# HW1 - MA1100 HW1 Solutions 1.24 1.28 1.50 1.54 1.61 2.59...

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1 MA1100 HW1 Solutions 1.24, 1.28, 1.50, 1.54, 1.61, 2.59, 2.62, 2.63, 2.68 (Grader will mark 1.24, 1.28, 1.54, 1.61, 2.62(a), 2.63) 1.24 (a) The sets ׎ , { ׎ } are elements of A. (b) |A| = 3. (c) All of ׎ , { ׎ }, { ׎ , { ׎ }} are subsets of A. (d) ׎ A = ׎ . (e) { ׎ } A = { ׎ }. (f) { ׎ , { ׎ }} ׎ , { ׎ }}. (g) ׎׫ A = A. (h) { ׎ } ׫ ) ׎ ׎ }} ׫ A=A (i) { , { }} A = A. Note: Students are only required to write down the answer to each part. Explanation is not required.

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2 1.28 » α א A S α = S 1 ׫ S 3 ׫ S 4 = [0, 3] ׫ [2, 5] ׫ [3, 6] = [0, 6]. α א A S α = S 1 S 3 S 4 = [0, 3] [2, 5] [3, 6] = {3}. Note: Students are only required to write down the answer to each part. Explanation is not required. 1.50 (a) |A| = 6 (direct counting) (b) |B| = 0 (B is empty since absolute value can’t be negative) negative) (c) |C| = 3 (C = {3,4,5} (d) |D| = 0 (D is empty since natural numbers are positive) (e) |E| = 10 (E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (f) |F| = 20 (F = { 1, 2, 3 , 4, 5, 6, 7, 8, 9, 10} 1.54 P(A) = { ׎ , {1}}, P(C) = { ׎ , {1}, {2},C}. So B can only be { ׎ , {1}, {2}}. Note: Students are required to list P(A) and P(C) to verify the answer for B.
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HW1 - MA1100 HW1 Solutions 1.24 1.28 1.50 1.54 1.61 2.59...

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