Answer to Question #174367 in Discrete Mathematics for Lam Zhi Onn

Question #174367

Q1. a) Let π‘ˆ = {π‘₯: π‘₯ ∈ 𝑍, 1 ≀ π‘₯ ≀ 12},

𝐴 = {2π‘₯: π‘₯ ∈ π‘ˆ π‘Žπ‘›π‘‘ π‘₯ 𝑖𝑠 π‘Ž π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘’ π‘œπ‘“ 4},

𝐡 = {π‘₯: π‘₯ ∈ π‘ˆ, π‘₯ 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 2} π‘Žπ‘›π‘‘ 𝐢 = {π‘₯: π‘₯ ∈ π‘ˆ, π‘₯2 ≀ 16}.

i) List the elements belong to the sets A, B and C respectively. (3 marks)

ii) Find 𝐢 βˆ’ (𝐴̅ ∩ 𝐡) (3 marks)

iii) Find π΅βŠ•πΆΜ…. (2 marks)


b) Prove by induction that 1 + 5 + 9 + … + (4n – 3) = n(2n – 1) for all n β‰₯ 1. (5 marks)

c) Let π‘₯ = 866 π‘Žπ‘›π‘‘ 𝑦 = 732.

(i) Find the greatest common divisor of x and y and then express it in the form of

ax + by, where π‘Ž, 𝑏 ∈ 𝑍. (5 marks)

(ii) Find the least common multiple of x and y. (2 marks)

[Total: 20 marks]


1
Expert's answer
2021-03-31T15:27:46-0400

Solution.

Q1.

a)

i)

"\ud835\udc48 = \\{\ud835\udc65: \ud835\udc65 \u2208 \ud835\udc4d, 1 \u2264 \ud835\udc65 \u2264 12\\}=\\{1,2,3,4,5,6,7,8,9,10,11,12\\},\\newline\n\ud835\udc34 = \\{2\ud835\udc65: \ud835\udc65 \u2208 \ud835\udc48 \\text{\ud835\udc4e\ud835\udc5b\ud835\udc51 \ud835\udc65 \ud835\udc56\ud835\udc60 \ud835\udc4e \ud835\udc5a\ud835\udc62\ud835\udc59\ud835\udc61\ud835\udc56\ud835\udc5d\ud835\udc59\ud835\udc52 \ud835\udc5c\ud835\udc53 } 4\\}=\\{8,16,24\\},\\newline\n\ud835\udc35 = \\{\ud835\udc65: \ud835\udc65 \u2208 \ud835\udc48, \ud835\udc65 \\text{\ud835\udc51\ud835\udc56\ud835\udc63\ud835\udc56\ud835\udc51\ud835\udc52\ud835\udc60 } 2\\}=\\{2,4,6,8,10,12\\},\\newline \n\ud835\udc36 = \\{\ud835\udc65: \ud835\udc65 \u2208 \ud835\udc48, \ud835\udc65^2 \u2264 16\\}=\\{1,2,3,4\\}."

ii)

"\\overline{A}=\\{1,2,3,4,5,6,7,9,10,11,12\\},\\newline\n\\overline{A}\\cap B=\\{2,4,6,10,12\\},\\newline\nC-(\\overline{A}\\cap B)=\\{1,3\\}."

iii)

"\\overline{C}=\\{5,6,7,8,9,10,11,12\\},\n\\newline\nB\\oplus \\overline{C}=\\{2,4,5,7,9,11\\}."

b)

"1 + 5 + 9 + \u2026 + (4n \u2013 3) = n(2n \u2013 1) \\text{for all } n \u2265 1."

Let

"P(n):1+5+9+ . . .+(4n-3)=n(2n-1), \\text{for all } n \u2265 1.\\newline\nP(1):1=1(2\\cdot 1\u22121), 1=1, \\text{which is true.} P(1) \\text{is true.}\\newline\n\\text{Assume that } P(n) \\text{ is true for } n=k.\\newline\nP(k):1+5+9+ . . .+(4k-3) = k(2k-1) \\text{ is true.}\\newline\n\\text{Prove } P(k+1) \\text{ is true.}\\newline\nP(k+1):1+5+9+. . . +(4k-3)+4(k+1)-3=\\newline\n=k(2k-1)+4(k+1)-3=2k^2\u2212k+4k+4\u22123=\\newline\n=2k^2+3k+1=2k^2+2k+k+1=(k+1)(2k+1)=\\newline\n=(k+1)(2k+1+1-1)=(k+1)(2(k+1)-1).\\newline\n\\text{So, } P(k+1) \\text{ is true, whenever } P(k) \\text{ is true, hence } P(n) \\text{is true.}"

c)

"\ud835\udc65 = 866 \\text{ and } \ud835\udc66 = 732."

i)

"866 = 2 \u00b7 433,\\newline\n732 = 2 \u00b7 2 \u00b7 3 \u00b7 61 = 2^2 \u00b7 3 \u00b7 61,\\newline"

from here the greatest common divisor of "x" and "y" is "2."

GCD"(866,732)=2."

"ax+by=-71\\cdot866+84\\cdot732=2."

ii)

"866 = 2 \u00b7 433,\\newline\n732 = 2 \u00b7 2 \u00b7 3 \u00b7 61 = 2^2 \u00b7 3 \u00b7 61,\\newline"

from here LCM"(866,732)=2\\cdot433\\cdot2\\cdot3\\cdot61=361956."


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