Solution to Q1. a) Let π‘ˆ = {π‘₯: π‘₯ ∈ 𝑍, 1 ≀ π‘₯ ≀ 12}, 𝐴 … - Sikademy
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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]

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Solution.

Q1.

a)

i)

π‘ˆ = \{π‘₯: π‘₯ ∈ 𝑍, 1 ≀ π‘₯ ≀ 12\}=\{1,2,3,4,5,6,7,8,9,10,11,12\},\newline 𝐴 = \{2π‘₯: π‘₯ ∈ π‘ˆ \text{π‘Žπ‘›π‘‘ π‘₯ 𝑖𝑠 π‘Ž π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘’ π‘œπ‘“ } 4\}=\{8,16,24\},\newline 𝐡 = \{π‘₯: π‘₯ ∈ π‘ˆ, π‘₯ \text{𝑑𝑖𝑣𝑖𝑑𝑒𝑠 } 2\}=\{2,4,6,8,10,12\},\newline 𝐢 = \{π‘₯: π‘₯ ∈ π‘ˆ, π‘₯^2 ≀ 16\}=\{1,2,3,4\}.

ii)

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

iii)

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

b)

1 + 5 + 9 + … + (4n – 3) = n(2n – 1) \text{for all } n β‰₯ 1.

Let

P(n):1+5+9+ . . .+(4n-3)=n(2n-1), \text{for all } n β‰₯ 1.\newline P(1):1=1(2\cdot 1βˆ’1), 1=1, \text{which is true.} P(1) \text{is true.}\newline \text{Assume that } P(n) \text{ is true for } n=k.\newline P(k):1+5+9+ . . .+(4k-3) = k(2k-1) \text{ is true.}\newline \text{Prove } P(k+1) \text{ is true.}\newline P(k+1):1+5+9+. . . +(4k-3)+4(k+1)-3=\newline =k(2k-1)+4(k+1)-3=2k^2βˆ’k+4k+4βˆ’3=\newline =2k^2+3k+1=2k^2+2k+k+1=(k+1)(2k+1)=\newline =(k+1)(2k+1+1-1)=(k+1)(2(k+1)-1).\newline \text{So, } P(k+1) \text{ is true, whenever } P(k) \text{ is true, hence } P(n) \text{is true.}

c)

π‘₯ = 866 \text{ and } 𝑦 = 732.

i)

866 = 2 Β· 433,\newline 732 = 2 Β· 2 Β· 3 Β· 61 = 2^2 Β· 3 Β· 61,\newline

from here the greatest common divisor ofΒ xΒ andΒ yΒ isΒ 2.

GCD(866,732)=2.

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

ii)

866 = 2 Β· 433,\newline 732 = 2 Β· 2 Β· 3 Β· 61 = 2^2 Β· 3 Β· 61,\newline

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

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Question ID: mtid-5-stid-8-sqid-3251-qpid-1950