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PcumP_Ravenclaw
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Alan F beardon, Algebra and Geometry
chapter 1
6. For any two sets A and B the symmetric difference AΔB of A and B is the
set of elements in exactly one of A and B; thus
AΔB = {x ∈ A ∪ B : x / ∈ A ∩ B} = (A ∪ B)\(A ∩ B)
. Let be a non-empty set and let G be the set of subsets
of (note that G includes both the empty set ∅ and ). Show that G with
the operation Δ is a group with ∅ as the identity element of G. What is
A−1? Now let = {1, 2, . . . , 7}, A = {1, 2, 3}, B = {3, 4, 5} and
C = {5, 6, 7}. By considering A−1 (inverse) and B−1 (inverse), solve the two equations for x
AΔX = B, and AΔXΔB = C.
My answer:
condition to get null set is
A\A = 0
make (A U B) same as (A n B) by changing B
so A^-1 (A inverse) must be A itself so that (A u A) \ (A n A) = Null set
Doubts : Are the two X the same or different and why do they want to consider A-1 and B-1. I have solved without considering them.
for AΔX = B
X = {1 2 4 5}
and for AΔXΔB = C
X = {1, 2, 4, 6, 7}
x is calculated backwards by AΔX = Y
so Y Δ B = C
Y = {3,4,6,7}
AΔX = Y
{1,2,3}ΔX = {3,4,6,7}
so X= {1,2,4,6,7}
chapter 1
6. For any two sets A and B the symmetric difference AΔB of A and B is the
set of elements in exactly one of A and B; thus
AΔB = {x ∈ A ∪ B : x / ∈ A ∩ B} = (A ∪ B)\(A ∩ B)
. Let be a non-empty set and let G be the set of subsets
of (note that G includes both the empty set ∅ and ). Show that G with
the operation Δ is a group with ∅ as the identity element of G. What is
A−1? Now let = {1, 2, . . . , 7}, A = {1, 2, 3}, B = {3, 4, 5} and
C = {5, 6, 7}. By considering A−1 (inverse) and B−1 (inverse), solve the two equations for x
AΔX = B, and AΔXΔB = C.
My answer:
condition to get null set is
A\A = 0
make (A U B) same as (A n B) by changing B
so A^-1 (A inverse) must be A itself so that (A u A) \ (A n A) = Null set
Doubts : Are the two X the same or different and why do they want to consider A-1 and B-1. I have solved without considering them.
for AΔX = B
X = {1 2 4 5}
and for AΔXΔB = C
X = {1, 2, 4, 6, 7}
x is calculated backwards by AΔX = Y
so Y Δ B = C
Y = {3,4,6,7}
AΔX = Y
{1,2,3}ΔX = {3,4,6,7}
so X= {1,2,4,6,7}
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