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For all sets A and B, (A – B) ∪ (A ∩ B) = A
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L . H . S . = ( A − B ) ∪ ( A ∩ B ) = [ ( A − B ) ∪ A ] ∩ [ ( A − B ) ∪ B ] = A ∩ ( A ∪ B ) = A = R . H . S . H e n c e , t h e g i v e n s t a t e m e n t i s ' T r u e ' .
For all sets A, B and C, show that (A – B) ∩ (C – B) = A –(B ∪ C) .
T o p r o v e ( A − B ) ∩ ( A − C ) = A − ( B ∪ C ) L . H . S . L e t x ∈ ( A − B ) ∩ ( A − C ) ⇒ x ∈ ( A − B ) a n d x ∈ ( A − C ) ⇒ ( x ∈ A a n d x ∉ B ) a n d ( x ∈ A a n d x ∉ C ) ⇒ x ∈ A a n d ( x ∉ B a n d x ∉ C ) ⇒ x ∈ A − ( B ∪ C ) S o , ( A − B ) ∩ ( A − C ) ⊂ A − ( B ∪ C ) … ( i ) R . H . S . L e t y ∈ A − ( B ∪ C ) ⇒ y ∈ A a n d y ∉ ( B ∪ C ) ⇒ y ∈ A a n d ( y ∉ B a n d y ∉ C ) ⇒ ( y ∈ A a n d y ∉ B ) a n d ( y ∈ A a n d y ∉ C ) ⇒ y ∈ ( A − B ) a n d y ∈ ( A − C ) S o , A − ( B ∪ C ) ⊂ ( A − B ) ∩ ( A − C ) … ( i i ) F r o m e q . ( i ) a n d ( i i ) , w e g e t A − ( B ∪ C ) = ( A − B ) ∩ ( A − C )
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Let U be the set of all boys and girls in a school, G be the set of all girls in the school, B be the set of all boys in the school, and S be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationships among sets U, G, B and S.
A, B and C are subsets of Universal Set U. If A = {2, 4, 6, 8, 12, 20}
B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.
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If Y = {1, 2, 3, . 10}, and a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.
(i) a ∈ Y but a2 ∉ Y
(ii) a + 1 = 6, a ∈ Y
(iii) a is less than 6 and a ∈ Y
G i v e n t h a t : Y = { 1 , 2 , 3 , … , 1 0 } ( i ) { a ∈ Y b u t a 2 ∉ Y } = { 4 , 5 , 6 , 7 , 8 , 9 , 1 0 } ( i i ) { a + 1 = 6 , a ∈ Y } = { 5 } ( i i i ) { a < 6 a n d a ∈ Y } = { 1 , 2 , 3 , 4 , 5 }
If X = {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by
(i) 4n
(ii) n + 6
(iii) n2
(iv) n – 1
W e a r e g i v e n t h a t : X = { 1 , 2 , 3 } ( i ) { 4 n | n ∈ X } = { 4 , 8 , 1 2 } ( i i ) { n + 6 | n ∈ X } = { 7 , 8 , 9 } ( i i i ) { n 2 | n ∈ X } = { 1 2 , 1 , 3 2 } ( i v ) { ( n − 1 ) | n ∈ X } = { 0 , 1 , 2 }
Given that N = {1, 2, 3, ., 100}. Then write
(i) The subset of N whose elements are even numbers.
(ii) The subset of N whose element are perfect square numbers.
W e a r e g i v e n t h a t : N = { 1 , 2 , 3 , 4 , 5 , … , 1 0 0 } (i)Required subset whose elements are even = { 2 , 4 , 6 , 8 , … , 1 0 0 } (ii)Required subset whose elements are perfect squares = { 1 , 4 , 9 , 1 6 , 2 5 , 3 6 , … , 1 0 0 }
If A and B are subsets of the universal set U, then show that
(i) A ⊂ A ∪ B
(ii) A ⊂ B ⇔ A ∪ B = B
(iii) (A ∩ B) ⊂ A
( i ) G i v e n t h a t : A ⊂ U a n d B ⊂ U L e t x ∈ A o r x ∈ B ⇒ x ∈ A ∪ B H e n c e , A ⊂ ( A ∪ B ) ( i i ) I f A ⊂ B T h e n l e t x ∈ A ∪ B ⇒ x ∈ A o r x ∈ B ⇒ A ∪ B ⊂ B … ( i ) B u t B ⊂ A ∪ B … ( i i ) F r o m e q . ( i ) a n d ( i i ) , w e g e t A ∪ B = B L e t y ∈ A ⇒ y ∈ ( A ∪ B ) ⇒ y ∈ B ⇒ y ∈ B ⇔ A ∪ B = B ( i i i ) L e t x ∈ A ∩ B ⇒ x ∈ A a n d x ∈ B ⇒ x ∈ A S o , A ∩ B ⊂ A
Given L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}
Verify that L – (M ∪ N) = (L – M) ∩ (L – N)
G i v e n , L = { 1 , 2 , 3 , 4 } , M = { 3 , 4 , 5 , 6 } a n d N = { 1 , 3 , 5 } ∴ M ∪ N = { 1 , 3 , 4 , 5 , 6 } L − M ∪ N = { 2 } N o w , L − M = { 1 , 2 } , L − N = { 2 , 4 } ∴ ( L − M ) ∩ ( L − N ) = { 2 } H e n c e , L − M ∪ N = ( L − M ) ∩ ( L − N ) .
State which of the following statements a true and which are false. Justify your answer.
(i) 35 ∈ {x | x has exactly four positive factors}.
(ii) 128 ∈ {y | the sum of all the positive factors of y is 2y}
(iii) 3 ∉ {x | x4 – 5x3 + 2x2 – 112x + 6 = 0}
(iv) 496 ∉ {y | the sum of all the positive factors of y is 2y}.
( i ) G i v e n t h a t : 3 5 ∈ { x | x h a s e x a c t l y f o u r p o s i t i v e f a c t o r } ∴ F a c t o r s o f 3 5 a r e 1 , 5 , 7 , 3 5 H e n c e , t h e s t a t e m e n t ( i ) i s ' T r u e ' . ( i i ) G i v e n t h a t : 1 2 8 ∈ { y | t h e s u m o f a l l p o s i t i v e f a c t o r s o f y i s 2 y } ∴ F a c t o r s o f 1 2 8 a r e 1 , 2 , 4 , 8 , 1 6 , 3 2 , 6 4 a n d 1 2 8 . S u m o f a l l f a c t o r s = 1 + 2 + 4 + 8 + 1 6 + 3 2 + 6 4 + 1 2 8 = 2 5 5 ≠ 2 * 1 2 8 H e n c e , t h e s t a t e m e n t ( i i ) i s ' F a l s e ' . ( i i i ) G i v e n t h a t : 3 ∈ { x | x 4 − 5 x 3 + 2 x 2 − 1 1 2 x + 6 = 0 } ∴ x 4 − 5 x 3 + 2 x 2 − 1 1 2 x + 6 = 0 N o w f o r x = 3 , w e h a v e ( 3 ) 4 − 5 ( 3 ) 3 + 2 ( 3 ) 2 − 1 1 2 ( 3 ) + 6 ⇒ 8 1 − 1 3 5 + 1 8 − 3 3 6 + 6 ⇒ − 3 6 6 ≠ 0 H e n c e , t h e s t a t e m e n t ( i i i ) i s ' T r u e ' . ( i v ) G i v e n t h a t : 4 9 6 ∉ { y | t h e s u m o f a l l p o s i t i v e f a c t o r s o f y i s 2 y } ∴ T h e p o s i t i v e f a c t o r s o f 4 9 6 a r e 1 , 2 , 4 , 8 , 1 6 , 3 1 , 6 2 , 1 2 4 , 2 4 8 a n d 4 9 6 . ∴ T h e s u m o f a l l p o s i t i v e f a c t o r s = 1 + 2 + 4 + 8 + 1 6 + 3 1 + 6 2 + 1 2 4 + 2 4 8 + 4 9 6 = 9 9 2 = 2 * 4 9 6 H e n c e , t h e s t a t e m e n t ( i v ) i s ' F a l s e ' .
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