第2單元 - 集合

(1) 

(a) 即求其解集合 (x-5)(x-4) = 0 , x = 4 or x =5 所以 A = { 4 , 5 }.

(b) m+n = 5 ; m , n 為自然數.  可能組合為: 

      m  = 1 , n = 4 

      m  = 2 , n = 3 

      m  = 3 , n = 2 

      m  = 4 , n = 1 所以, {1/4, 2/3, 3/2, 4/1 } 

(2) A = { 1 , {1}, {2} , {1,2} } , 則  ABE 是正確的 . 

(3) S = { {} , 1 , {1,2} , 3 } , 則 ABD 是正確的

(4) A = { f(x) | f(x) = 0 } , B = { g(x) | g(x) = 0 } , C = { h(x) | h(x) = 0 }   

(a) f(x)g(x) = 0 即 A∩B 

(b) g(x)/h(x) = 0 , g(x) = 0 , 但  h(x) ≠ 0 , 所以 B∩C' 

(c) f(x) = 0 且 g(x)/h(x) = 0 ,即 A∩(B∩C') 

(5) A = { x , y , z } , B = { x+1 , 2 , 3 } 

     若 A = B , 可能的情況 : 

      x = x + 1 , y = 2 , x = 3 (矛盾) 

      x = 2 , y = 3 , z = x+1 , z = 3 代回 { 2 , 3 , 3 } 

      x = 3 , y = 2 , z = x+1 , z = 4 代回 { 3 , 2 , 4 } 

      x = 2 , y = x + 1 , y = 3 , z = 3 同上

      x = 3 , y = x + 1 , y = 4 , z = 3 { 3 , 4 , 3 }

      故 共 { 2 , 3 , 3 } ,  { 3 , 2 , 4 } , { 3 , 4 , 3 } 

(6) n(A) = 7 , n(B) = 5  , 可能的情況 : 

      if B is super subset of A , B ⊂ A . then n(A∩B) = 5 , 

      if A∩B = {} , it means they have no same elements . n(A∩B) = 0 

(7) 在 1 與 1000 之間 , A 為7的倍數所成的集合 , B為2的倍數所成的集合 , C 為5的倍數所成的集合. 

n(A) = (994 - 7)/7+1 = 142 

n(B) = (1000 - 2)/2+1 = 500 

n(C) = (1000 - 5)/5+1 = 200 

(a) 142 

(b) n(A∪B∪C) '  = n(U) - n(A∪B∪C) 

      n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) +  n(A∩B∩C) = 142 + 500 + 200 - 71 - 100 - 28 + 14 = 657  

  n(A∩B) = (994 -14)/14 + 1 = 71 

  n(B∩C) = (1000 - 10 )/10 + 1 = 100 

  n(A∩C) = (980 - 35)/35 + 1 = 28 

  n(A∩B∩C) = (980 - 70)/70 + 1 = 14 

(c) n((B∪C)∩A') = n(B∪C)  -  n[(A∩C)∪(B∩A)] = n(B)+ n(C) - n(B∩C) -[ n(A∩C) + n (B∩A) - n(A∩B∩C) = 500 + 200 - 100 - (28+71-14) = 600 - 28 - 71 + 14 = 515 .  

(8) U = {1,2,3,4, ... , 10} , A = {1,3,5} , B= {2,3,5} . 

     (A∩B)' = A'∪B'  , A'  = U - A = { 2,4,6,7,8,9,10} , B'  = { 1,4,6,7,8,9,10} 

     A∩B = {3,5} 

     (A∩B)' = U - (A∩B) = {1,2,4,6,7,8,9,10} =  A' ∪B' 

     (A∪B)' = A'∩B' 

      A∪B = {1,2,3,5} 

     (A∪B)' = U - (A∪B) = {4,6,7,8,9,10} = A'∩B'

(9) S = { 1,2,3,4,5 } 

      假設 a,b,c,d,e,f 分別代表子集合個數為 0,1,2,3,4,5 的集合. 

      a = {{}} , n(a) = 1

      b = {{1},{2},{3},{4},{5}} , n(b) = 5  

      c = {{1,2}, {1,3},{1,4},{1,5},{2,3},{2,4},{2,5} ,{3,4},{3,5},{4,5} } , n(c) = 10 

      d = {{1,2,3},{2,3,4},{3,4,5},{1,2,4},{1,2,5},{2,3,5},{2,4,5},{1,4,5},{1,3,5},{2,4,5} }  , n(d) = 10  

      e = { {1,2,3,4} , {2,3,4,5} , {1,2,4,5}, {1,3,4,5} , {1,3,4,5} } , n(e) = 5  

      f = S , n(f) = 1 

(a) 1+5+10+10+5+1 = 32 = 2^5 

(b) 10 

(10)  P∩T = { 2 , 5 } 故 P, T 中必有兩個共同元素 2 , 5 . 但  P 中已經有一個元素為 2 ,則就將 5 代入 T 中任意個元素求其a值; 但是要帶入哪到哪一個式子? 這有些運氣存在嗎 ? , 假設我們從最簡單的式子帶入, 即 a^2 - 3 = 5 , a^2 = 8 ; 將 a 值帶回P中,但求出來非5,所以不合. 所以思考方向是錯的. 應該先由元素少的方向著手.即 a^2-2a-3 = 5 . 求出雙根 a = 4 , -2 再反代入其中, 能符合其解即可. 

(11) A'∩B' = {1,9,10} , A∩B = {3} , A'∩B = {2,5,8} . U = {1,2,3,4,5,6,7,8,9,10 } 

        A'∩B' = U - (A∪B) = {1,2,3,4,5,6,7,8,910} -  (A∪B) = {1,9,10} 

        (A∪B) = {2,3,4,5,6,7,8} ,  (A'∩B)∪(A∩B) = B = {2,3,5,8} 

        (A∪B) = A ∪ {2,3,5,8} = {2,3,4,5,6,7,8} 

        A = {2,3,4,5,6,7,8} - {2,3,5,8} = {4,7,8} As A∩B = {3} , A = {3,4,7,8}    

       

 (12) A∩B =   { x | x ∈ R , 0 < x < 3/4 }  

       (A∩B)' = { x | x ∈ R , x < 0 and  x > 3/4 }

       C' = { x |  x ∈ R , x < 0 and x > 1/2 } 

       ((A∩B)'∪C)'  = (A∩B)∩C = C 

       (A∩B)'∩(A'∪C)' = { x | x ∈ R , x < 0 and  x > 3/4 } ∩ { x | x ∈ R , x < 1/2 and  x > 3/4 } = { x | x ∈ R , x < 0 and  x > 3/4 }.      

           

 (13) A = { x |  x > 3 or x < -1 } , B = { x | |x-a| ≤ b } 

        A∪B = R ,  |x-a| ≤ b ,  - b ≤  x -a ≤ b , a - b ≤ x ≤ b+a

        A∩B = { x | 3 < x ≤ 4 } , b + a = 4 , a - b= -1