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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'pCIW 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 = f1 o 91. 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/” EZ'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 = 33+0'7, 10 = 1.3+17, and 12 = 43+07,
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.
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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: Ea + 72. W 9e 107/91: gee“ k: /7«3, rim é/va
___———""‘_'l' I l M A/cmoj 517/?ﬂ/ﬂ /(’” A//ﬂ/W 36; 55W IQ: 3ﬂ+7A I % ﬂ'ﬁJ 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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