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Unformatted text preview: EDCALJSOI You may choose to prove set identities using VEnn—Diagram regions. If
you need extra space, please start with the back of the page. Please put numeric answers near the questionE where they may easily be
found. Name I l. Prove any prime number p is either of the form p=8q+l, p=8q+2,
p=8q+3, p=8g+5.r or p=8q+?, where q is some integer. Solution: Suppose Egejz+ is a prime. By the Quotient—Remainder Theorem, ElgEZ and rE[ﬂ,l,2,3,4.5,6,T] such that p=8q+r {q a r depend on p}. This gives 8 possible cases for the remainder r. Now. Case r = D: p==8q is impossible because EJ=2b4q would be a
factorization of the prime p into factors each 21 {or
because En=23q would violate the Fundamental Theorem of
Arithmetic [that a number may be factored into primes
only one way, except for order]}. Case r = 4: p==8q44 is impossible because p==2{4g+2} would be a
factorization of the prime p into factors each 51 {or because F;=;f{2q+1) would violate the Fundamental Theorem
of Arithmetic}.
Case r = 6: p==8qF6 is impossible because p==214q+3} would be a
factorization of the prime p into factors each h 1.
Since we have excluded the 3 remainders r==U, 4.6, only 5 possibilities remain: p = 8q+r and r E {1, 2, 3, 5ir 7"} . Bi!
2. Calculate S = Z (312:2 — 311:: 1! like: S = 4 3f17. _ 8 9 + 9 18 + + 83 84 + 84 85
_ 26 29 29 32 I” 251 254 254 25? 8 85 2056—2218 154 [ 77 ]
———————— ==————— or ———— . )_ Your answer should be in a format Solution: 26 257__ 6682 6682 3341 3. f={[—3, 3), (—2, 2}, {—1, 13,110, 0), (2, 2)} is a function whose co—
domain and range are the same. {a} What is the range of f?
Solution: The range =[O, 1, 2, 3} . {b} What is the domain of f?
Solution: The domain ={—3,—2,l.0r2} {c} f is 1—1, True or False. {Circle one, no proofs}
[f is not l—l because f{2}=f(—2}=2.] {d} f is onto, True or False. {Circle one, no proofs)
[f is onto, by definition, since we are told the co—domain and range are the same]
4. Prove or disprove: V sets a, B, and C: A—{B—C}=[A—B}—C. Solution: The assertion is false, as the following ( 5. Suppose the sequence an,a1,a2, . .. =8, 31, 13?, counterexample shows: A = {1, 2, 4, 5}, A
B = {2, 3, 5, 6}. and C = {4, 5, 6, 7'}. Here, A—(B—c}=[1,2,4,5}—{2,3}=[1,4,5}, while [AB}C = [1, 4}—{4, 5, 5, 7]={1}. Hence, A—{B—c}=[1,4,5]¢[1}={A—s}—c. satisfies the recurrence relation a1 = land—10a,"2 V k E Z where k is 2 . Find a compact algebraic formula for the general term akV ICED. Solution: The characteristic equation t2—7t+10 ={t—2){t—5) has roots t=2, t=5. So EIA, BER such that ak=A2k+B5". Substituting k=0, 1 in the first two terms of the sequence, we get two equations _ 8: 115+ B —16=—2A—2B _
in two unknowns: —1 Adding, we get
31=2A+5B~ 3l= 2A+5B. 15:33, so B=5. Substituting B=5 into any equation, we get A=3, so ak=32"+55" or ak=32k+5""'1. 6. Observe: 497,2212 — 213,6582 = 247,228,722,841 — 45,735,244,164 = 201,493,478,677 = 54,267,029 * 3,713
Factor 54,267,029 in a non—trivial way. Solution: Find a factor by computing
GCD{54,267,029 ; (497,221 — 213,858)} = GCDIS4,267,029 ; 283,363}: 54,267,029 = 283,363 X 191 + 144,696
283,363 = 144,696 X 1 + 138,667
144,696 = 138,667 X 1 + 6,029
138,667 = 6,029 X 23 + 0 Now calculate: 54,267,029 + 6,029 = 9,001.
This gives the factorization: 54,267,029 = 6,029 x 9,001. Alternate solution: Find a factor by computing
GCD{54,267,029 ; (497,221 + 213,858}} = GCD{54,267,029 ; 711,079}: 54,267,029 2 711,079 X 76 + 225,025
711,079 = 225,025 X 3 + 36,004
225,025 2 36,004 X 6 + 9,001 36,004 = 9,001 X 4 + 0 Now calculate: 54,267,029 + 9,001 = 6,029.
This gives the factorization: 54,267,029 2 6,029 x 9,001. kEll 7. Calculate the sum 53: 2: 311.1}1 as a decimal fraction with 3 Ila5 digits of accuracy, e.g. like S = 215.434. 3{1.1}51— 3(1.1}5 _ 387.3898—4.8315 Solution: S=
1.1—1 .1 =3,825.583. 8. Express the repeating decimal r==18.412322322322 as a rational number, i.e., express r as the ratio of two integers. Solution:
1000r = 18412.32232
— r = — 18.41232 ...
9991 = 18393.91000 ...
r'— 18,393.91 _ 1,839,391 999 99,900 ...
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 Spring '08
 Ellis,W

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