Complete Solution Manual First Course in Abstract Algebra Rings Groups and Fields 3rd Edition Anderson
Solution Manual for First Course in Abstract Algebra Rings Groups and Fields 3rd Edition Anderson / All Chapters Full Complete 2023
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Solution Manual for First Course in Abstract Algebra A, 8th
Edition by John B. Fraleigh, Verified Chapters 1 - 56, Complete
Newest Version
set - ANSWER: a well defined collection of objects
elements/members - ANSWER: the objects that belong to a set
subset - ANSWER: A is a subset of B if every element of A is also an element of B.
proper subset - ANSWER: A is a proper subset of B if it is a subset of B but not equal
to B
equal sets - ANSWER: If A is a subset of B and B is a subset of A
empty set - ANSWER: a set with no elements
union of sets - ANSWER: the set of elements which are in A or in B
intersection of sets - ANSWER: the set of elements which are in both A and B
disjoint - ANSWER: when two sets have no elements in common
universal set - ANSWER: a fixed set U
complement - ANSWER: For any A a subset of the universal set, the complement of A
is the set of all elements of U which are not in A.
difference of two sets - ANSWER: A\B is the set of all elements of A which are not in
B
DeMorgan's Laws - ANSWER: Let A and B be sets, then the complement of their
union is the intersection of their complements, and the complement of their
intersection is the union of their complements.
Cartesian product - ANSWER: AxB={(a, b) | a in A, b in B}
relations - ANSWER: subsets of AxB
mapping/function - ANSWER: f is a subset of AxB from a set A to a set B is the special
relation where (a,b) is in f if for every element of A there is a unique b from B.
domain of f - ANSWER: A, given f a subset of AxB
range/image of f - ANSWER: f(A)={f(a) | a in A}, a subset of B
, well defined relation - ANSWER: if each element of the domain is assigned to a
unique element of the range
onto/surjective - ANSWER: If f:A to B is a map and the image of f is B, i.e., f(A)=B
one-to-one/injective - ANSWER: If a₁≠a₂ implies f(a₁)≠f(a₂)
bijective map - ANSWER: both one-to-one and onto
composition of f and g - ANSWER: Let f:A→B, g:B→C be mappings. (f₀g)(x)=(f(g(x)).
linear maps/transformations - ANSWER: maps from Rⁿ to Rⁿ' given by matrices
permutation - ANSWER: For any set S, a one-to-one, onto mapping π:S→S.
Let f:A→B, g:B→C, h:C→D.
1. The composition of mappings is:
2. If f and g are both one-to-one, then:
3. If f and g are both onto, then:
4. If f and g are both bijections, then: - ANSWER: 1. associative: (h₀g)₀f=h₀(g₀f).
2. the mapping g₀f is one-to-one.
3. the mapping g₀f is onto.
4. the mapping g₀f is a bijection.
identity mapping - ANSWER: If S is any set, id(s)=s for all s in S.
inverse mapping - ANSWER: A map g:B→A is an inverse of f:A→B if g₀f=id(A) and
f₀g=id(B)
invertible - ANSWER: a map that has an inverse, f⁻¹
A mapping is invertible iff - ANSWER: it is both one-to-one and onto.
equivalence relation on a set X - ANSWER: A relation R⊂X×X such that:
1. (reflexive) (x,x)∈R for all x in X.
2. (symmetric) (x,y)∈R implies (y,x)∈R.
3. (transitive) (x,y)∈R and (y,z)∈R implies (x,z)∈R.
similar matrices - ANSWER: A~B if there exists invertible P such that PAP⁻¹=B
partition P of a set X - ANSWER: A collection of nonempty sets X₁,X₂,... such that
Xi∩Xj=∅ for i≠j and ∪Xₙ=X
equivalence class of x - ANSWER: [x]={y in X | y~x}
1. Given an equivalence relation ~
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