COS1501 Assignment 3 2021
Question 1
𝑈 = {1, 2, 3, 4, 5, 𝑎, 𝑏, 𝑐} 𝑎𝑛𝑑 𝐴 = {𝑎, 𝑏, 𝑐, 1, 2, 3, 4}
𝑅: 𝐴 → 𝐴
𝐴 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑖𝑓 𝑒𝑣𝑒𝑟𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒
𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛.
1. {(1, 3), (𝑏, 3), (1,4), (𝑏, 2), (𝑐, 2)}
1 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 3 𝑎𝑛𝑑 1 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 4.
1 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
2. {(𝑎, 𝑐), (𝑏, 𝑐), (𝑐, 𝑏), (1, 3), (2, 3), (3, 𝑎)}
𝑎, 𝑏, 𝑐, 1, 2 𝑎𝑛𝑑 3 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
3. {(𝑎, 𝑎), (𝑐, 𝑐), (2, 2), (3, 3), (4, 4)}
𝑎, 𝑐, 2, 3 𝑎𝑛𝑑 4 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
4. {(𝑎, 𝑐), (𝑏, 𝑐), (1, 3), (3, 3)}
𝑎, 𝑏, 1 𝑎𝑛𝑑 3 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
Question 1 ONE
,Question 2
𝑈 = {1, 2, 3, 4, 5, 𝑎, 𝑏, 𝑐} 𝑎𝑛𝑑 𝐴 = {𝑎, 𝑏, 𝑐, 1, 2, 3, 4}
𝑅: 𝑈 → 𝐴 𝑈 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑎𝑛𝑑 𝐴 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛.
𝑇ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑈 𝑎𝑟𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒𝑠 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝐴 𝑎𝑟𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 𝑖𝑚𝑎𝑔𝑒𝑠.
𝐴 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑓 𝑖𝑡 𝑖𝑠 𝑎 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝑎𝑛𝑑 𝒇𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒄𝒐𝒅𝒐𝒎𝒂𝒊𝒏, 𝒕𝒉𝒆𝒓𝒆
𝒊𝒔 𝒂𝒏 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒐𝒎𝒂𝒊𝒏 𝒕𝒉𝒂𝒕 𝒊𝒔 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐 𝒊𝒕.
𝑇ℎ𝑎𝑡 𝑖𝑠, 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑦 ∈ 𝐴, 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑥 ∈ 𝑈 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 (𝑥, 𝑦) ∈ 𝑅.
𝐸𝑣𝑒𝑟𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴 𝑚𝑢𝑠𝑡 ℎ𝑎𝑣𝑒 𝑎 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒 𝑓𝑜𝑢𝑛𝑑 𝑖𝑛 𝑈.
1. {(1, 4), (2, 𝑏), (3, 3), (4, 3), (5, 𝑎), (𝑎, 𝑐), (𝑏, 1), (𝑐, 𝑏)}
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑈, 𝑡ℎ𝑎𝑡 𝑖𝑠, 1, 2, 3, 4, 5, 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴. 𝑆𝑜, 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝐴𝑙𝑠𝑜,
2 ∈ 𝐴 𝑏𝑢𝑡 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 (? , 2) 𝑝𝑎𝑖𝑟
2 ℎ𝑎𝑠 𝑛𝑜 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
2. {(𝑎, 1), (𝑏, 2), (𝑐, 𝑎), (1, 4), (2, 𝑏), (3, 3), (4, 𝑐)}
𝑇ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 5 𝑖𝑠 𝑖𝑛 𝑈, ℎ𝑜𝑤𝑒𝑣𝑒𝑟, 5 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑎𝑛𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
3. {(1, 𝑎), (2, 𝑐), (3, 𝑏), (4, 1), (𝑎, 𝑐), (𝑏, 2), (𝑐, 3)}
𝑇ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 5 𝑖𝑠 𝑖𝑛 𝑈, ℎ𝑜𝑤𝑒𝑣𝑒𝑟, 5 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑎𝑛𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
4. {(1, 𝑎), (2, 𝑏), (3, 4), (4, 3), (5, 𝑐), (𝑎, 𝑎), (𝑏, 1), (𝑐, 2)}
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑈, 𝑡ℎ𝑎𝑡 𝑖𝑠, 1, 2, 3, 4, 5, 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴. 𝑆𝑜, 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐴, 𝑡ℎ𝑎𝑡 𝑖𝑠, 1, 2, 3, 4, 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 ℎ𝑎𝑣𝑒 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒𝑠 𝑖𝑛 𝑈.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
Question 2 FOUR
, Question 3
𝐴 = {1, 2, 3, 4}
𝐺: 𝐴 → 𝐴 𝑎𝑛𝑑 𝐿: 𝐴 → 𝐴
𝐺 = {(1, 2), (2, 3), (4, 3)}
(1, 2) ⇒ 𝐺(1) = 2
(2, 3) ⇒ 𝐺(2) = 3
(4, 3) ⇒ 𝐺(4) = 3
𝐿 = {(2, 2), (1, 3), (3, 4)}
(2, 2) ⇒ 𝐿(2) = 2
(1, 3) ⇒ 𝐿(1) = 3
(3, 4) ⇒ 𝐿(3) = 4
𝐿(𝐺(1)) = 𝐿(2) = 2 ⇒ (1, 2) ∈ 𝐿 ∘ 𝐺
𝐿(𝐺(2)) = 𝐿(3) = 4 ⇒ (2, 4) ∈ 𝐿 ∘ 𝐺
𝐿(𝐺(3)) =? 𝐺(3) 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑. 𝑆𝑜, 𝐿(𝐺(3)) 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑.
𝐿(𝐺(4)) = 𝐿(3) = 4 ⇒ (4, 4) ∈ 𝐿 ∘ 𝐺
𝐿 ∘ 𝐺 = {(1, 2), (2, 4), (4, 4)}
Question 3 TWO
Question 1
𝑈 = {1, 2, 3, 4, 5, 𝑎, 𝑏, 𝑐} 𝑎𝑛𝑑 𝐴 = {𝑎, 𝑏, 𝑐, 1, 2, 3, 4}
𝑅: 𝐴 → 𝐴
𝐴 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑖𝑓 𝑒𝑣𝑒𝑟𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒
𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛.
1. {(1, 3), (𝑏, 3), (1,4), (𝑏, 2), (𝑐, 2)}
1 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 3 𝑎𝑛𝑑 1 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 4.
1 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
2. {(𝑎, 𝑐), (𝑏, 𝑐), (𝑐, 𝑏), (1, 3), (2, 3), (3, 𝑎)}
𝑎, 𝑏, 𝑐, 1, 2 𝑎𝑛𝑑 3 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
3. {(𝑎, 𝑎), (𝑐, 𝑐), (2, 2), (3, 3), (4, 4)}
𝑎, 𝑐, 2, 3 𝑎𝑛𝑑 4 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
4. {(𝑎, 𝑐), (𝑏, 𝑐), (1, 3), (3, 3)}
𝑎, 𝑏, 1 𝑎𝑛𝑑 3 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙.
Question 1 ONE
,Question 2
𝑈 = {1, 2, 3, 4, 5, 𝑎, 𝑏, 𝑐} 𝑎𝑛𝑑 𝐴 = {𝑎, 𝑏, 𝑐, 1, 2, 3, 4}
𝑅: 𝑈 → 𝐴 𝑈 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑎𝑛𝑑 𝐴 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛.
𝑇ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑈 𝑎𝑟𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒𝑠 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝐴 𝑎𝑟𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 𝑖𝑚𝑎𝑔𝑒𝑠.
𝐴 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑓 𝑖𝑡 𝑖𝑠 𝑎 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝑎𝑛𝑑 𝒇𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒄𝒐𝒅𝒐𝒎𝒂𝒊𝒏, 𝒕𝒉𝒆𝒓𝒆
𝒊𝒔 𝒂𝒏 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒐𝒎𝒂𝒊𝒏 𝒕𝒉𝒂𝒕 𝒊𝒔 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐 𝒊𝒕.
𝑇ℎ𝑎𝑡 𝑖𝑠, 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑦 ∈ 𝐴, 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑥 ∈ 𝑈 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 (𝑥, 𝑦) ∈ 𝑅.
𝐸𝑣𝑒𝑟𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴 𝑚𝑢𝑠𝑡 ℎ𝑎𝑣𝑒 𝑎 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒 𝑓𝑜𝑢𝑛𝑑 𝑖𝑛 𝑈.
1. {(1, 4), (2, 𝑏), (3, 3), (4, 3), (5, 𝑎), (𝑎, 𝑐), (𝑏, 1), (𝑐, 𝑏)}
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑈, 𝑡ℎ𝑎𝑡 𝑖𝑠, 1, 2, 3, 4, 5, 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴. 𝑆𝑜, 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝐴𝑙𝑠𝑜,
2 ∈ 𝐴 𝑏𝑢𝑡 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 (? , 2) 𝑝𝑎𝑖𝑟
2 ℎ𝑎𝑠 𝑛𝑜 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
2. {(𝑎, 1), (𝑏, 2), (𝑐, 𝑎), (1, 4), (2, 𝑏), (3, 3), (4, 𝑐)}
𝑇ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 5 𝑖𝑠 𝑖𝑛 𝑈, ℎ𝑜𝑤𝑒𝑣𝑒𝑟, 5 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑎𝑛𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
3. {(1, 𝑎), (2, 𝑐), (3, 𝑏), (4, 1), (𝑎, 𝑐), (𝑏, 2), (𝑐, 3)}
𝑇ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 5 𝑖𝑠 𝑖𝑛 𝑈, ℎ𝑜𝑤𝑒𝑣𝑒𝑟, 5 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑎𝑛𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
4. {(1, 𝑎), (2, 𝑏), (3, 4), (4, 3), (5, 𝑐), (𝑎, 𝑎), (𝑏, 1), (𝑐, 2)}
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑈, 𝑡ℎ𝑎𝑡 𝑖𝑠, 1, 2, 3, 4, 5, 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 𝑎𝑟𝑒 𝑒𝑎𝑐ℎ 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝐴. 𝑆𝑜, 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐴, 𝑡ℎ𝑎𝑡 𝑖𝑠, 1, 2, 3, 4, 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 ℎ𝑎𝑣𝑒 𝑝𝑟𝑒 − 𝑖𝑚𝑎𝑔𝑒𝑠 𝑖𝑛 𝑈.
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
Question 2 FOUR
, Question 3
𝐴 = {1, 2, 3, 4}
𝐺: 𝐴 → 𝐴 𝑎𝑛𝑑 𝐿: 𝐴 → 𝐴
𝐺 = {(1, 2), (2, 3), (4, 3)}
(1, 2) ⇒ 𝐺(1) = 2
(2, 3) ⇒ 𝐺(2) = 3
(4, 3) ⇒ 𝐺(4) = 3
𝐿 = {(2, 2), (1, 3), (3, 4)}
(2, 2) ⇒ 𝐿(2) = 2
(1, 3) ⇒ 𝐿(1) = 3
(3, 4) ⇒ 𝐿(3) = 4
𝐿(𝐺(1)) = 𝐿(2) = 2 ⇒ (1, 2) ∈ 𝐿 ∘ 𝐺
𝐿(𝐺(2)) = 𝐿(3) = 4 ⇒ (2, 4) ∈ 𝐿 ∘ 𝐺
𝐿(𝐺(3)) =? 𝐺(3) 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑. 𝑆𝑜, 𝐿(𝐺(3)) 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑.
𝐿(𝐺(4)) = 𝐿(3) = 4 ⇒ (4, 4) ∈ 𝐿 ∘ 𝐺
𝐿 ∘ 𝐺 = {(1, 2), (2, 4), (4, 4)}
Question 3 TWO
