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MAT2611 ASSIGNMENT 1 SEMESTER 1 2022
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- University Of South Africa (Unisa)
This document contains MAT2611 ASSIGNMENT 1 SEMESTER 1 2022 SOLUTIONS. EXPLANATIONS AND CLEAR CALCULATIONS ARE SHOWN.
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JC TUTORIALSMAT2611
ASSIGNMENT 1
SEMESTER 1
2022
, Exercise 2.8:
Solution:
Part 1.
𝐴 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝒑 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑤ℎ𝑒𝑟𝑒 𝒑 > 1 𝑎𝑛𝑑 ℎ𝑎𝑠 𝑜𝑛𝑙𝑦 𝑡𝑤𝑜 𝑓𝑎𝑐𝑡𝑜𝑟 1 𝑎𝑛𝑑 𝒑, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝐴 = {2, 3,5,7,11,13,17,19,23, . . . . . }
𝐴𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑖𝑛𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝐵 = {1,3,5,7,9,11,13,15,17,19,21,23, . . . . . . . }
𝑂𝑛 𝑜𝑛𝑒 ℎ𝑎𝑛𝑑 , 𝑤𝑒 ℎ𝑎𝑣𝑒 2 ∈ 𝐴 𝑏𝑢𝑡 2 ∉ 𝐵 , ℎ𝑒𝑛𝑐𝑒
𝐴 ⊈ 𝐵
𝑂𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 ℎ𝑎𝑛𝑑 , 𝑤𝑒 ℎ𝑎𝑣𝑒 1 ∈ 𝐵 𝑏𝑢𝑡 1 ∉ 𝐴 , ℎ𝑒𝑛𝑐𝑒
𝐵 ⊈ 𝐴
Part 2.
𝐴 = 𝐵 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑎 ∈ 𝐴 , 𝑡ℎ𝑒𝑛 𝑎 ∈ 𝐵 𝑎𝑛𝑑 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑏 ∈ 𝐵, 𝑡ℎ𝑒𝑛 𝑏 ∈ 𝐴
𝑆𝑖𝑛𝑐𝑒 2 ∈ 𝐴 𝑏𝑢𝑡 2 ∉ 𝐵 𝑎𝑛𝑑 1 ∈ 𝐵 𝑏𝑢𝑡 1 ∉ 𝐴, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝐴≠𝐵
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