10fe1s - Prof Girardi Math 554 Fall 2010 09.23.10 Exam 1...

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Unformatted text preview: Prof. Girardi Math 554 Fall 2010 09.23.10 Exam 1 MARK BOX PROBLEM POINTS - --- instructions INSTRUCTIONS: (1) For the fill—in-the—blank/ box problems, write your directly on this exam paper. (2) For the proof part of the proof pmblems, write a neat formal proof on (the front side only of) your homemade Blue Book. Start each new problem on a new page. Do NOT recopy the statement of the problem but on each page, in the upper right hand cornor, put the problem number. (3) When finished, hand in this exam paper and your proofs on your lined paper. Please put all in order and staple. Do your scratch work on the scratch paper provided and do not hand it in. (4) The MARK BOX indicates the problems along with their points. This test is copied 2-sided. Check that your copy of the exam has all of the problems. (5) You may not use any electronic device books, or personal notes. (6) During this exam, do not leave your seat. If you have a question, raise your hand. When you finish: turn your exam over, put your pencil down, and raise your hand. (7) This exam covers (from Introduction to Real Analysis by William F. Trench, Free Edition 1): Sections 1.1 and 1.2. . Honor Code Statement I understand that it is the responsibility of every member of the Carolina community to uphold and maintain the University of South Carolina’s Honor Code. As a Carolinian, I certify that I have neither given nor received unauthorized aid on this exam. Furthermore, I have not only read but will also follow the above Instructions. Signature : (frank/m 11/15?”th 1 Cl lel Exam 1 Bid, 1. Fields. 121. A field F is a set F on which the operations of (J) metal and mull/QM ‘cm/ are defined so that every pair of elements from F has a unique sum and product, both in F, with the following properties. TMM (A) ml): +a. am echebcu EWWfllML ' (ela)c:o.(lao a: ‘ adad‘jxfi 6, (E) ll“ 1b. An ordered field is a field F along with a relation < which has the following properties. (F) do L2. F ml} 45!" (g) I! <x<lo Cwol l)‘ C , Klee 0x 4 ‘" l1”; £31;va / ’ l 1:: Mb a+c<lorc v’e ‘ 0c <loe , (H) lJML 1c. An ordered field is complete provided it has the following property. / All. 1‘3 LOW/VIM dill—5*)”fl A35 0. ‘ Cl r“ 1d. Give an example of a field S for which it is impossible to define an order relation < that would V Lqub‘w lam/é make it an ordered. field. No proof necessary but do give how the 2 operations indicated in part (la-J) are defined. ‘ C’ —_ ' + c = x . 2 = or = phi o Wflfi ”(1% “Ii 1 3+3 ml lg (“B/K k /X li+h=§5¥x=v . ,, 3 <2 d” 3.. , 5: 1e. Give an example of an ordered field that is not a complete ordered field. No proof necessary but do clearly indicate why the field is not complete. i 6:) be [Eff Wu (54" ”if. l{’ c , a , 7, r+ l / Wolff 'll'kfl M S ’ l’l/V E Q l X A 3 Woblem Ins/fl/‘a/llofl/f Clasg de/Ul: 2. Let R be the set of real numbers, lli be the extended real numbers and S C R. 2a. Another term for the supremum of S, i.e. for sup S, is find? W v. S gr lu’l’s . 2b. Another term for the infimum of S, i.e. for the inf S is wlést \O‘Nfl’ lDGWV‘Ol " S W LS . 2c. Let S be a nonempty set that is bounded above. Then the supS is the unique real number fl 6 R such that (1) for each a: E S, .___%—_—_X '4‘ _ _________. f” . (2) supS = 00 if and only if US W1“ “ M lDOleM OJW’I‘T: _ (3) supS z —00 if and only if _—g_g :- . __ . 2e. Let S be a nonempty set that is bounded below. Then the inf S is the unique real number oz 6 R such that 4 (1) foreachmES, og —?C III X8<D<~‘l—8 (2 ) if a > 0, then 3.735 E S such that 1 4 A [filldflwflowafw M X£*°<1l’5 2f. We know that inf S e R. More specifically, C, p i l l ,I V (1)infS’ElRifandonlyif Q M WWW»! 7W Owfi‘ berm/wide lat/{WT ' - .r 0 (2) inf S = —oo if and only if S S M "T‘WMiwir-u (Mi/L "Yet?f l‘JWM/x M low ‘ ' \. i r ’ . (3) infS= 00 if and only if 5" é ‘ ?rol3lcm Iflsbiroul’lfimz 3 WWé % I,’ _‘y: 5 3. Fill in the chart. No proof necessary. I—I-— __—I II_ 22/; __—| _Im-n __|- TS x I 0" /\ H ’{I‘ :2 l (y /\ s l“ p $3 (/3 i l ”13‘ \N L) x \_ {a \M ”is, S: PM} VF”, 4. Let r be a positive real number. Let {mn}§°=0 be a sequence of real numbers satisfying Iran“ — mnl S 7‘ lacn —— mn_1| for each n E N . (4) Let Pn be the proposition that mm Wind. Inch—mg s W1 lml—mol . . (Pn) Show that Pn is true for each n E N by math induction. Use strong induction only if you must do so. 5. In class we proved Archimedean’s Property, which is the following. If p > 0 and e > 0, then there exists n E Z such that me > p. Provide a proof of Archimedean’s property. 6. Let T be a nonempty subset of R that is bounded above and let ,6 = sup T. Let c be a positive constant (i.e. c 6 R and c > 0). Define the subset cT by cT := {ctEletGT}. Show that sup (CT) = c,8. (6) . 1%on Inspirahm : 53th \-2.3 ’hram whiLmré 4. Let r be a positive real number. Let {xn}r°f=0 be a sequence of real numbers satisfying lmn+1 —— mnl S r Ixn — asnnll for each n E N . (4) Let Pn be the proposition that ‘ lmn — 2311-1] S r”‘1 lacl — 3:0] . . ' (Pn) Show that Pn is true for each n E N by math induction. Use strong induction only if you must do so. ”Problem IflSflmllm " dial in Cl 153) Very slow/lg ./ Archimedean’s Property Archimedean’s Property If p > 0 and e > 0, then there exists n E Z such that 718 > p. (AP) Proof. Note that (AP) can symbolically be written as (Vp>0) (V5>0) (ElnEZ) [ne>p]. We shall show that (AP) holds by contradiction. Thus, assume that (AP) does not hold. So we are assuming that ' (3p > 0) (35 > 0) (Vn E Z) [n5 5 p] » (1) and we want to find a contradiction. Let p > 0 and e > 0 be as given in (1), thus (Vn E Z) [718 S p] . (2) Consider the set S := {zeR:x=ne,n€Z} = {neelenEZ}. Then S is nonempty (e.g., 8 E S). Also, (2) implies that S is bounded above by p. So by the completeness axiom of IR, S has a supremum in R. Let [3 := supS . By definition of supremum, if s E S then s 3 fl and so ianN,thennegfl. (3) If n E N then 71 + 1 E N and so (3) implies ianN,then (n+1)e$fi, which is equivalent to (by some algebra) ianN,thenna_<_fl—s. (4) Now (4) implies that fl — a is a_n upper bound of S. However, 3 is the least upper bound of S. This is a contradiction. So (AP) holds. El l l h m \1 Prolaluem Insp‘mtm U) SISMI‘ch‘ +0 WWI/E § [I 4L— [0 (2) (Magda 03 an Em... 1a: a; Supremum of a Set Let T be a nonempty subset of lit-that is bounded above and let ,6 = sup T. Let c be a positive constant (i.e. c E R and c > 0). Define the subset cT by cT := {ctER:tET} . Show that sup (cT) = cfl . (WTS) Proof. Let the givens be given. Note that fl 6 1R since T is nonempty and bounded above. We will show that equation (WTS) holds. Step 1: First we will show that cfi is _a_n upper bound of cT. Fix :1: 6 CT. By definition of the set cT, there exists t E T such that x = ct . (1) Since t E T and fl = supT t g ,6 . (2) Since c > O, inequality (2) implies that at < cfl . (3) Combining (l) and (3) now give :1: < cfl So cfl is indeed fl upper bound of cT. Step2: Next we show that cfl is the least upper bound of cT. Fix 5 > 0. Way # 1 for Step 2 <WTS cfl — 5 is not an upper bound of cT so WTF .1; 6 cT s.t. cfl - a < $2-> Note that % > 0 since 5 > 0 and c > 0. Thus, since fl = sup T, there is a t5 6 T such that 5 fl _ Z < t5 - ‘ (4) Since c > 0, inequality (4) implies that cfl — E < cte . By definition of the set cT, since tE E T we know that ctE 6 CT. Way # 2 for Step 2 <Since c > o, WTS cfi — as is not an upper bound of cT so WTF :25 5 0T s.t. c5 — ca < z,.> Since fl = sup T, there is a t5 6 T such that )6 — 5 < t5 . (5) Since 0 > 0, inequality (5) implies that cfl — c5 < ct5 . By definition of the set cT, since t5 6 T we know that ate 6 cT. Step 3: By definition of supremum of a set cfi = sup(cT). El ...
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