2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.

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1 CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively. Field Axioms of R The real umbers are a field (as are the ratioal umbers Q ad the complex umbers C). That is, there are biary operatios + ad defied o R 3 (A) a + b = b + a 8a, b 2 R. (A2) (a + b) + c = a + (b + c) 8a, b, c 2 R. (A3) R a = a ad a + 0 = a 8a 2 R. (A4) 8a 2 R, 9 a 2 R 3 a + ( a) = 0 ad ( a) + a = 0. (M) ab = ba 8a, b 2 R. (M2) (ab)c = a(bc) 8a, b, c 2 R. (M3) 9 2 R, 6= 0, 3 a = a ad a = a 8a 2 R. (M4) 8a 2 R, a 6= 0, 9 a 2 R 3 a = ad a =. a a (D) a(b + c) = ab + ac ad (b + c)a = ba + ca 8a, b, c 2 R.

2 2 2. THE REAL NUMBERS Some Properties of R Theorem (2). If z, a 2 R 3 z + a = a, the z = 0. (i.e., the umber 0 guarateed by (A3) is uique.) Proof. By (A4), 9 a 2 R 3 a + ( a) = 0. The z = z + 0 = z + a + ( a) = (z + a) + ( a) A3 A2 = a + ( a) = 0. Theorem (3). Let a, b 2 R. The a + x = b has the uique solutio x = ( a) + b. (i.e., we are defiig b so ( Proof. a.) a + ( a) + b = A2 a + ( a) + b =A4 0 + b = A3 b, a) + b is a solutio. For uiqueess, suppose y is ay solutio of the equatio, i.e., a + y = b. The y = 0 + y A3 ( a) + a + y = A4 = A2 ( a) + (a + y) = ( a) + b

3 2.. THE ALGEBRAIC AND ORDER PROPERTIES OF R 3 Note. N Z Q {z R C } all fields The last three all satisfy the field axioms, so the field axioms do ot characterize R. Recall that Q is closed uder + ad, i.e., if a, b 2 Q, the a + b 2 Q ad a b 2 Q. Homework Pages 30-3 #4 (Hit: assume a 6= 0 ad prove a = ), 5 (Hit: /(ab) = (/a) (/b) if (/a) (/b) does what /(ab) is supposed to do), 8b (st part) Order Properties of R R is a ordered field, i.e., the followig properties are satisfied: () (Trichotomy) For a, b 2 R, exactly oe of the followig is true: a < b, a = b, or a > b. (2) (Trasitive) For a, b, c 2 R, if a < b ad b < c, the a < c. (3) For a, b, c 2 R, if a < b, the a + c < b + c. (4) For a, b, c 2 R, if a < b ad c > 0, ac < bc. Some Order Properties Theorem (4). If a, b 2 R, the a < b () a > b. Proof. a < b () a + ( a) + ( b) < b + ( a) + ( b) () a + ( a) + ( b) < b + ( b) + ( a) () 0 + ( b) < b + ( b) + ( a) () b < 0 + ( a) () b < a () a > b

4 4 2. THE REAL NUMBERS Theorem (5). If a, b, c 2 R, the a < b ad c < 0 =) ac > bc Proof. c < 0 =) TH4 c > 0. The a( c) < b( c) =) ac < bc =) TH4 ac > bc. Theorem (2..9). If a 2 R 3 0 apple a < 8 > 0, the a = 0. Proof. Suppose a > 0. Sice 0 < 2 <, 0 < 2 a < a. Let 0 = 2 a. The 0 < 0 < a, cotradictig our hypothesis. Thus a = 0. Problem (Page 3 #8). Let a, b 2 R, ad suppose a apple b + (or a apple b) 8 > 0. The a apple b. Proof. By way of cotradictio, suppose b < a. [Need to fid a that gives a cotradictio.] Let 0 = 2 (a b). The a 0 = a 2 (a b) = 2 a + 2 b > 2 b + 2 b = b, cotradictig our hypothesis. Thus a apple b.

5 2.. THE ALGEBRAIC AND ORDER PROPERTIES OF R 5 Theorem (Arithmetic-Geometric Mea Iequality). Suppose a, b > 0. The p ab apple (a + b) 2 with equality holdig () a = b. Proof. () Suppose a 6= b. The p a > 0, p b > 0, ad p a 6= p b. Thus p a p b 6= 0 =) p a p b 2 > 0 =) a 2 p a p b + b > 0 =) 2 p a p b > (a + b) =) p ab < (a + b) 2 (2) If a = b, p p ab = a2 = a = a = 2 (2a) = 2 (a + a) = (a + b). 2 (3) If p ab = (a + b), 2 ab = 4 a + b 2 =) 4ab = a 2 + 2ab + b 2 =) 0 = a 2 2ab + b 2 =) 0 = (a b) 2 =) 0 = a b =) a = b.

6 6 2. THE REAL NUMBERS Theorem (Beroulli s Iequality). If x > Proof. [We use MI to prove this.] Let S N for which ( + x) 2 S sice ( + x) = + x. Suppose k 2 S, i.e., ( + x) k, the ( + x) + x 8 2 N. + x. + kx. The ( + x) k+ = ( + x) k ( + x) Thus S = N by MI. ( + kx)( + x) = + (k + )x + kx 2 + (k + )x Note. We ow have N Z Q R {z } ordered fields C.

7 2.. THE ALGEBRAIC AND ORDER PROPERTIES OF R 7 Problem (Page 30 #6d). Fid all x 2 R 3 x < x2. Solutio. x < x2 () x 2 x > 0 () x (x3 ) > 0 () o x > 0 ad x3 > 0o or x < 0 ad x3 < 0 () o o x > 0 ad x 3 > or x < 0 ad x 3 < () o o x > 0 ad x > or x < 0 ad x < () x > or x < 0. Homework Page 3 # 3 (Hit: for =), suppose, WLOG, a 6= 0, the reach a cotradictio), 6c, 20

8 8 2. THE REAL NUMBERS 2.2. Absolute Value ad the Real Lie Defiitio. Theorem (2.2.2). (a) ab = a b 8a, b 2 R. (b) a 2 = a 2 8a 2 R. 8a 2 R, a = (c) If c 0, the a apple c () c apple a apple c. ( a if a 0 a if a < 0 (c ) If c > 0, the a < c () c < a < c. (d) a apple a apple a 8a 2 R. Theorem (2.2.3 Triagle Iequality). Proof. [We wish to use Theorem 2.2.2(c)] By Theorem 2.2.2(d), 8a, b 2 R, a + b apple a + b. a apple a apple a ad b apple b apple b. The a + b = a b apple a + b apple a + b =) a + b apple a + b by Theorem 2.2.2(c)

9 (a) 2.2. ABSOLUTE VALUE AND THE REAL LINE 9 Corollary (2.2.4). If a, b 2 R, the (a) a b apple a b (b) a b apple a + b Note. These are also referred to as triagle iequalities. Proof. [We use a smugglig techique.] Thus by Theorem (c) a = a b + b apple a b + b =) a b apple a b. b = b a + a apple b a + a =) b a apple b a =) a b apple a b. a b apple a b apple a b =) a b apple a b (b) Just replace b by ( b) i the triagle iequality. Corollary (2.2.5). 8a, a 2,..., a 2 R, a + a a apple a + a a.

10 20 2. THE REAL NUMBERS Problem (Page 36 #0a). Fid all x 2 R 3 x > x +. Solutio. First, cosiderig the values of x that make oe of the absolute values 0, If x <, so x is a solutio. If apple x <, so x < or apple x < or x. x > x + =) x + > x =) >, x > x + =) x + > x + =) 0 > 2x =) x < 0, apple x < 0 are solutios. If x >, x > x + =) x > x + =) >, which is impossible. Therefore, {x : x < 0} is the solutio set. Recall. a b give the distace from a to b o the umber lie. Defiitio (2.2.7). Let a 2 R ad > 0. The the -eighborhood of a is the set V (a) = {x 2 R : x a < }. Corollary. x 2 V (a) () < x a < e () a < x < a +.

11 2.2. ABSOLUTE VALUE AND THE REAL LINE 2 Problem (Page 36 # 7). If a, b 2 R ad a 6= b, the 9 -eighborhoods U of a ad V of b 3 U \ V = ;. Proof. WLOG (without loss of geerality), suppose a < b. Now Choose The a < a + 3 (b a) < a (b a) = b (b a) < b. 3 Homework Page 36 # 2, 3 U = V 3 (b a) (a) ad V = V 3 (b a)(b) U \ V = ;.

12 22 2. THE REAL NUMBERS 2.3. The Completeess Property of R GOAL to characterize the real umbers Defiitio (2.3.). Let ; 6= S R. (a) S is bouded above if 9 u 2 R 3 s apple u 8s 2 S. The u is a upper boud (u.b.) of S. (b) S is bouded below if 9 w 2 R 3 w apple s 8s 2 S. The w is a lower boud (l.b.) of S. (c) S is bouded if it is both bouded above ad below. Otherwise it is ubouded. Corollary. (a) 2 R is ot a u.b. of S if (b) z 2 R is ot a l.b. of S if Example. () S = {x 2 R : x 5}. 9 s 0 2 S 3 v < s 0. 9 s 00 2 S 3 s 00 < z. S is ot bouded above sice, if v were a upper boud, a cotradictio. max{5, v + } > v ad max{5, v + } 2 S, S is bouded below by ay w apple 5.

13 (2) (from Page 39 #4) S 4 = upper boud for S 4. Solutio. 0 < 2.3. THE COMPLETENESS PROPERTY OF R 23 ( ) apple ad apple apple ( ) o : 2 N. Fid a lower boud ad < N =) apple 8 2 N =) apple ( ) apple 8 2 N =) ( ) 0 apple apple N. Thus 0 is a lower boud ad 2 is a upper boud of S 4. Defiitio (2.3.2). Let ; 6= S R. (a) If S is bouded above, the u is a supremum (or least upper boud) of S, writte u = sup S, if () u is a upper boud of S; (2) if v is ay upper boud of S, u apple v. (b) If S is bouded below, the w is a ifimum (or greatest lower boud) of S, writte w = if S, if () w is a lower boud of S; (2) if t is ay lower boud of S, t apple w.

14 24 2. THE REAL NUMBERS Lemma (2.3.3). Suppose ; 6= S R. (a) u = sup S () () s apple u 8s 2 S, (2) if v < u, The 9 s 0 2 S 3 v < s 0. (b) w = if S () () w apple s 8s 2 S, (2) if w < z, the 9 s 00 2 S 3 s 00 < z. Lemma (2.3.4). (Property S) Let ; 6= S R. u = sup S () () u is a u.b. for S; (2) 8 > 0, 9 s 2 S 3 u < s. (Property I) Let ; 6= S R. w = if S () () w is a l.b. for S; (2) 8 > 0, 9 s 2 S 3 w + > s. Proof. (of Property S) (=)) Assume u = sup S. The, by defiitio, u is a u.b. for S. Let > 0 be give. The u < u, so by Lemma s 0 2 S 3 u < s 0. Let s = s 0. ((=) () u is a u.b. for S =) s apple u 8s 2 S. (2) Suppose v < u. Let = u v. The 9 s 2 S 3 u (u v) < s =) v < s. Let s 0 = s =) v < s 0. The u = sup S by Lemma

15 Example. () If ; 6= S R is a fiite set, sup S is the largest elemet of S ad if S is the least elemet of S. (2) S = x 2 R : 2 < x apple 5. (a) sup S = 5 2 S. Proof. () 8x 2 S, x apple 5 =) 5 is a u.b. of S THE COMPLETENESS PROPERTY OF R 25 (2) Let > 0 be give. 5 < 5 2 S. Let s = 5. The 5 = sup S by Property S. (b) if S = 2 /2 S. Proof. () 8 x 2 S, 2 < x =) 2 apple x =) 2 is a l.b. of S. (2) Let > 0 be give. If > 3, 2 + > = 5 2 S, so let s = 5. If apple 3, 2 < apple = 7 2 apple 5, so S. The 2 + > S, so let s = Thus 2 = if S by Property I. Axiom (Completeess Property of R or Supremum Property of R). Every o-empty set of real umbers that has a upper boud also has a supremum i R. Note. Thus R is a complete ordered field, while Q is ot.

16 26 2. THE REAL NUMBERS Theorem (Ifimum Property of R). Every oempty set of real umbers that is bouded below has a ifimum i R. Proof. Suppose ; 6= S R is bouded below. The ; 6= S 0 = s : s 2 S is bouded above. This is true sice w a lower boud of S =) w apple s 8s 2 S =) s apple w 8s 2 S =) w is a upper boud of S 0. By the Completeess Property, u = sup S 0 exists. We claim u = if S. () u = sup S 0 =) s apple u 8s 2 S =) u apple s 8s 2 S =) u is a lower boud of S. (2) Let > 0 be give. By Property S, 9 ( s ) 2 S 0, where s 2 S, 3 u < s =) u + > s. Thus u = if S = sup s : s 2 S by Property I.

17 Problem (Page 39 #4). Let S 4 = Fid if S 4 ad sup S 4. Solutio THE COMPLETENESS PROPERTY OF R 27 ( ) : 2 N. We showed earlier that 0 is a lower boud of S 4 ad 2 is a upper boud sice ( ) 0 apple apple N. Now S4 = 2, 2, 4 3, 3 4, 6 5, 5 6, 8 7, 7 8,.... We claim if S 4 = 2. () If is odd, ( ) = = + = + > > 2. If is eve, 2, so ( ) = = 2 = 2 = 2. Thus 2 is a lower boud for S 4. (2) Give > 0, 2 2 S 4 ad 2 + > 2, so choose s = 2. By Property I, 2 = if S 4. [Fidig ad provig sup S 4 is Homework] Homework Pages #2, 4 (prove sup S 4 =?), 5bc.

18 28 2. THE REAL NUMBERS 2.4. Applicatios of the Supremum Property Two questios Is there a largest atural umber? Is N bouded above i R? Theorem (2.4.3 Archimedea Property). If x 2 R, the 9 x 2 N 3 x < x. Proof. Suppose apple x The x is a u.b. of N, so N has a supremum. Let u = sup N. By Property S, 9m 2 N 3 u < m. The u < m + 2 N, cotradictig that u is a upper boud of N. 8 2 N. [We are usig cotradictio.] Thus 9 x 2 N 3 x < x. Note. The ext 3 corollaries ca also be referred to as Archimedea. o Corollary (2.4.4). If S = : 2 N, if S = 0. Proof. Clearly, 0 is a lower boud of S, so if S exists. Let w = if S, so w 0. 8 > 0, 9 2 N 3 0 apple w apple < =) < (Archimedea) =) <. Thus w = 0 by Theorem Thus if S = 0.

19 2.4. APPLICATIONS OF THE SUPREMUM PROPERTY 29 Corollary (2.4.5). If t > 0, the 9 t 2 N 3 0 < t < t. o Proof. Sice if : 2 N = 0 ad t > 0, o t is ot a lower boud of : 2 N, so 9 t 2 N 3 0 < t < t. Corollary (2.4.6). If y > 0, the 9 y 2 N 3 y apple y < y. Proof. Let E y = {m 2 N : y < m}. By the Archimedea property, E y 6= ;. By Well-Orderig, E y has a least elemet, say y. The y 62 E y, so y apple y < y.

20 30 2. THE REAL NUMBERS Theorem (Desity Theorem). If x, y 2 R 3 x < y, the 9 r 2 Q 3 x < r < y. Proof. WLOG, assume x > 0. Suppose the theorem is true for x > 0. The, if x apple 0, the Archmedea Property says that 9 2 N 3 x <, so x + > 0. Sice x + < y +, 9r 2 Q 3 x + < r < y + =) x < r < y ad r 2 Q. Now 0 < x < y =) y x > 0 =) (by Corollary 2.4.5) 9 2 N 3 < y x =) x + < y. Applyig Corollary to x > 0, 9 m 2 N 3 m apple x < m. The m apple x + < y =) x < m < y =) x < m < y. Let r = m. Corollary (2.4.9). If x, y 2 R with x < y, the 9 a irratioal z 2 R 3 x < z < y. Proof. x < y =) x p 2 < y p 2. By desity, 9 r 2 Q 3 x p 2 < r < y p 2 ad r 6= 0 (why?). The x apple p 2r < y. Note that p 2r is irratioal as the product of a ratioal with a irratioal. Let z = p 2r.

21 Homework 2.5. INTERVALS 3 Pages # (Use Property S. You are likely to eed the Archimedea Property i Part (2)), 2 (Just fid ad prove sup S - same hit as for #). Page 46 #9 (Look at proof of Corollary as a model) Itervals Theorem (2.5. Characterizatio of Itervals). If S is a subset of R that cotais at least two poits ad has the property the S is a iterval. if x, y 2 S ad x < y, the [x, y] S, (a, b) = {x 2 R : a < x < b} [a, b] = {x 2 R : a apple x apple b} (a, b] = {x 2 R : a < x apple b} (a, ) = {x 2 R : x > a} (, b] = {x 2 R : x apple b} I = [0, ] is called the uit iterval. Defiitio. A sequece of itervals I, 2 N, is ested if the followig chai of iclusios holds: I I 2 I 3 I I + Example. h () I = 0, i 8 2 N. \ I = {0}. =

22 32 2. THE REAL NUMBERS (2) J = 0, 8 2 N. \ J = ;. = (3) K = (, ) 8 2 N. \ K = ;. = Theorem (2.5.2 Nested Itervals Property). If I = [a, b ], 2 N, is a ested sequece of closed, bouded itervals, the 9 2 R 3 2 I 8 2 N. Proof. By estig, I I 8 2 N, so a apple b 8 2 N. Thus is bouded above. Let = sup A, so a apple 8 2 N. [To show apple b 8 2 N.] ; 6= {a : 2 N} = A Let 2 N be give (so b is arbitrary, but fixed). [To show b is a u.b. of A, so the apple b.] () Suppose apple k. The I I k, so a k apple b k apple b. (2) Suppose > k. The I k I, so a k apple a apple b. Thus, 8 k 2 N, a k apple b =) b is a u.b. of A =) apple b.

23 2.5. INTERVALS 33 Theorem (2.5.3). If I = [a, b ], 2 N, is a ested sequece of closed, bouded itervals 3 if{b a : 2 N} = 0, the 9 a uique 2 R 3 2 I 8 2 N. Proof. Let = if{b : 2 N}. Usig a argumet similar to that of the previous theorem, we have a apple 8 2 N, so = sup{a : 2 N} apple. Thus, x 2 I 8 2 N () apple x apple. Let > 0 be give. [To show = 0.] By Property I, sice if{b a : 2 N} = 0, 9 m 2 N > b m a m. The 0 apple apple b m a m <. Thus 0 apple < 8 > 0. By Theorem 2..9, = 0, so = is the oly poit belogig to I 8 2 N.

Math 299 Supplement: Real Analysis Nov 2013

Math 299 Supplement: Real Analysis Nov 2013 Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality