Algebraic-Structures with one operation as well as two operations and its variations like semigroup, groups etc

Algebraic Structures
•Algebraic systems
•Semi groups
•Monoids
•Groups
•Sub groups
•Homomorphism
•Isomorphism
•Congruence Relation
•Ring
• Integral Domains and Fields
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Algebraic systems
• N = {1,2,3,4,….. } = Set of all natural numbers.
Z = { 0,  1,  2,  3,  4 , ….. } = Set of all integers.
Q = Set of all rational numbers.
R = Set of all real numbers.
• Binary Operation: The binary operator * is said to be a binary operation
(closed operation) on a non empty set A, if
a * b  A for all a, b  A (Closure property).
Ex: The set N is closed with respect to addition and multiplication
but not w.r.t subtraction and division.
• Algebraic System: A set ‘A’ with one or more binary(closed) operations
defined on it is called an algebraic system.
Ex: (N, + ), (Z, +, – ), (R, +, . , – ) are algebraic systems.
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Properties
• Associativity: Let * be a binary operation on a set A.
The operation * is said to be associative in A if
(a * b) * c = a *( b * c) for all a, b, c in A
• Identity: For an algebraic system (A, *), an element ‘e’ in A is said to be
an identity element of A if
a * e = e * a = a for all a  A.
• Note: For an algebraic system (A, *), the identity element, if exists, is
unique.
• Inverse: Let (A, *) be an algebraic system with identity ‘e’. Let a be an
element in A. An element b is said to be inverse of A if
a * b = b * a = e
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Semi groups
• Semi Group: An algebraic system (A, *) is said to be a semi group if
1. * is closed operation on A.
2. * is an associative operation, for all a, b, c in A.
• Ex. (N, +) is a semi group.
• Ex. (N, .) is a semi group.
• Ex. (N, – ) is not a semi group.
• Monoid: An algebraic system (A, *) is said to be a monoid if the
following conditions are satisfied.
1) * is a closed operation in A.
2) * is an associative operation in A.
3) There is an identity in A.
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Monoids
• Ex. Show that the set ‘N’ is a monoid with respect to multiplication.
• Solution: Here, N = {1,2,3,4,……}
1. Closure property : We know that product of two natural numbers is
again a natural number.
i.e., a.b = b.a for all a,b  N
 Multiplication is a closed operation.
2. Associativity : Multiplication of natural numbers is associative.
i.e., (a.b).c = a.(b.c) for all a,b,c  N
3. Identity : We have, 1  N such that
a.1 = 1.a = a for all a  N.
 Identity element exists, and 1 is the identity element.
Hence, N is a monoid with respect to multiplication.
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Groups
• Group: An algebraic system (G, *) is said to be a group if the following
conditions are satisfied.
1) * is a closed operation.
2) * is an associative operation.
3) There is an identity in G.
4) Every element in G has inverse in G.
• Abelian group (Commutative group): A group (G, *) is
said to be abelian (or commutative) if
a * b = b * a a, b  G.
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Algebraic systems
Groups
Abelian groups
Monoids
Semi groups
Algebraic systems
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Properties
• In a Group (G, * ) the following properties hold good
1. Identity element is unique.
2. Inverse of an element is unique.
3. Cancellation laws hold good
a * b = a * c  b = c (left cancellation law)
a * c = b * c  a = b (Right cancellation law)
4. (a * b) -1
= b-1
* a-1
• In a group, the identity element is its own inverse.
• Order of a group : The number of elements in a group is called order of
the group.
• Finite group: If the order of a group G is finite, then G is called a finite
group.
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Ex. Show that, the set of all integers is an abelian group with
respect to addition.
• Solution: Let Z = set of all integers.
Let a, b, c are any three elements of Z.
1. Closure property : We know that, Sum of two integers is again an
integer.
i.e., a + b  Z for all a,b  Z
2. Associativity: We know that addition of integers is associative.
i.e., (a+b)+c = a+(b+c) for all a,b,c  Z.
3. Identity : We have 0  Z and a + 0 = a for all a  Z .
 Identity element exists, and ‘0’ is the identity element.
4. Inverse: To each a  Z , we have – a  Z such that
a + ( – a ) = 0
Each element in Z has an inverse.
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Contd.,
• 5. Commutativity: We know that addition of integers is commutative.
i.e., a + b = b +a for all a,b  Z.
Hence, ( Z , + ) is an abelian group.
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Ex. Show that set of all non zero real numbers is a group with respect to
multiplication .
• Solution: Let R*
= set of all non zero real numbers.
Let a, b, c are any three elements of R*
.
1. Closure property : We know that, product of two nonzero real numbers
is again a nonzero real number .
i.e., a . b  R*
for all a,b  R*
.
2. Associativity: We know that multiplication of real numbers is
associative.
i.e., (a.b).c = a.(b.c) for all a,b,c  R*
.
3. Identity : We have 1  R*
and a .1 = a for all a  R*
.
 Identity element exists, and ‘1’ is the identity element.
4. Inverse: To each a  R*
, we have 1/a  R*
such that
a .(1/a) = 1 i.e., Each element in R*
has an inverse.
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Contd.,
• 5.Commutativity: We know that multiplication of real numbers is
commutative.
i.e., a . b = b . a for all a,b  R*
.
Hence, ( R*
, . ) is an abelian group.
• Note: Show that set of all real numbers ‘R’ is not a group with respect to
multiplication.
• Solution: We have 0  R .
The multiplicative inverse of 0 does not exist.
Hence. R is not a group.
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Example.
• Ex. Show that set of all non zero rational numbers is a group with
respect to multiplication
• Home work
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Example
• Ex. Let (Z, *) be an algebraic structure, where Z is the set of integers
and the operation * is defined by n * m = maximum of (n, m).
Show that (Z, *) is a semi group.
Is (Z, *) a monoid ?. Justify your answer.
• Solution: Let a , b and c are any three integers.
Closure property: Now, a * b = maximum of (a, b)  Z for all a,b  Z
Associativity : (a * b) * c = maximum of {a,b,c} = a * (b * c)
 (Z, *) is a semi group.
Identity : There is no integer x such that
a * x = maximum of (a, x) = a for all a  Z
 Identity element does not exist. Hence, (Z, *) is not a monoid.
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Example
• Ex. Show that the set of all strings ‘S’ is a monoid under the operation
‘concatenation of strings’.
Is S a group w.r.t the above operation? Justify your answer.
• Solution: Let us denote the operation
‘concatenation of strings’ by + .
Let s1, s2, s3 are three arbitrary strings in S.
Closure property: Concatenation of two strings is again a string.
i.e., s1+s2  S
Associativity: Concatenation of strings is associative.
(s1+ s2 ) + s3 = s1+ (s2 + s3 )
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Contd.,
• Identity: We have null string ,   S such that s1 +  = S.
•  S is a monoid.
• Note: S is not a group, because the inverse of a non empty string does
not exist under concatenation of strings.
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Example
• Ex. Let S be a finite set, and let F(S) be the collection of all functions f: S
 S under the operation of composition of functions, then show that
F(S) is a monoid.
Is S a group w.r.t the above operation? Justify your answer.
• Solution:
Let f1, f2, f3 are three arbitrary functions on S.
Closure property: Composition of two functions on S is again a function
on S.
i.e., f1o f2  F(S)
Associativity: Composition of functions is associative.
i.e., (f1 o f2 ) o f3 = f1 o (f2 o f3 )
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Contd.,
• Identity: We have identity function I : SS
such that f1 o I = f1.
 F(S) is a monoid.
• Note: F(S) is not a group, because the inverse of a non bijective function
on S does not exist.
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Ex. If M is set of all non singular matrices of order ‘n x n’.
then show that M is a group w.r.t. matrix multiplication.
Is (M, *) an abelian group?. Justify your answer.
• Solution: Let A,B,C  M.
1.Closure property : Product of two non singular matrices is again a non
singular matrix, because
AB = A . B  0 ( Since, A and B are nonsingular)
i.e., AB  M for all A,B  M .
2. Associativity: Marix multiplication is associative.
i.e., (AB)C = A(BC) for all A,B,C  M .
3. Identity : We have In  M and A In = A for all A  M .
 Identity element exists, and ‘In’ is the identity element.
4. Inverse: To each A  M, we have A-1
 M such that
A A-1
= In i.e., Each element in M has an inverse.
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Contd.,
•  M is a group w.r.t. matrix multiplication.
We know that, matrix multiplication is not commutative.
Hence, M is not an abelian group.
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Ex. Show that the set of all positive rational numbers forms an abelian
group under the composition * defined by
a * b = (ab)/2 .
• Solution: Let A = set of all positive rational numbers.
Let a,b,c be any three elements of A.
1. Closure property: We know that, Product of two positive rational
numbers is again a rational number.
i.e., a *b  A for all a,b  A .
2. Associativity: (a*b)*c = (ab/2) * c = (abc) / 4
a*(b*c) = a * (bc/2) = (abc) / 4
3. Identity : Let e be the identity element.
We have a*e = (a e)/2 …(1) , By the definition of *
again, a*e = a …..(2) , Since e is the identity.
From (1)and (2), (a e)/2 = a  e = 2 and 2  A .
 Identity element exists, and ‘2’ is the identity element in A.
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Contd.,
• 4. Inverse: Let a  A
let us suppose b is inverse of a.
Now, a * b = (a b)/2 ….(1) (By definition of inverse.)
Again, a * b = e = 2 …..(2) (By definition of inverse)
From (1) and (2), it follows that
(a b)/2 = 2
 b = (4 / a)  A
 (A ,*) is a group.
• Commutativity: a * b = (ab/2) = (ba/2) = b * a
• Hence, (A,*) is an abelian group.
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Example
• Ex. Let R be the set of all real numbers and * is a binary operation
defined by a * b = a + b + a b. Show that (R, *) is a monoid.
Is (R, *) a group?. Justify your answer.
• Try for yourself.
identity = 0
inverse of a = – a / (a+1)
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Finite groups
Ex. Show that G = {1, -1} is an abelian group under multiplication.
Solution: The composition table of G is
.
. 1 – 1
• 1 1 – 1
• – 1 – 1 1
1. Closure property: Since all the entries of the composition table are the
elements of the given set, the set G is closed under multiplication.
2. Associativity: The elements of G are real numbers, and we know that
multiplication of real numbers is associative.
3. Identity : Here, 1 is the identity element and 1 G.
4. Inverse: From the composition table, we see that the inverse elements of
1 and – 1 are 1 and – 1 respectively.
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Contd.,
Hence, G is a group w.r.t multiplication.
5. Commutativity: The corresponding rows and columns of the table are
identical. Therefore the binary operation . is commutative.
Hence, G is an abelian group w.r.t. multiplication..
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Ex. Show that G = {1, , 2
} is an abelian group under multiplication.
Where 1, , 2
are cube roots of unity.
• Solution: The composition table of G is
•
• . 1  2
•
• 1 1  2
•   2
1
• 2
2
1 
1. Closure property: Since all the entries of the composition table are the elements of
the given set, the set G is closed under multiplication.
2. Associativity: The elements of G are complex numbers, and we know that
multiplication of complex numbers is associative.
4. Inverse: From the composition table, we see that the inverse elements of
1 , 2
are 1, 2
,  respectively.
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Contd.,
• Hence, G is a group w.r.t multiplication.
• 5. Commutativity: The corresponding rows and columns of the table are
identical. Therefore the binary operation . is commutative.
• Hence, G is an abelian group w.r.t. multiplication.
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Ex. Show that G = {1, –1, i, –i } is an abelian group under multiplication.
•
• . 1 –1 i -i
• 1 1 -1 i - i
• -1 -1 1 -i i
• i i -i -1 1
• -i -i i 1 -1
1. Closure property: Since all the entries of the composition table are the
elements of the given set, the set G is closed under multiplication.
2. Associativity: The elements of G are complex numbers, and we know
that multiplication of complex numbers is associative.
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Contd.,
• 4. Inverse: From the composition table, (wherever the identity element
1 is present, the corresponding column element is the inverse of the
row element) we see that the inverse elements of
1 -1, i, -i are 1, -1, -i, i respectively.
identical. Therefore the binary operation . is commutative. Hence, (G, .)
is an abelian group.
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Modulo systems.
• Addition modulo m ( +m )
let m is a positive integer. For any two positive integers a and b
a +m b = a + b if a + b < m
a +m b = r if a + b  m where r is the remainder obtained
by dividing (a+b) with m.
• Multiplication modulo p ( p )
let p is a positive integer. For any two positive integers a and b
a p b = a b if a b < p
a p b = r if a b  p where r is the remainder obtained
by dividing (ab) with p.
• Ex. 3 5 4 = 2 , 5 5 4 = 0 , 2 5 2 = 4
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Ex.The set G = {0,1,2,3,4,5} is a group with respect to addition modulo 6.
•
• +6 0 1 2 3 4 5
• 0 0 1 2 3 4 5
• 1 1 2 3 4 5 0
• 2 2 3 4 5 0 1
• 3 3 4 5 0 1 2
• 4 4 5 0 1 2 3
• 5 5 0 1 2 3 4
• 1. Closure property: Since all the entries of the composition table are
the elements of the given set, the set G is closed under +6 .
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Contd.,
• 2. Associativity: The binary operation +6 is associative in G.
for ex. (2 +6 3) +6 4 = 5 +6 4 = 3 and
2 +6 ( 3 +6 4 ) = 2 +6 1 = 3
• 3. Identity : Here, The first row of the table coincides with the top row.
The element heading that row , i.e., 0 is the identity element.
• 4. . Inverse: From the composition table, we see that the inverse
elements of 0, 1, 2, 3, 4. 5 are 0, 5, 4, 3, 2, 1 respectively.
identical. Therefore the binary operation +6 is commutative.
• Hence, (G, +6 ) is an abelian group.
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Ex.The set G = {1,2,3,4,5,6} is a group with respect to multiplication
modulo 7.
•
• 7 1 2 3 4 5 6
• 1 1 2 3 4 5 6
• 2 2 4 6 1 3 5
• 3 3 6 2 5 1 4
• 4 4 1 5 2 6 3
• 5 5 3 1 6 4 2
• 6 6 5 4 3 2 1
• 1. Closure property: Since all the entries of the composition table are
the elements of the given set, the set G is closed under 7 .
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Contd.,
• 2. Associativity: The binary operation 7 is associative in G.
for ex. (2 7 3) 7 4 = 6 7 4 = 3 and
2 7 ( 3 7 4 ) = 2 7 5 = 3
• 4. . Inverse: From the composition table, we see that the inverse
elements of 1, 2, 3, 4. 5 ,6 are 1, 4, 5, 2, 5, 6 respectively.
identical. Therefore the binary operation 7 is commutative.
• Hence, (G, 7 ) is an abelian group.
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Ex. Show that the set G = {1,3,5,7} is a group with respect to
multiplication modulo 8.
•Solution: The composition table of G is
•
• 8 1 3 5 7
• 1 1 3 5 7
• 3 3 1 7 5
• 5 5 7 1 3
• 7 7 5 3 1
•1. Closure property: Since all the entries of the
composition table are the elements of the given set,
the set G is closed under 8 .
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Contd.,
• 2. Associativity: The binary operation 8 is associative in G.
for ex. (3 85) 8 7 = 78 7 = 1 and
3 8 ( 5 8 7 ) = 3 8 3= 1
• 4. Inverse: From the composition table, we see that the inverse
elements of 1, 3, 5 ,7 are 1, 3, 5, 7 respectively.
identical. Therefore the binary operation 8is commutative.
• Hence, (G, 8) is an abelian group.
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Sub groups
• Def. A non empty sub set H of a group (G, *) is a sub group of G,
• if (H, *) is a group.
Note: For any group {G, *}, {e, * } and (G, * ) are trivial sub groups.
• Ex. G = {1, -1, i, -i } is a group w.r.t multiplication.
H1 = { 1, -1 } is a subgroup of G .
H2 = { 1 } is a trivial subgroup of G.
• Ex. ( Z , + ) and (Q , + ) are sub groups of the group (R ,+).
• Theorem: A non empty sub set H of a group (G, *) is a sub group of G
iff
• i) a * b  H  a, b  H
• ii) a-1
 H  a  H
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Homomorphism and Isomorphism.
• Homomorphism : Consider the groups ( G, *) and ( G1
, )
A function f : G  G1
is called a homomorphism if
f ( a * b) = f(a)  f (b)
• Isomorphism : If a homomorphism f : G  G1
is a bijection then f is
called isomorphism between G and G1
.
Then we write G  G1
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Example
• Ex. Let R be a group of all real numbers under addition and R+
be a
group of all positive real numbers under multiplication. Show that the
mapping f : R  R+
defined by f(x) = 2x
for all x  R is an
isomorphism.
• Solution: First, let us show that f is a homomorphism.
• Let a , b  R .
• Now, f(a+b) = 2a+b
• = 2a
2b
• = f(a).f(b)
•  f is an homomorphism.
• Next, let us prove that f is a Bijection.
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Contd.,
• For any a , b  R, Let, f(a) = f(b)
•  2a
= 2b
•  a = b
•  f is one.to-one.
• Next, take any c  R+
.(then c= 2x
so x=log2c)
• Then log2 c  R and f (log2 c ) = 2 log2 c
= c.
•  Every element in R+
has a pre image in R.
• i.e., f is onto.
•  f is a bijection.
• Hence, f is an isomorphism.
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Example
• Ex. Let R be a group of all real numbers under addition and R+
be a
group of all positive real numbers under multiplication. Show that the
mapping f : R+
 R defined by f(x) = log10 x for all x  R is an
isomorphism.
• Solution: First, let us show that f is a homomorphism.
• Let a , b  R+
.
• Now, f(a.b) = log10 (a.b)
• = log10 a + log10 b
• = f(a) + f(b)
•  f is an homomorphism.
• Next, let us prove that f is a Bijection.
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Contd.,
• For any a , b  R+
, Let, f(a) = f(b)
•  log10 a = log10 b
•  a = b
•  f is one.to-one.
• Next, take any c  R.
• Then 10c
 R and f (10c
) = log10 10c
= c.
•  Every element in R has a pre image in R+
.
• i.e., f is onto.
•  f is a bijection.
• Hence, f is an isomorphism.
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Congruence Relation
• If a and b are integers and m is positive integer, then a is said to be congruent to
b modulo m, if m divides a − b or a − b is multiple of m. This is denoted as
• a≡ b(mod m)
• m is called the modulus of the congruence, b is called the residue of a(mod m). If
a is not congruent to b modulo m, then it is denoted by a ≡b
̸ (mod m).
• Example:
• 89 ≡ 25(mod 4), since 89-25=64 is divisible by 4. Consequently 25 is the residue
of 89(mod 4) and 4 is the modulus of the congruent.
• 153 ≡ −7(mod 8), since 153-(-7)=160 is divisible by 8. Thus -7 is the residue of
153(mod 8) and 8 is the modulus of the congruent.
• 24 ≡
̸ 3(mod 5), since 24-3=21 is not divisible by 5. Thus 24 and 3 are incon-gruent
• modulo 5
• Note: If a ≡ b(mod m) ⇔ a − b = mk, for some integer k
• - a = b + mk, for some integer k.
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• Properties of Congruence
• Property 1: The relation ‖Congruence modulo m‖ is an equivalence relation. i.e., for all integers
a, b and c, the relation is
• Reflexive: For any integer a, we have a ≡ a(mod m)
• Symmetric: If a ≡ b(mod m), then b ≡ a(mod m)
• Transitive: If a ≡ b(mod m) and b ≡ c(mod m), then a ≡ c(mod m).
• Proof: (i). Let a be any integer. Then a − a = 0 is divisible by any fixed positive integer m. Thus a
≡ a(mod m).
∴ The congruence relation is reflexive.
• Given a ≡ b(mod m)
• ➙ a − b is divisible by m ➙ −(a − b) is divisible by m ➙ b − a is divisible by m
• i.e., b ≡ a(mod m).
• Hence the congruence relation is symmetric.
• Given a ≡ b(mod m) and b ≡ c(mod m)
• ➙ a − b is divisible of m and b − c is divisible by m. Hence (a − b) + (b − c) = a − c is divisible by m
• i.e., a ≡ c(mod m)
• ➙ The congruence relation is transitive.
• Hence, the congruence relation is an equivalence relation.
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• Property 2: If a ≡ b(mod m) and c is any integer, then
• a ± c ≡ b ± c(mod m)
• ac ≡ bc(mod m). Proof:
• Since a ≡ b(mod m) ➙ a − b is divisible by m. Now (a ± c) −
(b ± c) = a − b is divisible by m.
• ∴ a ± c ≡ b ± c(mod m).
• Since a ≡ b(mod m) ➙ a − b is divisible by m. Now, (a − b)c =
ac − bc is also divisible by m.
• ∴ ac ≡ bc(mod m).
• Note: The converse of property (2) (ii) is not true always.
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• Property 3: If ac ≡ bc(mod m), then a ≡ b(mod m) only if gcd(c,m) = 1. In fact, if c is an
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Rings
 Rings
 A ring (R,+, ・ ) is a set R, together with two binary operations +
and ・ on R satisfying the following axioms. For any elements a, b,
c R,
∈
(i) (a + b) + c = a + (b + c). (associativity of addition)
(ii) a + b = b + a. (commutativity of addition)
(iii) there exists 0 R, called the zero, such that
∈
a + 0 = a. (existence of an additive identity)
(iv) there exists (−a) R such that a + (−a) = 0.(existence of an
∈
additive inverse)
(v) (a ・ b) ・ c = a ・ (b ・ c). (associativity of multiplication)
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Rings
(vi) there exists 1 R such that
∈
1 ・ a = a ・ 1 = a. (existence of multiplicative identity)
(vii) a ・ (b + c) = a ・ b + a ・ c
and (b + c) ・ a = b ・ a + c ・ a.(distributivity)
 Axioms (i)–(iv) are equivalent to saying that (R,+) is an abelian
group.
 The ring (R,+, ・ ) is called a commutative ring if, in addition,
(viii) a ・ b = b ・ a for all a, b R. (commutativity of
∈
multiplication)
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Commutative Rings
 The integers under addition and multiplication satisfy all of
the axioms above,so that (,+, ・ ) is a commutative ring.
Also, (, +, ・ ), (,+, ・ ), and (,+, ・ ) are all commutative
rings. If there is no confusion about the operations, we write
only R for the ring (R,+, ・ ). Therefore, the rings above would
be referred to as ,,, or . Moreover, if we refer to a ring R
without explicitly defining its operations, it can be assumed
that they are addition and multiplication.
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Integral Domains and Fields
 Integral Domains and Fields
 One very useful property of the familiar number systems is the fact
that if ab = 0, then either a = 0 or b = 0. This property allows us to
cancel nonzero elements because if
ab = ac and a  0, then a(b − c) = 0, so b = c. However, this property
does not hold for all rings. For example, in 4, we have [2] ・ [2] =
[0], and we cannot always cancel since
[2] ・ [1] = [2] ・ [3], but [1][3].
 If (R,+, ・ ) is a commutative ring, a nonzero element a R is called
∈
a zero divisor if there exists a nonzero element b R such that a
∈ ・
b = 0. A nontrivial commutative ring is called an integral domain if
it has no zero divisors.
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Integral Domains and Fields
 A field is a ring in which the nonzero elements form
an abelian group under multiplication. In other
words, a field is a nontrivial commutative ring R
satisfying the following extra axiom.
(ix) For each nonzero element a R there exists a
∈ −1
R such that a
∈ ・ a−1
= 1.
 The rings ,, and  are all fields, but the integers do
not form a field.
 Proposition:- Every field is an integral domain; that
is, it has no zero divisors.
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References
• Discrete Mathematical Structures with applications to computer
science J.P. Tremblay , R. Manhoar
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