MATH3283W LECTURE NOTES: WEEK 3 2//200 Proof without words: picture depicts What is beig proved from Fig.3.? Addig more ad more dots gives bigger ad bigger squares. It is too vague ad it is ot actually a math statemet. 2 Each cosecutive lie has two more dots tha the previous lie. Nothig to prove. 3 The sum of cosecutive odd umbers gives a square umber. It ca be proved by iductio: Observatio: P () : + 3 + + (2 + ) =? =, + ( + 2) = + ( + 2 ) = 4 = 2 2 = 2, + ( + 2) + ( + 4) = + ( + 2 ) + ( + 2 2) = 9 = 3 2 So we guess that P () is + 3 + + (2 + ) = ( + ) 2 ad prove it by iductio. = 0, OK. Assume P () is true, we wat to show that P ( + ) is true. + 3 + 5 + + (2 + ) + (2( + ) + ) P () = ( + ) 2 + 2 + 3 = 2 + 2 + + 2 + 3 = 2 + 4 + 4 = ( + 2) 2 So P ( + ) is true. = (( + ) + ) 2 Upper ad lower bouds Suppose A( ) R has a upper boud (bouded above). Let B = {r R r : upper boud for A}
2 MATH3283W LECTURE NOTES: WEEK 3 Suppose B has a smallest elemet w. The w is called the least upper boud of A, or the supremum of A, write w = luba or w = sup A. Thus w = sup A if () w is a upper boud for A ad (2) if r is a upper boud for A, the r w. Aother form of (2), usig cotrapositive: (2 ) r R(r < w a A, r < a) Note that ay r < w is ot a upper boud. Facts If r > w = sup A, the r B. Sice w is also a upper boud, B is a ray [w, ). Let ε > 0, the w ε < w ad by (2 ), a A, w ε < a, so we also have a equivlat coditio: (2 ) ε > 0, a A, w ε < a. Similar for lower bouds: Suppose A( ) R is bouded below ad w is the greatest lower boud for A, write w = glba or w = if A (ifemum of A). Thus w = if A if () w is a lower boud for A ad (2) if s is a lower boud for A, the w s. Agai, usig cotrapositive of (2) (2 ) r R(r > w a A, a < r) Facts If s < w = if A, the S is a lower boud ad s C : set of lower bouds of A (C is the ray (, w]) We also have (2 ) ε > 0, a A, a < w + ε. Note: If A has a maximal value w, the w = sup A. If A has a miimum value s, the s = if A. sup s ad if s geeralize max/mi values. Examples:
MATH3283W LECTURE NOTES: WEEK 3 3 () A = (0, ). The sup(0, ) = ad if(0, ) = 0. (Observe: is a upper boud. If r <, we have to show that r is ot a upper boud. Or we wat to fid s (0, ) such that r < s <. Choose average +r +r, the r < = s <.) 2 2 A = [0, ] also has sup[0, ] = ad if[0, ] = 0. So sup A ad if A may or may ot be a elemet of A. (, 0] (0, ) [, ) set of lower bouds A set of upper bouds (2) (3) A = {, 2, 3, } = { N} Sice is decreasig, is the largest elemet of A ad sup(a) =. calculus: 0 as x + x replacig x by (iteger values): 0 So if(a) = 0. This meas: ε > 0, N s.t. < ε. A = { 2, 2 3, 3 4,, +, } A is bouded above by. What is sup(a)? Note that + = + + Let ε > 0, N s.t. + = + < ε. The ε < + = By (2 ), = sup A. (4) Let A = {x x 3 < 4} + Now [x > 0 ad x 3 < 4] iff 0 < x < 4 3. If x < 0, the x 3 < 0 ad so x 3 < 4. Hece A = (, 4 3 ). sup A = 4 3 ad if A DOES NOT exist. (5) A = {x cos x 0 x π}
4 MATH3283W LECTURE NOTES: WEEK 3 Observe: f(0) = 0, f( π ) = 0, f(π) = π. The graph may look 2 like Fig.3.2. By calculus, f has a max ad mi value o [0, π]. f (x) = x si x + cos x = 0 cos x = x si x, ta x = x The x.87 ad x cos x.56. So sup A =.56 (check by graph), f(π) = π is the miimum value ad if A = π. (6) A = {x x 2 + x 6 < 0} x 2 + x 6 = (x + 3)(x 2) = 0 whe x = 3 or x = 2. For x = 0, we ca get x 2 + x 6 = 6 < 0. So A = ( 3, 2) ad 2/3/200 Q: Is 2 2k + prime for ay k N? A: No! sup A = 2, if A = 3 k = 5, 2 25 + = 2 32 + = 4294967297 = 64 670047 Examples: () Let A R ad suppose sup A = if A. What ca we say about A? Let w = sup A = if A. If a A, the w = sup A w a w = if A w a So w = a ad A = {w}. (2) Let A R ad B A. What ca we say about sup A ad sup B? Assumig both A ad B are bouded. Let w = sup A, the w is a upper boud for A. Sice B A, w is also a upper boud for B. Hece w sup B. Exercise: What ca you prove about if A ad if B? (3) A = { 3, 2 4, 3 5, 4 6, } a = = ( 2 whe ) +2 + 2 or = +2 2 = 2. +2 +2 +2 So sup A = ad if A =. Note that sup A is ot a elemet 3 of A but if A A.
(4) (5) (6) MATH3283W LECTURE NOTES: WEEK 3 5 B = {, 2, 3, 4, 5, 6, } sup A = ad if A = 2. A = {x R x 2 + x > 0, x > 0} x 2 + x = 0 = x(x + ), x = 0,. x = 2, ( 2 )2 + ( 2 ) = 4 2 = 2 < 0 So A = (0, + ). No sup A ad if A = 0. B = {, 3, 9, 27,, 3, } sup A =, if A = 0 because 0. 3 I order to prove if A = 0, we eed to show that ε > 0, 0+ε = ε is ot a lower boud. So we eed to fid some such that < ε or < 3 ε 3. Take logs: l( l 3 ) < l(3), l(ε) < l(3), >. ε l ε The existece of such iteger is give by the ext topic. The Least upper boud axiom Math statemet that the reals R have o holes. Equivaletly, if we approach a umber as a l.u.b, the that umber exists. Least upper boud/complete axiom Every o-empty set of real umbers that is bouded above has a least upper boud. From this, we get a versio of the well-orderig theorem for the reals. Theorem 0.. Let A, A R ad A bouded below. The glba exists. Proof. Cosider B = { a a A}. Sice A is bouded below, x R, a A, a x. The a A, a x ad x is a upper boud for B. By LUB axiom, B has a l.u.b., say y = lub(b). Claim: y = glb(a). First, we wat to show that y is a lower boud. a A, a y a A, a y ad y is a lower boud. Secod, we have to show that y is the greatest oe. Suppose y < r, the y > r. Sice y = lub(b), a A, y > a > r. The a < r ad r is ot a lower boud for A. So y = glb(a).
6 MATH3283W LECTURE NOTES: WEEK 3 A importat cosequece is: The atural umbers N ad i fact the set A r = {r N} for ay positive real r are ubouded above. Theorem 0.2 (Archimedea Property of Reals). Let a, b be positive real umbers, the N, a > b. Proof. By cotradictio. Suppose N, a b. The A = {a N} is bouded above. Let b = lub(a). Sice a > 0, b a is ot a upper boud. So m N, b a < ma. This implies b = (b a) + a < ma + a = (m + )a cotradicts that b is a upper boud for A. So N, a > b. Corollary 0.3. () N is ubouded above. (2) glb{ N} = 0 Proof. () N = { N} is ubouded by A.P. (a = ). (2) Foe ay r > 0, we wat to show, < r. Sice < r < r This follows from A.P. (b =, a = r). boud. Exercises () Let a > 0. The glb{ a N} = 0 (2) Prove the followig variat of A.P.: Let a, b > 0, the N, a < b So 0 is the greatest lower This meas { a N} ad { N} are ubouded i NEG SENSE (goes to ). (3) Prove: If a > 0, the lub{ a N} = 0 2/5/200 Theorem 0.4. There is a real umber x such that x 2 = 2.
MATH3283W LECTURE NOTES: WEEK 3 7 Proof. Let S = {s R s > 0 ad s 2 < 2}. Sice S, S is ot empty. Moreover, 2 is a upper boud. This ca be proved by cotrapositive: If r 2, the r 2 2 2 = 4 > 2 r is ot i S. By LUB axiom, sup S exists. Let x = sup S >. Claim: x 2 2 ad x 2 2, which says x 2 = 2. Suppose x 2 < 2, the b = 2 x 2 > 0. Set a = 2x +. By exercise (), N, a < b i.e. (2x + ) < 2 x2 (2x + ) (2x + ) < 2 x2 x 2 + 2 + ( )2 < 2, (x + )2 < 2 ad x + S. This cotradicts to the fact that x = sup S, so x 2 2. A similar argumet shows that if x 2 > 2, we ca fid N with (x )2 > 2, cotradictig that x is the smallest upper boud. So we also have 2 x 2 ad hece x 2 = 2. Now we wat to show that there are ratioal umbers everywhere. Theorem 0.5. Let a, b be real umbers with 0 < a < b <, the r Q with a < r < b. Proof. Sice b > a, b a > 0. Sice glb{ N} = 0, we have, 2 N with < b a ad 2 < a. Let = 2, the < b a ad < a (see Fig.3.3). Let B = { j j ad j a} B sice B ad boud above by.. By LUB axiom, B has a max elemet j 0. (sice B is fiite, lub(b) B). The j 0+ > a. Also j 0 + = j 0 + < a + (b a) = b. So we ca choose r = j 0+. Theorem 0.6 ( th roots of positive umbers). Let N ad y > 0. The x > 0 such that x = y i.e. x = y = y. Examples: fid lub ad glb if they exist: () A = {x x 2 < 4} = {x x < 2} = ( 2, 2) lub(a) = 2, glb(a) = 2. (2) B = {x x 5 > 9} (x egative x 5 egative) x > 0 If 9 x 5, the 9 5 (x 5 ) 5 = x (Why? Check it). So B = {x x > 9 5 } = [9 5, + ) No lub, glb(b) = 9 5 B. (3) C = {2, 2, 2, } = {2 + 2} 2 3 4 lub(c) = 2 C. 2 glb(c) = 2 + glb{ 2} = 2 + 0 = 2 ot i C.
8 MATH3283W LECTURE NOTES: WEEK 3 (4) D = {x x > 0 ad l x < }. l x = x = e. So D = (0, e), glb(d) = 0, lub(d) = e. (5) (6) A = { 2, 2, 2 3, 3, 3 4, 4, } sup(a) =, if(a) = 2 A = {0, 2, 2, 3 4, 4, 7 8, 8, 5 6, 6,, 2, 2 2, } sup(a) =, if(a) =. 2 Fid a A with a >.99: = 7, 27 = > =.99 2 7 28 00 Fid a A with a >.999: = 0, 20 = > =.999 2 0 024 000 (7) A = {x x 3 + x > 0} = {x x > 0} x 3 + x = x(x 2 + ) = 0 x = 0 No sup B, if B = 0 (see Fig.3.4). (8) B = {x x 3 x > 0} x 3 x = 0 = x(x )(x + ) B = {x x > or < x < 0} = (, 0) (, + ) No sup B, if B = (see Fig.3.5).