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1、离散数学习题参考答案第一章集合1分别用穷举法，描述法写出以下集合(1) 偶数集合(2) 36的正因子集合(3) 自然数中3的倍数(4) 大于1的正奇数(1) E=., 6,4,-2, 0, 2, 4, 6, .=2 i I ie I (2) D= 1, 2, 3, 4, 6, = xo | xl36 (3) N3= 3, 6, 9, 、 = 3n I nN (4) Ad= 3, 5, 7, 9, = 2n+1 | nN 2.确定以下结论正确与否(1) 66 X(2) 6 6 /(3) 6 g 6V(4)6 g 6 V(5) 6 a X(6) 6 g a V(7) (a,bG(a,b,c,(a,
2、b,cX(8) a,b ga,b,c,a,b,cV(9) a,bG(a,b,a,b X(10) (a,b c a,b, a,b/3.写出以下集合的幂集(1) a如 a(2 ) 6E(3) 如6E,(4) 如如a,e,a, a E ,ba, a,b ,a b , 6, a,如e 6 , 4 a, b a ,6, a, b ,(5)P(P(6)m 6,4. 对任意集合A,B,C,确定以下结论的正确与否(1) 假设A6B，且B g C,则 AGC V(2) 假设 AEB，且B g C,则A g C X(3) 假设A g B,且 BGC，则 AGC X(4) 假设A g B,且 BGC，则A g C
3、X5. 对任意集合A,B,C，证明(1) A - (B - C) = (A - B) U (A A C)左差 a n(B - C)差 a n(B n c)D a n( b u c)分配( a n b)u( a n c)=右(2) A - (B n C) = (A - B) U (A - C)1) 左差 A - (B - C)(1)的结论(A - B) U (A A C)差=(A - B) U (A - C)=右2) 左差 a n(B n C)D a n( b u c)分配 (A n B) U (A n C)差(A - B) U (A - C)=右(3) A - (B U C) = (A - B
4、) n (A - C)左差 a n(B u C)D a n( b n c)幂等(a n a)n( b n c)结合交换(a n b)n( a n c)=右(4) (A + B) B = A U B左差(A n B) B 对称差(A n B) U B) - (A A B) A B)分配, 结合(A u b)n(百 u b)一( a n( b)n b)(a u B)n u)-( a n 饥S=(A u B) - e = (A U B)=右(5) (A - B) - C = A - (B U C)者姑合 -左差(a n b)n c 结合 a n( b n c)D A A (B U C)差 A - (
5、B U C)(6) (A - B) - C = (A - C) - B左差(a n b)n c 结合 a n( b n c)交换 a n( c n b)结合(a n c)n b差.=(A - C) - B =右(7) (A - B) - C = (A - C) - (B - C).(5), 差右马 A - (C U (B - C)至 A - (C U (B n C)分配 a -C u bn c u c互补 a -C u bn u零一 A - (C U B)交换 A - (B U C)(5)4=(A - B) - C =左6. 问在什么条件下，集合A,B,C满足以下等式 a nB u C= a
6、n bu c左=a n Bu a n Cc 右假设要右c左，须c g a n b u c,C c A时等式成也2A - B = A左c右是显然的A G A - B = A n 瓦 A G B, a n b=e时等式成立。(3) A - B = BAB = B,B c B ,B = e,代入原式徽-e=e,.a=b=e时等式成立。(4) A - B = B - Aa n b=b n a ,只能=eA - B = 0, A c B,B - A = 0,B c A,A = B时等式成立。(5) A B = AB =。，假设B 壬。盘 b e B,当b e A, b史A B =人矛盾;当b右A,b g
7、 A B =人矛盾。(6)A U B = A A B右G左是显然的A U B c A n B, vA * n B, A c B 认=B、B G A n B,B G A 尸A = B时等式成立。(7) (A - B) U (A - C) = A左=(a nB) u( a n。=a n(页 uc)=a n(B nc)=a -( b n c)=aa n b n c=o时等式成立。(8) ( A - B) U (A - C) = 8左=(An0)u( AnC)=a n(BUc)=An(BC) = A - (B A C) = eA j (B A C),A c B, A c C时等式成立。(9) ( A
8、- B) A (A - C) = 8左=(a n b)n( a n C)=a n(B n C)=A A (B U C) = A - (B U C) = eA c (B U C)时等式成立。(10) ( A - B)(A - C) = 8(A - B) U (A - C) - (A - B) A (A - C)=。(A - B) U (A - C) = (A - B) A (A - C)由(6)知, (A - B) = (A - C), A - B = A - C,A n B = A n C时等式成立。(11) A U (B - A) = BA U (B n A) = (A U B) n (A
9、U A)=(A U B) n U = (A U B) = BA c B时等式成立。7. 设A=a,b,a,b,，求以下各式(1) en。 =e(2) 。白。=。(3) 。,。= 。，。(4) 。，。 。= 。(5) 。，。 。 = 。(6 )Aa,b =a, b,。(7)A。= A(8)A。 = a, b,a,b) (9 )6 A =。(10) 0A =。8. 在以下条件下，一定有B = C吗？(1) A U B = A U C否，例：A=1,2,3, B= 4, C= 3, 4),A U B = A U C = 1,2,3,4,而 B 壬 C。(2)anb = anc否，例:A= 1,2,
10、3, B= 2, 3,C=2, 3, 4A nB = A n C = 2,3,而 B 主 C。(3) AB = AC对,假设B丰C,不妨,3a g B, a宅C,假设a g A, a e A U B, a g A A B, a 史 A B,a e A U C, a 史 A A C, a e A C;假设a 成 A, a e A U B, a 史 A A B, a g A B,a 史 A U C, a 关 A A C, a 史 A C矛盾。(4) a n b = a n c 且 a n b=a n cVb g B, 假设b e A, b e A A B = A A C, b g C,假设b 金
11、A, b 金 A A B = A A C, b e C, B c C,同理,C g B，二 B = C。9. (1) (A U B) n (B U C) c A n B证:Va e 左，a 宅(B U C), a 电 B, a e B;a g (A U B),而 a w B, a e A, a g A A B。(2) 假设A g (B U C)且 B c (A U C),则 B 主奴假设B。奴 Ba g B c (A U C) = (A A C),a e A c (B U C),、a 宅 C, a e B 即 a w B,矛盾 10 .化简(A U B U C) n (A U B) - (A
12、U (B U C) A A)=(A U B) - A = (A U B )HA=(An A )U (Bn&) = 8 U (B - A) = B - A11. 设 A= 2, 3, 4, B=1, 2, C=4, 5, 6,求(1) A B = 1,3,4(2) A B C = 1,3,5,6(3) (A B) (B C) = 2,3,5,612. 设 A=1,2, 3, 4, B= 1, 2, 5,求(1) P(A) A P(B) = 如1, 2, 1, 2(2) P( A ) U P( B )=如1, 2, 3, 4, 1,2 ,1, 3 ,1, 4,2,3, 2, 4, 3,4,1,2,3,1,2,4, 1,3,4,2,3,4 ,1, 2,3,4, 5 ,1,5,2,5 ,1, 2 (3) P( A ) - P( B )=3,4,1,3, 1,4,2,3,2,4,3,4,1,2, 3,1,2,4 , 1,3,4,2, 3,4, 1, 2,3,4 (4) P(A) P(B)=3 ,4, 1, 3, 1,4 , 2, 3 , 2,4 ,3, 4, 1, 2时1,2, 4, 1,3,4,2,3, 4, 1,2,3, 4 , 5, 1, 5, 2,5, 1,2,5
