)+x < ( x)+0 = x, and hence x 2 = ( x)( x) > 0. In particular, 1 = 1 2 > 0. 2

An ordered field F is said to be archimedean if for every pair of positive elements x and y in F, there is a positive integer k such that kx > y. The ordered field Q is an archimedean field. Indeed, if x and y are two positive elements in Q, then x = m/n and y = p/q for some m, n, p, q IN. There exists a positive integer k such that kmq > np. Consequently, kx = km/n > p/q = y. Although the rational numbers form a rich algebraic system, they are inadequate for the purpose of analysis because they are, in a sense, incomplete. For example, there is no rational number r such that r 2 = 2. In order to prove this statement, it suffices to show that m 2 2n 2 for any pair of positive integers m and n. Write m = 2 j p and n = 2 k q, where j, k IN 0 and p and q are odd numbers. Then we have m 2 = 2 2j p 2 and 2n 2 = 2 2k+1 q 2. Note that p 2 and q 2 are odd numbers. Moreover, 2j 2k + 1, since one is an even number and the other is an odd number. There are two possibilities: 2j > 2k + 1 or 2j < 2k + 1. If 2j > 2k + 1, then 2 2j 2k 1 p 2 is an even number and q 2 is an odd number. Hence, 2 2j 2k 1 p 2 q 2. It follows that m 2 = 2 2j p 2 2 2k+1 q 2 = 2n 2. If 2j < 2k + 1, then p 2 2 2k+1 2j q 2, and thereby m 2 2n 2. Let A be the set {r Q : r 2 2}. Then A has no least upper bound in Q. 2. Real Numbers The absolute value of a rational number r is defined by r := { r if r 0, r if r < 0. For r, s Q, we have rs = r s and r + s r + s. A sequence (r n ) n=1,2,... of rational numbers is said to converge to a rational number r if, for every rational number ε > 0, there exists a positive integer N(ε) such that r n r < ε for all n N(ε). In this case, we say that r is the limit of the sequence and write lim r n = r. If a sequence of rational numbers converges, then it has a unique limit. convergent sequence is bounded. Also, a A sequence (r n ) n=1,2,... of rational numbers is called fundamental if, for every rational number ε > 0, there exists a positive integer N(ε) such that r m r n < ε whenever m, n N(ε). A sequence of rational numbers that converges (to a rational number) is fundamental. Also, a fundamental sequence of rational numbers is bounded. 3

Two fundamental sequences (r n ) n=1,2... and (s n ) n=1,2... are said to be equivalent if lim (r n s n ) = 0. We write (r n ) n=1,2... (s n ) n=1,2... when (r n ) n=1,2... and (s n ) n=1,2... are equivalent. It is easy to verify that this is indeed an equivalence relation. An equivalence class of a fixed fundamental sequence of rational numbers is called a real number. Thus, a real number is represented by a fundamental sequence (r n ) n=1,2... of rational numbers. Such a representation is not unique. If (s n ) n=1,2... is another representation for the same real number, then we must have lim (r n s n ) = 0. Let IR denote the set of real numbers. We define addition and multiplication on IR as follows. Suppose x and y are real numbers represented by fundamental sequences (r n ) n=1,2,... and (s n ) n=1,2,... of rational numbers, respectively. Then x + y is defined to be the real number represented by (r n + s n ) n=1,2,... and xy is the real number represented by (r n s n ) n=1,2,.... If (r n ) n=1,2... (r n) n=1,2... and (s n ) n=1,2... (s n) n=1,2..., then lim (r n r n) = 0 and lim (s n s n) = 0. It follows that [ lim (rn + s n ) (r n + s n) ] = 0 and lim (r ns n r ns n) = 0. Therefore, the addition and multiplication are well defined. Evidently, the addition and multiplication are associative and commutative. For r Q, let ϕ(r) be the real number represented by the sequence (r n ) n=1,2..., where r n = r for all n IN. Then we have ϕ(r + s) = ϕ(r) + ϕ(s) and ϕ(rs) = ϕ(r)ϕ(s) for r, s Q. Clearly, ϕ is an injective mapping from Q to IR. We identify ϕ(r) with r. Thus, Q is embedded into IR via the mapping ϕ. It is easily seen that 0 is the identity for addition. Suppose x is a real number represented by a fundamental sequence (r n ) n=1,2,... of rational numbers. Let x be the real number represented by ( r n ) n=1,2,.... Then ( x) + x = 0. This shows that (IR, +) is an abelian group. For three real numbers x, y, z IR, we have (x + y)z = xz + yz. In other words, the multiplication is distributive with respect to addition. Thus, (IR, +, ) is a commutative ring with 1 as its (multiplicative) identity. Suppose x IR \ {0} and x is represented by a fundamental sequence (r n ) n=1,2,... of rational numbers. Then there exists a rational number ε > 0 and a positive integer N such that r n ε for all n N. Indeed, if this is not true, then there exists a sequence (n k ) k=1,2,... of positive integers such that n k < n k+1 and r nk < 1/k for all k IN. It follows that lim k r nk = 0 and thereby x = 0. Thus, r n ε for all n N. We define x 1 to be the real number represented by the sequence (s n ) n=1,2,..., where s n = 1 for n < N and s n = rn 1 for n N. Clearly, (s n ) n=1,2,... is a fundamental sequence. We have x 1 x = 1. This shows that (IR, +, ) is a field. 4

A sequence (r n ) n=1,2,... of rational numbers is said to be eventually non-negative (nonpositive) if there exists a positive integer N such that x n 0 (x n 0) for all n N. A real number z is said to be non-negative and written z 0, if z is represented by a fundamental sequence of rational numbers that is eventually non-negative. For two real numbers x and y, we write x y if y x 0. The relation is a linear ordering. Indeed, this relation is reflexive, since x x for all x IR. Let x, y, z IR. If x y and y z, then y x 0 and z y 0. It follows that z x = (z y) + (y x) 0. Hence, is transitive. Moreover, if x y and y x, then both y x and x y are non-negative. Consequently, y x and x y are represented by fundamental sequences (r n ) n=1,2... and (s n ) n=1,2,... of rational numbers respectively such that both sequences are eventually non-negative. In particular, (r n ) n=1,2... and ( s n ) n=1,2,... are equivalent. It follows that lim (r n + s n ) = lim (r n ( s n )) = 0. There exists a positive integer N such that 0 r n r n + s n for all n N. Hence, lim r n = 0. This shows x y = 0, i.e., x = y. Thus, the relation is antisymmetric. It remains to prove that any two real numbers x and y are comparable. Suppose that y x is represented by a fundamental sequence (r n ) n=1,2... of rational numbers. If the sequence is eventually non-negative, then x y; if the sequence is eventually non-positive, then y x. If the sequence is neither eventually non-negative nor eventually non-positive, then both the sets {n IN : r n 0} and {n IN : r n 0} are infinite. Since (r n ) n=1,2... is fundamental, for every rational number ε > 0, there exists a positive integer N such that r m ε < r n < r m + ε for all m, n > N. But there exist m 1, m 2 > N such that r m1 0 and r m2 0. Therefore, ε r m1 ε < r n < r m2 + ε ε. Thus, ε < r n < ε for all n > N. It follows that y x = 0 and x = y. We have proved that the relation is a linear ordering. Theorem 2.1. The field (IR, +, ) with the ordering is an Archimedean ordered field. Proof. Let x, y, z IR. If x y, then (y + z) (x + z) = y x 0; hence x + z y + z. If, in addition, z 0, then yz xz = (y x)z 0. Hence, xz yz for all z 0. This shows that (IR, +, ) with the ordering is an ordered field. Suppose x > 0 and y > 0. Then u := x/y > 0. The real number u is represented by a fundamental sequence (r n ) n=1,2,... of rational numbers. By what has been proved before, there exists a rational number ε > 0 and a positive integer N such that r n ε for all n N. Since u > 0, the set {n IN : r n < 0} is finite. Hence, the sequence (r n ε) n=1,2,... is eventually non-negative. This shows u ε. For the rational number ε > 0, there exists a positive integer k such that kε > 1. Consequently, ku kε > 1. It follows that kx > y. Therefore, (IR, +, ) with the ordering is an Archimedean ordered field. As a corollary of the above theorem, there exists a rational number between any 5

two distinct real numbers. Let x and y be two real numbers such that x < y. By the Archimedean property, there exists a positive integer k such that k(y x) > 2. It follows that y x > 2/k. Let m be the least integer such that m kx. Then x m/k < (m+1)/k. On the other hand, (m 1)/k < x implies (m + 1)/k < x + 2/k < y. Thus, r := (m + 1)/k is a rational number such that x < r < y. Theorem 2.2. Every nonempty subset of IR bounded above has the least upper bound. Proof. Suppose that X is a nonempty subset of IR bounded above. Let B := {b Q : b x for all x X} and A := Q \ B. Then both A and B are nonempty. Moreover, a < b whenever a A and b B. We define two sequences (a n ) n=1,2,... and (b n ) n=1,2,... recursively as follows. Choose a 1 A and b 1 B. Suppose n IN. If (a n + b n )/2 A, set a n+1 := (a n + b n )/2 and b n+1 := b n ; if (a n + b n )/2 B, set a n+1 := a n and b n+1 := (a n + b n )/2. Then a n A, b n B, and b n+1 a n+1 = (b n a n )/2. By induction on n, we obtain b n a n = (b 1 a 1 )/2 n 1 for all n IN. Consequently, lim (b n a n ) = 0. Moreover, we have a n+1 a n (b n a n )/2. It follows that a n+1 a n (b 1 a 1 )/2 n. Hence, for n, k IN we have k 1 a n+k a n a n+j+1 a n+j j=0 k 1 j=0 b 1 a 1 2 n+j < b 1 a 1 2 n 1. For every rational number ε > 0, there exists some N IN such that (b 1 a 1 )/2 N 1 < ε. Hence, for n > N and k 1 we obtain a n+k a n < ε. This shows that (a n ) n=1,2,... is a fundamental sequence. For the same reason, (b n ) n=1,2,... is also a fundamental sequence. These two sequences are equivalent. Let u be the (unique) real number represented by these two sequences of rational numbers. We claim that u is the least upper bound of X. To justify our claim, it suffices to prove the following two statements: (1) u x for all x X; (2) if y is an upper bound for X, then u y. Suppose that (1) is not true. Then u < x for some x X. There exists a rational number r such that u < r < x. We have r b n for all n IN. Hence, (b n r) n=1,2,... is a non-negative sequence of rational numbers. Consequently, u r 0, i.e., u r. This contradiction verifies statement (1). Suppose that (2) is not true. Then u > y for some upper bound y for X. There exists a rational number s such that y < s < u. We have a n s for all n IN. Hence, (s a n ) n=1,2,... is a non-negative sequence of rational numbers. Consequently, s u 0, i.e., u s. This contradiction verifies statement (2). In light of (1) and (2), we conclude that u is the least upper bound of X. 6

For a pair of real numbers a and b, we define (a, b) := {x IR : a < x < b}, [a, b) := {x IR : a x < b}, [a, b] := {x IR : a x b}, (a, b] := {x IR : a < x b}. The set (a, b) is clled an open interval, the set [a, b] is called a closed interval, and the sets [a, b) and (a, b] are called half-open (or half-closed) intervals. We introduce two symbols and. The ordering in IR can be extended to IR := IR {, } by defining < a < for all a IR. Then we have (, ) = IR and (a, ) = {x IR : x > a}, (, b) = {x IR : x < b}, [a, ] = {x IR : x a}, (, b] = {x IR : x b}. 3. Limits of Sequences The absolute value of a real number x is defined by x := { x if x 0, x if x < 0. For x, y IR, we have xy = x y and x + y x + y. A sequence (x n ) n=1,2,... of real numbers is said to converge to a real number x if for every real number ε > 0 there exists a positive integer N(ε) such that x n x < ε for all n > N(ε). The real number x is called the limit of the sequence and we write lim x n = x. If a sequence does not converge to a real number, it is said to diverge. A sequence (x n ) n=1,2,... of real numbers is said to diverge to, and we write lim x n = provided that for every M IR there exists a positive integer N such that x n > M for all n > N. Similarly, A sequence (x n ) n=1,2,... of real numbers is said to diverge to, and we write lim x n = provided that for every M IR there exists a positive integer N such that x n < M for all n > N. 7

If a sequence converges, its limit is unique. bounded. Moreover, a convergent sequence is Suppose that (x n ) n=1,2,... and (y n ) n=1,2,... are convergent sequences of real numbers with lim x n = x and lim y n = x. Then (1) lim (x n + y n ) = x + y; (2) lim (ax n ) = ax for all a IR; (3) lim (x n y n ) = xy; (4) lim (x n /y n ) = x/y, provided y n 0 for all n IN and y 0; (5) If x n y n for all n IN, then x y; (6) If x = y and x n z n y n for all n IN, then lim z n = x. A sequence (x n ) n=1,2,... of real numbers is said to be increasing if x n x n+1 for all n IN, and decreasing if x n+1 x n for all n IN. A monotone sequence is either increasing or decreasing. Theorem 3.1. Every monotone bounded sequence of real numbers is convergent. Proof. Let (x n ) n=1,2,... be a bounded and increasing sequence of real numbers. By Theorem 2.2, the set {x n : n IN} has a least upper bound. Let x := sup{x n : n IN}. We claim that lim x n = x. For ε > 0, x ε is not an upper bound for the set {x n : n IN}, because x is the least upper bound. Hence, there exists a positive integer N such that x ε < x N. Since the sequence (x n ) n=1,2,... is increasing, we have This shows lim x n = x. x n x = x x n x x N < ε for all n > N. The proof for the decreasing case is similar. Let (x n ) n=1,2,... be a bounded sequence of real numbers. For k = 1, 2,..., let s k := sup{x n : n k} and t k := inf{x n : n k}. Then (s k ) k=1,2,... is a decreasing sequence and (t k ) k=1,2,... is an increasing sequence. We define Clearly, s k t k for all k IN; hence, lim sup x n := lim s k and lim inf x n := lim t k. k k lim sup x n lim inf x n. 8

For example, the sequence (( 1) n ) n=1,2,... is bounded. For k = 1, 2,..., we have Consequently, s k = sup{( 1) n : n k} = 1 and t k = inf{( 1) n : n k} = 1. lim sup ( 1) n = 1 and lim inf ( 1)n = 1. Theorem 3.2. Let (x n ) n=1,2,... be a bounded sequence of real numbers. Then the sequence converges to a real number x if and only if lim inf x n = lim sup x n = x. Proof. For k = 1, 2,..., let t k := inf{x n : n k} and s k := sup{x n : n k}. Suppose lim x n = x. Then for every real number ε > 0 there exists a positive integer N(ε) such that x n x < ε for n > N(ε), that is, x ε < x n < x + ε for all n > N(ε). It follows that x ε t k s k x + ε for all k > N(ε). Hence, lim k t k = lim k s k = x. Conversely, suppose lim inf x n = lim sup x n = x. Then we have lim t k = lim s k = x. k k Consequently, for every ε > 0, there exists a positive integer N(ε) such that x ε < t k s k < x + ε for all k > N(ε). It follows that x ε < x n < x + ε for all n > N(ε), because t k x n s k for n k. Therefore, lim x n = x. A sequence (x n ) n=1,2,... of real numbers is said to be a Cauchy sequence (or a fundamental sequence) if for every ε > 0 there exists a positive integer N(ε) such that x n x m < ε whenever m, n > N(ε). Clearly, a Cauchy sequence is bounded. Moreover, a convergent sequence is a Cauchy sequence. 9

Theorem 3.3. Every Cauchy sequence of real numbers converges. Proof. Let (x n ) n=1,2,... be a Cauchy sequence of real numbers. It is bounded. Let s k := sup{x n : n k} and t k := inf{x n : n k}, k = 1, 2,.... For ε > 0, there exists a positive integer N(ε) such that x n x m < ε whenever m, n > N(ε). Moreover, for k > N(ε), there exist m k and n k such that s k < x m + ε and t k > x n ε. Hence, 0 s k t k < (x m x n ) + 2ε < 3ε for k > N(ε). This shows that lim k (s k t k ) = 0. Consequently, lim inf x n = lim t k = lim s k = lim sup x n. k k By Theorem 3.2, the sequence (x n ) n=1,2,... converges. 4. Infinite Series Given a sequence (a n ) n=1,2,... of real numbers, we define the sequence (s n ) n=1,2,... recursively by s 1 := a 1 and s n+1 := s n + a n+1, n IN. Then s n is called the nth partial sum of the sequence (a n ) n=1,2,... and we write n s n = a k = a 1 + + a n, k=1 n IN. We also refer to the sequence (s n ) n=1,2,... of partial sums as the infinite series n=1 a n. If (s n ) n=1,2,... converges to a real number s, we say that the series n=1 a n converges and we write a n = s. n=1 The real number s is called the sum of the infinite series n=1 a n. If the sequence (s n ) n=1,2,... diverges, then we say that the series n=1 a n diverges. If lim s n =, we 10

say that the series n=1 a n diverges to and write n=1 a n =. If lim s n =, we say that the series n=1 a n diverges to and write n=1 a n =. It is easily seen that a n = s n s n 1 for n 2. If the series n=1 a n converges, then (s n ) n=1,2,... converges to a real number s. It follows that lim a n = lim (s n s n 1 ) = s s = 0. Thus, if a sequence (a n ) n=1,2,... diverges or lim a n 0, then the series n=1 a n diverges. If a, r IR and a n = ar n 1 for n IN, then the series a n = ar n 1 n=1 n=1 is called a geometric series. The case a = 0 is trivial. Suppose a 0. If r 1, then the sequence (ar n 1 ) n=1,2,... diverges or converges to a nonzero real number. Hence, the geometric series n=1 arn 1 diverges for r 1. Suppose r < 1. Then s n = n k=1 ar k 1 = a(1 + r + + r n 1 ) = a(1 rn ), n IN. 1 r For r < 1 we have lim r n = 0. Consequently, lim s n = a 1 r. Therefore, for r < 1, the geometric series n=1 arn 1 converges and its sum is a/(1 r). Theorem 4.1. Let (a n ) n=1,2,... be a sequence of real numbers with a n 0 for all n IN. Then the series n=1 a n converges if and only if the sequence (s n ) n=1,2,... of partial sums is bounded. Proof. We have s n = a 1 + + a n. Since a n 0 for all n IN, s n+1 s n for all n IN. Thus, (s n ) n=1,2,... is an increasing sequence. If this sequence is bounded, then it converges, by Theorem 3.1. Hence, the series n=1 a n converges if (s n ) n=1,2,... is bounded. If (s n ) n=1,2,... is unbounded, then the sequence diverges. Consequently, the series n=1 a n diverges. Theorem 4.2. Let (a n ) n=1,2,... and (b n ) n=1,2,... be two sequences of real numbers such that 0 a n b n for all n IN. If the series n=1 b n converges, then the series n=1 a n converges. Proof. For n IN, let s n := a 1 + + a n and t n := b 1 + + b n. Since 0 a n b n for all n IN, we have s n t n for all n IN. If the series n=1 b n converges, then the sequence 11

(t n ) n=1,2,... is bounded. Consequently, the sequence (s n ) n=1,2,... is bounded. Therefore, the series n=1 a n converges, by Theorem 4.1. Let us investigate convergence or divergence of the p-series n=1 1 n p, where p is a real number. For n IN, let a n := 1/n p and s n := a 1 + + a n. Suppose p > 1. For n IN have s 2 n 1 = n k=1 2 k 1 m=2 k 1 1 m p n 2 k 1 (2 k 1 ) p = n k=1 k=1 (2 1 p ) k 1 < 1 1 2 1 p. This shows that the sequence (s n ) n=1,2,... is bounded. Hence, the p-series converges for p > 1. For p = 1 and n IN we have s 2 n = 1 + n 2 k k=1 m=2 k 1 +1 1 n m 1 + k=1 2 k 1 2 k = 1 + n 2. This shows that the sequence (s n ) n=1,2,... is unbounded. Hence, the harmonic series n=1 1/n diverges. For p 1, we have 1/n 1/np for all n IN. By Theorem 4.2, the p-series diverges for p 1. For a real number a, let a + := max{a, 0} and a := max{ a, 0}. We call a + the positive part of a and a the negative part of a, respectively. Evidently, a = a + + a and a = a + a. Theorem 4.3. Let (a n ) n=1,2,... be a sequence of real numbers. If the series n=1 a n converges, then the series n=1 a n converges. Proof. We observe that 0 a + n a n and 0 a n a n for all n IN. If the series n=1 a n converges, then both n=1 a+ n and n=1 a n converge, by Theorem 4.2. But a n = a + n a n for all n IN. We conclude that the series n=1 a n converges. If n=1 a n converges, then we say that the series n=1 a n converges absolutely. If n=1 a n converges, but n=1 a n diverges, then we say that n=1 a n converges conditionally. For example, the alternating harmonic series converges conditionally. ( 1) n n=1 12 n

Theorem 4.4. The set IR of real numbers is uncountable. Proof. We use {0, 1} IN to denote the set of all mappings from IN to {0, 1}. For u {0, 1} IN, we define ϕ(u) to be the ternary series n=1 u(n) 3 n. Since 0 u(n)/3 n 1/3 n for all n IN, the above series converges. Let us show that ϕ is an injective mapping. Suppose that u, v {0, 1} IN and u v. Let k be the least integer such that u(k) v(k). Without loss of any generality we may assume that u(k) = 0 and v(k) = 1. Then ϕ(v) ϕ(u) 1 3 k n=k+1 1 3 n > 0. This shows that ϕ is an injective mapping from {0, 1} IN to IR. If IR were countable, then there would exist an injective mapping ψ from IR to IN. Consequently, ψ ϕ would be an injective mapping from {0, 1} IN to IN. But {0, 1} IN is uncountable. Hence, there is no injective mapping from {0, 1} IN to IN. This contradiction demonstrates that IR is uncountable. 13