The Real Number System

MATH 337 The Real Number System Sets of Numbers Dr. Neal, WKU A set S is a well-defined collection of objects, with well-defined meaning that there is a specific description from which we can tell precisely whether or not an element is in S. The collection of large numbers does not constitute a well-defined set. The set P of prime numbers is well-defined: P is the collection of integers n that are greater than 1 for which the only positive divisors of n are 1 and n. So we can tell precisely whether or not an object is in P. For instance, 13 P but 14 P because 14 = 7 2. The set of rational numbers Q is also well-defined. A rational number is an expression of the form a / b, where a and b are integers with b 0. Two rational numbers a / b and c / d are said to be equal if and only if ad = bc. It is an easy exercise to show that the sum and product of two rational numbers are also rational numbers. The set of integers Z is a subset of Q because an integer n can be written as n / 1. Proposition 1.1. Let x and y be rational numbers with x < y. There exists another rational number z such that x < z < y. Proof. Let z = (x + y) / 2. Because x, y, and 1/2 are rational, x + y is rational, thus so is 1 2 ( x + y) = z. Moreover, x = 1 2 (x + x ) < 1 2 (x + y) < 1 (y + y) = y ; thus, x < z < y. QED 2 Existence of Irrational Numbers The Pythagorean Theorem states: In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. So if we let c be the length of the hypotenuse of a right triangle with two other sides of length 1, then c 2 = 1 2 + 1 2 = 2. Thus, c is a positive number such that c 2 = 2; that is, c = 2, which exists as a physical length. But we have already proven using only facts about integers that 2 cannot be written as a ratio of integers. Thus, c is not a rational number; therefore we call it irrational. The set of real numbers R is now the union of the Rationals and Irrationals. Ordered Sets Definition 1.1. An order on a set S is a relation on S that satisfies two properties: (i) (Trichotomy) For all x, y S, exactly one of the following is true: x y or x = y or y x. (ii) (Transitive) For all x, y, z S, if x y and y z, then x z. We write x y to denote that either x y or x = y. Clearly is also transitive. That is, if x y and y z, then x z.

In general, x y is read as x precedes y thus indicating an ordering to the set. For example, let S be the set of standard lower-case English letters. Then a c, c f, and a f. We are most concerned with the ordering on the set of real numbers R, and its various subsets such as the natural numbers ℵ, the integers Z, and the rational numbers Q. The ordering used for these sets is the relation < (less than). For any two real numbers x and y, we say x < y if and only if 0 < y x (i.e., y x is positive). Then the relation < satisfies the two properties of an order. With numbers, we read x < y as x is less than y and x y is read as x is less than or equal to y. Definition 1.2. Let E be a non-empty subset of an ordered set S. We say that x is the least (first) element of E provided x E and x y for all other y E. We say that x is the greatest (last) element of E provided x E and y x for all other y E. Proposition 1.2. Let E = { x 1,..., x n } be a finite subset of n distinct elements from an ordered set. There exists a first (least) element and a last (greatest) element in E. Proof. If n = 1, then x 1 is both the least and greatest element in E. If n = 2, then the result follows from the Trichotomy axiom of order. Now assume the result holds for some n 2, and assume that x 1 is the least element and x n is the greatest element in E. Now consider the set E = { x 1,..., x n, x n+1 } with a new distinct element x n+1. If x n+1 x 1, then x n+1 is the least element and x n is the greatest element in E. If x 1 x n+1 x n, then x 1 is the least element and x n is the greatest element in E. If x n x n+1, then x 1 is the least element and x n+1 is the greatest element in E. With all cases being exhausted, we see that the result holds for a set of cardinality n + 1 provided it holds for a set of cardinality n. By induction, the proposition holds for all n 1. QED Upper and Lower Bounds Definition 1.3. Let E be a subset of an ordered set S. (a) We say that E is bounded above if there exists an element β in S such that x β for all x E. Such an element β is called an upper bound of E. (b) An element β in S is called the least upper bound of E or supremum of E, denoted by lub E or sup E, if (i) β is an upper bound of E and (ii) if λ is a different upper bound of E, then β λ. Note: Suppose β = sup E and λ β. Then λ cannot be an upper bound of E. Thus there must be an element x E such that λ x. { } Example 1.1. Let E = 9 + 1 2, 9 + 2 3, 9 + 3 4,...,9 + n n + 1,... sup E = 10. R. We claim that

First, 9 + n n +1 < 9 + n + 1 = 10 for all integers n 1; thus, 10 is an upper bound of E. n + 1 Next, suppose λ is a different upper bound of E but λ < 10. Obviously 9 < λ or else it would not be an upper bound of E. Then 9 < λ < 10, which makes 0 < λ 9 < 1. n Because lim = 1, we can choose an integer n large enough such that n n + 1 λ 9 < n n < 1. But then λ < 9 + < 10, which contradicts the fact that λ is an n + 1 n + 1 upper bound of E. So we must have 10 < λ which makes 10 = sup E. Proposition 1.3. Let β be an upper bound of a subset E of an ordered set S. If β E, then β = sup E. Proof. Let λ be a different upper bound of E. Then either λ β or β λ due to the Trichotomy. But if λ β, then λ would not be an upper bound of E because β E. So we must have β λ. Therefore, β is the least upper bound of E. QED Definition 1.4. Let E be a subset of an ordered set S. (a) We say that E is bounded below if there exists an element α in S such that α x for all x E. Such an element α is called a lower bound of E. (b) An element α in S is called the greatest lower bound of E or infimum of E, denoted by glb E or inf E, if (i) α is a lower bound of E and (ii) if λ is a different lower bound of E, then λ α. Note: Suppose α = inf E and α λ. Then λ cannot be a lower bound of E. Thus there must be an element x E such that x λ. Example 1.2. Let E = { 1 / n n ℵ} = {1, 1/2, 1/3, 1/4,... } R. Then sup E = 1, which is in E (see Proposition 1.3), but inf E = 0 which is not in E. Proposition 1.4. Let α be a lower bound of a subset E of an ordered set S. If α E, then α = inf E. (We leave the proof as an exercise.) The Least Upper Bound Property Definition 1.5. An ordered set S is said to have the least upper bound property if the following condition holds: Whenever E is a non-empty subset of S that is bounded above, then sup E exists and is an element in S.

Proposition 1.5. The set of integers Z has the least upper bound property. Proof. Let E be a non-empty subset of the integers Z that is bounded above. Then there exists an integer N such that n N for all n E. Because E is non-empty, there is an integer n E with n N. By Prop. 1.3, if n is an upper bound of E, then n = sup E which exists as an element in Z. If n is not an upper bound of E, then by Prop. 1.2. we pick the greatest element in E from among the finite number of integers in E that are from n to N. By Prop. 1.3, this greatest element in E is sup E and is an integer because it was chosen from E. QED In the proof of Proposition 1.5, we found the least upper bound of the non-empty subset E and in fact sup E is an element of E. If E were bounded below, then we also would have inf E E. We therefore can state: Proposition 1.6. (The Well-Ordering Principle of Z ). (a) Let E be a non-empty subset of the integers that is bounded above. Then E has a greatest element within the set E. (b) Let E be a non-empty subset of the integers that is bounded below. Then E has a least element within the set E. The Well-Ordering Principle is essential for the study of limits of sequences, and we often shall use it implicitly in the manner illustrated in the following example: Example of the Well-Ordering Principle: Let {a n } n =1 be a sequence of real numbers that increases to infinity. Then we can choose the smallest integer N such that a N 100. Because the sequence is increasing to infinity, we have a 1 < a 2 < a 3 <...< a n <... and at some point we must have 100 a k < a k+1 < a k +2 <.... The Well-Ordering Principle is then applied to the set of indices, not to the actual sequence values. Let E = {k : 100 a k }, which is the set of indices k for which a k 100. Then E is a non-empty subset of the integers that is bounded below by 1 (the first index). By the Well-Ordering Principle, E has a least element within E. Thus, there exists a smallest index N such that a N 100. Example 1.3. The set of rational numbers Q does not have the least upper bound property. As an example, let E Q be defined as follows: E = { 1. 4, 1. 41, 1. 414, 1. 4142, 1.41421,...} = 1 + a 1 10 1 +...+ a k 10 k k 1, where a i is the ith decimal place value in 2. Then E is a non-empty subset of the rational numbers and E is bounded above by 2. In fact 2 = sup E, but 2 is not a rational number. Thus, Q does not have the least upper bound property.

The Greatest Lower Bound Property Definition 1.6. An ordered set S is said to have the greatest lower bound property if the following condition holds: Whenever E is a non-empty subset of S that is bounded below, then inf E exists and is an element in S. Example 1.4. The set of integers Z has the greatest lower bound property, but the set of rational numbers Q does not. These results follow from Proposition 1.5, Example 1.3, and the following theorem: Theorem 1.1. Let S be an ordered set. Then S has the greatest lower bound property if and only if S has the least upper bound property. Partial Proof. Assume S has the least upper bound property and let E be a non-empty subset of S that is bounded below. Let L be the set of all elements in S that are lower bounds of E. Then L because E is bounded below. Because E is non-empty, there exists an element x 0 in E. By definition of L, x 0 is an upper bound of L; so L is bounded above. By the least upper bound property of S, α = sup L exists and is an element of S. We claim that α = inf E. sup L = α = inf E L (All lower bounds of E) E First, let x E. By definition of L, x is an upper bound of L; thus, α x because α is the least upper bound of L. Thus α is also a lower bound of E. But suppose λ is another (different) lower bound of E. Then λ L. But then λ α because α = sup L. But because λ is different than α, we must have λ α due to the properties of an order. Thus, α is in fact the greatest lower bound of E and it exists in S. Thus, S has the greatest lower bound property. (We leave the proof of the converse as an exercise.) S Decimal Expansions: Rational vs. Irrational A commonly used result is that a rational number has a decimal expansion that terminates, such as 4.0 (an integer) or 2.375 (a regular number), or has a decimal expansion that recurs such as 5.342. This property characterizes rational numbers and therefore provides another distinction between rational and irrational numbers. Theorem 1.2. A real number is rational if and only if it has a decimal expansion that either terminates or recurs.

Proof. Assume first that we have a real number x with a decimal expansion that either terminates or recurs. If it terminates, then x has the form x = ±(n. a 1 a 2...a k ), where n,a 1,..., a k are all non-negative integers, and 0 a i 9 for 1 i k. Thus, x = ± n + a 1 10 1 +... + a k 10 k is the sum of rational numbers which is rational. (If all a i are 0, then x = ±n is an integer and thus is a rational number.) If the decimal expansion is recurring, then x has the form x = ±n. a 1...a j a j+1...a j+k a j+1... a j+ k a j+1... a j+k... = ± n + a 1 10 1 +...+ a j 10 j + 1 a j+1 10 j 10 +... + a j+k 10 k + 1 a j+1 10 j 10 k +1 +...+ a j+k 10 2k +... = ± n + a 1 10 1 +.. + a j 10 j + a j+1 1 10 j 10 + 1 10 k+1 + 1 +.. +.. + a j+k 1 10 2k+1 10 j 10 k + 1 10 2k + 1 +.. 10 3k = ± n + a 1 10 1 +...+ a j 10 j + a i j+1 1 10 j+1 10 k +...+ a i j+k 1 i = 0 10 j+k 10 k i= 0 = ± n + a 1 10 1 +...+ a j 10 j + a j+1 10 j+1 +... + a j+k i 1 10 j+ k 10 k i= 0 = ± n + a 1 10 1 +...+ a j 10 j + 1 a j+1 10 j 10 1 +... + a j+k 10 k 10k 10 k, which is rational. 1 Conversely, let x = c / d be a rational number. We must obtain its decimal expansion. By dividing d into c, we may assume that x is of the form x = ±(n + a / b), where n,a,b are non-negative integers with b 0 and 0 a < b. If a = 0, then x = ±n is a terminating decimal expansion. If a 0, then let r 0 = a and divide b into 10r 0 to obtain 10r 0 = a 1 b + r 1 where 0 r 1 < b and 0 a 1 9 because 0 r 0 < b. Then divide b into 10r 1 to obtain 10r 1 = a 2 b + r 2 where 0 r 2 < b and 0 a 2 9 because 0 r 1 < b. Continue dividing b into 10r j to obtain 10r j = a j+1 b + r j+1 where 0 r j+1 < b and 0 a j+1 9 because 0 r j < b. Because all remainders r i must be from 0 to b 1, there can be at most b distinct remainders. At some point, we will have 10r k = a k+1 b + r k +1, where r k +1 = r i for some previous remainder r i. Because the next step of dividing 10r i by b yields the same quotient a i+1 and remainder as before, the pattern recurs for the successive divisions. We now claim that x = ±(n. a 1 a 2... a i a i+1..a k +1 a i+1.. a k+1 a i+1.. a k+1... ) To see this result, we first use 10a = 10r 0 = a 1 b + r 1. Dividing by 10b gives

a b = a 1 10 + 1 10 r 1 b But then 10r 1 = a 2 b + r 2, which gives r 1 b = a 2 10 + 1 10. r 2 b ; hence, a b = a 1 10 + 1 10 a 2 10 + 1 r 2 10 b = a 1 10 + a 2 10 2 + 1 r 2 10 2 b. Continuing, we obtain a b = a 1 10 + a 2 10 2 +...+ a i 10 i + a i+1 10 i+1 +... + a k +1 10 k+1 + 1 r k +1 10 k +1 b, where r k +1 = r i. But then we have a b = a 1 10 + a 2 10 2 +...+ a i 10 i + a i+1 10 i+1 +... + a k +1 10 k+1 + a i+1 10 k +2 + 1 r i+1 10 k +2 b = a 1 10 + a 2 10 2 +...+ a i 10 i + a i+1 10 i+1 +... + a k +1 10 k+1 + a i+1 10 k +2 + a i+ 2 10 k +3 + 1 r i+3 10 k +3 b = a 1 10 + a 2 10 2 +...+ a i 10 i + a i+1 10 i+1 +... + a k +1 10 k+1 + a i+1 10 k +2 + a i+ 2 10 k +3 +... + a k +1 10 2(k+1) i 1 r + k+1 10 2(k+1) i b By recursion, we obtain a recurring decimal expansion of x that terminates if b ever divides evenly into a remainder. QED Example 1.5 (a) Find the rational form of 5.86342342342... expansion of 3/8. (c) Find the decimal expansion of 3/14. (b) Find the decimal Solution. (a) By Theorem 1.2, we have 5.86342 = 5 + 8 10 + 6 100 + 1 103 0.342 100 10 3 1 = 5 + 8 10 + 6 100 + 1 100 342 999 = 5 + 8 10 + 6 100 + 342 99900 = 16271 2775. (b) For a / b = 3 / 8, we first divide 8 into 10 3 to obtain the next remainder r 1, then continue dividing 8 into 10r i until we obtain a 0 remainder or a repeated remainder. The decimal terms of 3/8 come from the successive quotients:

30 = 3 8 + 6 60 = 7 8 + 4 40 = 5 8 + 0 0 = 0 8 + 0 Thus, 3/8 = 0.375. (c) For a / b = 3 / 14, we first divide 14 into 10 3 to obtain the next remainder r 1, then continue dividing 14 into 10r i until we obtain a 0 remainder or a repeated remainder. The decimal terms of 3/14 come from the successive quotients: 30 = 2 14 + 2 20 = 1 14 + 6 60 = 4 14 + 4 40 = 2 14 +12 120 = 8 14 + 8 80 = 5 14 +10 100 = 7 14 + 2 repeated remainder 20 = 1 14 + 6 repeated quotient Thus, 3/14 = 0.2142857142857... Note: The decimal expansion of a rational number is not unique. An expansion that terminates also can be written as an expansion that ends in a string of all 9's. Specifically, the decimal x = n.a 1 a 2... a k, where 1 a k 9, can be re-written as x = n.a 1 a 2... (a k 1) 999999.... This result follows from the geometric series 1 9 i 10 i= k +1 = 9 (1 / 10)k+1 1 1 / 10 = 9 (1 / 10)k+1 9 / 10 1 = k. 10 For example, with k = 3 we have 0.000 9999...= 0.001. As another example, we have 2.4837 = 2.48369999. By convention, we always shall assume that a rational number never ends in a string of all 9's.

An immediate consequence to Theorem 1.2 is the following result: Dr. Neal, WKU Corollary 1.1. A number is irrational if and only if it has a decimal expansion that neither terminates nor recurs. We have already seen that 2 is irrational. An example of another irrational number is x = 0.123456789112233445566778899111222333... Another Construction of the Reals Theorem 1.2 and Corollary 1.1 give a method of constructing the real numbers based on the natural numbers. A brief outline of the steps follows: I. Let 0 denote no length and let 1 denote a fixed unit of length. II. By extending the initial length of 1 by incremental successive lengths of 1, we obtain the natural numbers ℵ = {1, 2, 3,...}. Including 0 gives us the whole numbers W. III. By considering all quotients a / b of natural numbers, we obtain the positive rational numbers Q +. Also allowing a = 0 gives us the non-negative rational numbers. IV. Every element in Q + a can be written uniquely in the form n + i i= 110 i, where n W, 0 a i 9 for all i, the sequence {a i } either terminates by ending in all 0's or is recurring, and the sequence {a i } does not end in a string of all 9's. V. Let the non-negative irrational numbers I + be defined by all values of the form a n + i i= 110 i, where n W, 0 a i 9 for all i, and the sequence {a i } neither terminates nor recurs. (In particular, irrational numbers also cannot end in a string of all 9's.) VI. Let the positive real numbers be defined by R + = Q + I +. a VII. Negative numbers can be written in the form n ± i i=1 10 i. A real number is therefore either rational or irrational. By comparing the decimal expansions of any two real numbers x and y, we can see that either x < y, x = y, or y < x. Thus, the Reals are ordered. The next results prove that both the rational numbers and the irrational numbers are dense within the Reals.

Theorem 1.3. Let x and y be any two real numbers with 0 x < y. (a) There exists a rational number z such that x < z < y. (b) There exists an irrational number w such that x < w < y. Proof. Let x = m. a 1 a 2 a 3... and y = n. b 1 b 2 b 3... be the decimal forms of x and y where 0 m n. Suppose first that m < n. Because x cannot end in a string of all 9's, let a i be the first decimal place value such that a i < 9. Then change a i to a i + 1 and let z = m. a 1 a 2 a 3...(a i + 1), which is rational, and let w = m. a 1 a 2 a 3...(a i + 1)c 1 c 2 c 3..., where the sequence {c i } neither terminates nor recurs, so that w is irrational. We then have x < z < y and x < w < y. Next assume that m = n. Because x < y, we must have a 1 b 1. We now choose the smallest index i such that a i < b i. Then we must have a j = b j for 1 j < i or else x would be larger than y. Now because x cannot end in a string of all 9's, choose the least k i such that a k < 9. Now let z = m. a 1 a 2 a 3...a i...(a k + 1). Then z is rational and x < z < y. Finally, let w = m. a 1 a 2 a 3...a i... (a k + 1)c 1 c 2 c 3... where the sequence {c i } neither terminates nor recurs, so that w is irrational. Then x < w < y. QED Corollary 1.2. Let x and y be any two real numbers with x < y. (a) There exists a rational number z such that x < z < y. (b) There exists an irrational number w such that x < w < y. Proof. If 0 x < y, then the result is proved in Theorem 1.3. If x < 0 < y, then by Theorem 1.3, we can find such z and w such that x < 0 < z < y and x < 0 < w < y. If x < y 0, then 0 y < x. So we can apply Theorem 1.3 to find a rational z and an irrational w such that y < z < x and y < w < x. But then z is rational and w is irrational with x < z < y and x < w < y. QED Example 1.6. Apply Theorem 1.3 to x = 8.9942... and y = 9.01 assuming x is rational and assuming x is irrational. Solution. Let z = 8.995, which is rational. Then x < z < y regardless of whether x is rational or irrational. If x is irrational, then let w = x + 0.001, which is irrational. Then x < w < y. However if x is rational, then let w = 8.995 + ( 2 1) / 1000 = 8.9954142..., which is irrational. (This expression simply appends the decimal digits of 8.995.) Then x < w < y. 2 to the end of We continue with two of the most important property of real numbers:

Theorem 1.4. The set of real numbers has the greatest lower bound property. Dr. Neal, WKU Proof. Let E a non-empty subset of R that is bounded below. Let r R be a lower bound of E and let F be the set of all integers k such that k r. Then r is an upper bound of F ; thus, by Prop. 1.6, we can choose a greatest element n in F. Then n r, so n is also a lower bound of E. By Prop. 1.4, if n E, then n = inf E. Otherwise, we continue: Choose the greatest integer a 1 from 0 to 9 such that n + a 1 / 10 is a lower bound of E. If n + a 1 / 10 E, then n + a 1 / 10 = inf E. Otherwise, choose the greatest integer a 2 from 0 to 9 such that n + a 1 / 10 + a 2 / 10 2 is a lower bound of E. Continuing, we thereby obtain a real number α = n + i=1a i / 10 i with α n + 1. k We claim α = inf E. So let x E. By construction, n + i=1a i / 10 i x for all k 1. In particular, n x. If n + 1 x, then α n + 1 x. But suppose x < n + 1 with x = n + i=1b i / 10 i where the string {b i } does not end in all 9's. If we had x < α, then we must have b k < a k for some k and we can choose the smallest integer k for which k this occurs. But then we would have x < n + i=1a i / 10 i, which is a contradiction. Thus, α x ; that is, α is a lower bound of E. Next, suppose λ is a different lower bound of E. We must show that λ < α. So write λ = m + i=1c i / 10 i Then m λ, so m is also a lower bound of E. By the choice of n above, we must have m n. If m < n, then λ < α. So suppose m = n. Now if a k < c k for some k and we choose the smallest k where this occurs, then we would k have n + c i / 10 i λ x for all x E. But then this c k contradicts the choice of a k. i= 1 So we must have c k a k for all k. That is, λ α. But because λ is different that α, we must have λ < α. That is, α = inf E. QED From Theorems 1.1 and 1.4, we obtain: Corollary 1.2. The set of real numbers has the least upper bound property. That is, any non-empty subset of R that is bounded above has a least upper bound in R. An important consequence of the glb and lub properties of the real line is the fact that the intersection of a nested decreasing sequence of closed intervals is non-empty: [a 1, b 1 ] [a 2, b 2 ] [a 3, b 3 ]... [sup{a i }, inf{b i }] a 1 a 2 a 3... a i... sup{a i } inf{b i }... b i... b 3 b 2 b 1 Theorem 1.5. (Nested Interval Property) Let [a i, b i ] be a nested decreasing sequence of closed intervals. Then [sup{a i }, inf{b i }] [a i, b i ]. In particular, [a i, b i ]. i=1 i=1

Proof. Because the intervals are nested decreasing, we have a i a i+1 b i+1 b i for i 1. So the set of points {a i } is bounded above by each b i and the set of points {b i } is bounded below by each a i ; hence a = sup{a i } and b = inf{b i } exist by the least upper bound and greatest lower bound properties of R. Because each b i is an upper bound of {a i } and a is the least upper bound of {a i }, we have a b i for all i 1. But then a becomes a lower bound of the {b i }; so a b because b is the greatest lower bound of the {b i }. For all i 1, we then have a i a b b i ; thus, [a, b] [a i, b i ]. In particular, [a i, b i ] contains a and b. QED i=1 i=1 Exercises 1. Prove: If c is rational, c 0, and x is irrational, then c x and c + x are irrational. 2. Prove that 12 is irrational. 3. Let E be a non-empty subset of an ordered set. Suppose α is a lower bound of E and β is an upper bound of E. Prove that α β. 4. Let E be a non-empty subset of an ordered set S that has the least upper bound property. Suppose E is bounded above and below. Prove that inf E sup E. 5. Let S be a non-empty subset of the real numbers that is bounded below. Define S by S = { x x S }. Prove: (i) S is bounded above. (ii) lub( S) = glbs. 6. Prove: Let α be a lower bound of a subset E of an ordered set S. If α E, then α = inf E. 7. Let S be an ordered set with the greatest lower bound property. Prove that S has the least upper bound property. 8. Find a rational number and an irrational number between x = 8.9949926 and y = 8.995. 9. Let a,b R with a < b, (i) For the open interval E = (a, b), prove that inf E = a and sup E = b. (ii) For the closed interval F = [a, b], prove that inf F = a and sup F = b.