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# exam3-answers - NAME Sets and Logic MHF3202(4628 Third Hour...

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Unformatted text preview: NAME Sets and Logic: MHF3202 (4628) Third Hour Exam W. Mitchell Wednesday March 31, 2010 Start your answers on the worksheet; if you need more space, 1. (10) continue them on the back. Be sure to show your work to receive 2_ (10) w full credit. 3 (10) _~. No calculators or notes are allowed on this test. ' 4. (10) Tot. (40) 1. (10 points) Prove the following theorem: Theorem. Suppose that f : A —> B, and them exists a function g: B —) A such that fog: 2'3. Then f is onto. / git“ 707(1) :— may) 2/ ﬁr 6: 7w W W {[ﬁj;0/. Cw/MZLLZ/J’ M2 ﬂffW/é‘p; W ) (7494 f(”/?‘5, M 775’” Wﬂﬁoa'é/[an sat/5W Vila Mi, ﬂ 2. (10 points) (a) Give an example of a function which is 1—1, but does not have an inverse. (“I f ’3 Mr W21; far ygw MZ’Zﬂv'p-CIW x3; inc/51% 479797. (b) Give a counterexample to the following (false) claim, and verify that your answer is a counterexample: Claim. Suppose that the functions f: R —> R and g: R —> R each have an inverse. Then (f o grl = f-1 o 9-1. W JDCX) 72X W fY/r/eﬁ 5.” 6rd;)ﬂ(x/.~ z.X+.2. Mg (fog) "m: LEE. 2" i u '1' ’1 a / WW 7‘ 9 : —»-— ﬁzz; — j 2- . 5M " X ’l X t xi —-1 2/” E-Z'ﬁé"! ’ ’57: (c) Deﬁne a sequence (u:n : n < w) by recursion: a0 = 17 and ' is WWWhat is (25?. an (15 = Z 19 Note: The Collatz conjecture asserts that, no matter w at you choose for ate, there is some n such that an : 1. Paul Erdt’js said that "Mathematics is not yet ready or such confusing, trou— bling, and hard problems,” and OEeIed US\$500 for ' 5 solution. an at, 65 a J: I 7 an H : / _ ,_ 312., +1 .‘ 4, al’ 3‘/7 +/ l 9 2- a,, ; 5 2/2. z 2 1; 1/251 "5 3. (10 points) Use strong induction to prove the following theorem: W Theorem. For every natural number n > 11 there are a and b in W such that n = 3a + 7?). NOTE: we have, for example, 9 = 3-3+0'7, 10 = 1.3+1-7, and 12 = 4-3+0-7, but 11 cannot be written in this form. Your proof will show that 11 is the largest which cannot be written in this form. Pﬂf" é/g ﬁzz/yeﬂ 179g ﬁv/é'o/frm/ 6% “W461 W ’72”: #ﬂw éze/L. jﬂﬁaf 3 a/ 6% 4: 35: +7.5, A/e Wy/ pm , If) V” [Ivy Aykco(n7//«7,9/UJ*7/{ﬁ)’ /‘ 7%” % 6/7” we A: Mimi We. 5:5ij 7% x2497?” é/xféx/Lr ‘ Vk[k;//4£<,) 7/40) 7:) ﬁrm Aqf, WC ems/afar- 4/ (W, I CE ’7’42' 7%” 53'? +7,0. £9353; n;/3, 71”:— ﬂ: 3:2 + 7,/ w n; Hf, ﬂ“ I4: E-a + 7-2. W 9e 107/91: gee“ k: /7«3, rim é/v-a ___———-""‘_'l' I l M A/cmoj 517/?ﬂ/ﬂ /(’” A//ﬂ/W 36; 55W IQ: 3ﬂ+7A I % ﬂ'ﬁ-J 5’ £2 W/i/Mjsﬂj ﬁr ﬂ~37£=3”(*75' ﬂfm x9 4m 729157; 35am +75“, 2/ L7: al‘f/ M 534’] 5.474425} W lﬂL/J/ SIM W V/KW'” ﬂo’) VII? 51,55 W We WW)” fé Ibo/qcﬂ‘ﬂ é/y/zj/‘J. S/M ﬂ 60 1/9/43 W 6/ N‘w7 xkdaor’x‘m owe m MGM Via/[4; nyof 'm WW7? Aer/M, 3 ﬂ you I“! 4. (10 points) Consider the follovsdng (invalid) theorem, and claimed proof: The lines have been numbered so you can refer to them. Theorem. For any n > 0 m N, and any function f with. domain Nﬂ 3 {1,1, ..,n} is a constant fenction. That is, there 13 some a seek that f r- o for alii e N“. Pmo f. 1, We prove the theorem by induction on n, and Let P(n} be the statement that any function f with domain Nu is a constant function. 2, [:1 Base case: The statement; PU) is clearly true, since N1 =2 {1} has oniy one member. 3. Induction step: Suppose that the statement P{n) is true, and let f be an arbitrary function with domain Nag. Consider the following two functions with domain Nﬂ: 4. D g(i) = for i E N“. 5. D :: f(i+1) for i E N”. 6. Then {3(a) implies that each of g(z') and is a constant function. 7’ 13 Let (29 be such that giz‘) = mg for allz' 6 NR, and let ah be such that 2 oh for 3.11 2' E N”. 8. I claim that (zg = oh. 9. X To see this, pick any I: such that 1 < k < n ~i— 1. (lag? is line ('1 1o. :1 Then ag = gﬂs) = ﬁle) = mm = oh.°°/“’-’.’ ’ “""0 11. D Thus = : :19 for i g n,whi1e f{n+ 1) :2 h((n+ 1) - 1) = 2 oh = G9. 12. Thus P(n + 1) is true. ' 13. It follows by mathematical induction that P(n) is true for all n 3 1, and the theorem follows. C! (a) What is the ieast m 2 1 in N such that the statement P(m) is not true? 2-— f/zfeif 95°24} (b) Give a. counterexample to Pm) for this n. 791’ a ,9 k 0” 4! 70C 7) «7; / (c) For each line of the proof which is not veﬁd, based on the proceeding lines, put a. cheekmark in the box. NOTES: i) There may be one §nvalid line, or more than one. ii) Every line without; a box is valid. iii) The invalid line or lines are outright wrong. Don’t check a line just because it doesn’t follow the eiass style. ((1) Explain your mower to (o). , 33‘]: J‘1};‘1’ 7%»: fén’ {Bf/41’ <m‘/’/ w” W/<k<2” M ﬁre V9 WJM/éﬂgygn y/n 6M 1”" [ﬁfe/é ’ f1» fé/é’, n’mméw Mi; ’puof ’ I'4 0/055 —’ as 0a.{( ﬁance”; J'th col 0",” ...
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