Some more
properties of a binary operation
Any
binary operation ∗
defined on a nonempty set S is said
to satisfy the commutative
property,
if
a ∗ b = b ∗ a ∀a ,
b ∈S .
Any
binary operation ∗
defined on a nonempty set S is said
to satisfy the associative property, if
a ∗ (b
∗c )
= ( a ∗b )
∗ c ∀a , b, c ∈S .
An
element e ∈ S
is said to be the Identity Element of S under the binary operation ∗ if for all a ∈ S we have that a ∗ e = a and e ∗ a = a .
If an
identity element e exists and if for
every a ∈ S
, there exists b in S such that a ∗ b = e
and b ∗ a = e then b ∈ S is said to be the Inverse Element of a . In such
instances, we write b = a^{−}^{1} .
Note
a^{–1} is an element of S.
It should be read as the inverse of a and
not as 1/a .
Note
(i) The multiplicative
identity is 1in ℤ and
it is the one and only one element with the property n ⋅ 1 = 1⋅ n = n, ∀n ∈ ℤ.
(ii) The
multiplicative
inverse of any element, say 2 in ℚ is ½ and no other nonzero rational number x has the property that 2 ⋅ x
=
x ⋅ 2 = 1.
Whenever
a mathematical statement involves ‘for every’ or ‘ for all’ , it has to be
proved for every pair or three elements. It is not easy to prove for every pair
or three elements. But these types of definitions may be used to prove the
negation of the statement. That is, negation of “for every” or “for all” is
“there exists not”. So, produce one such pair or three elements to establish
the negation of the statement.
The
questions of existence and uniqueness of identity and inverse are to be
examined. The following theorems prove these results in the more general form.
In an algebraic structure the identity element (if exists) must
be unique.
Let (S, ∗) be an algebraic structure. Assume
that the identity element of S exists
in S .
It is to
be proved that the identity element is unique. Suppose that e_{1} and e_{2} be any two identity elements of S .
First
treat e_{1} as the identity
and e_{2} as an arbitrary
element of S .
Then by the existence of identity property, e_{2} ∗ e_{1} = e_{1} ∗ e_{2} = e_{2} . ... (1)
Interchanging the role of e_{1} and e_{2} , e_{1} ∗ e_{2} = e_{2} ∗ e_{1} = e_{1} . …(2)
From (1)
and (2), e_{1} =
e_{2} . Hence the identity
element is unique which completes the proof.
In an algebraic structure the inverse of an element (if exists)
must be unique.
Let (S, ∗) be an algebraic structure and a ∈ S
. Assume that the inverse of a exists
in S . It is to be proved that the
inverse of a is unique. The existence
of inverse in S ensures the existence
of the identity element e in S .
Let a ∈ S
. It is to be proved that the inverse a
(if exists) is unique.
Suppose
that a has two inverses, say, a_{1} , a_{2} .
Treating
a_{1} as an inverse of a gives a ∗ a_{1} =
a_{1} ∗ a
=
e
…(1)
Next
treating a_{2} as the inverse
of a gives a ∗ a_{2} =
a_{2} ∗ a
=
e
…(2)
a_{1} = a_{1} ∗ e = a_{1} ∗ (
a ∗ a_{2} ) = ( a_{1} ∗ a)
∗ a_{2} = e ∗ a_{2} = a_{2 }(by (1) and (2)).
So, a_{1} =
a_{2} . Hence the inverse of a is unique which completes the proof.
Example 12.2
Verify
the (i) closure property, (ii) commutative property, (iii) associative property
(iv) existence of identity and (v) existence of inverse for the arithmetic
operation + on ℤ.
Solution
(i) m + n ∈ ℤ,
∀m,
n ∈ ℤ. Hence + is a binary operation on
ℤ.
(ii) Also
m + n
=
n + m,
∀m, n ∈ ℤ. So the commutative property is
satisfied
(iii) ∀m,
n, p ∈ ℤ,
m + ( n + p) = ( m + n) + p . Hence the associative property is
satisfied.
(iv) m + e = e + m = m ⇒ e = 0. Thus ∃ 0
∈ ℤ ⋺ (
m + 0) = ( 0 + m) = m . Hence the existence of identity is assured.
(v) m + m '
= m '+ m = 0 ⇒ m '
= − m. Thus ∀m ∈ ℤ,
∃ − m ∈ ℤ ⋺ m + ( − m)
= ( − m )
+ m = 0 . Hence, the
existence of inverse property is also assured.
Thus we see that the usual addition +
on ℤ satisfies all the above five
properties.
Note
that the additive identity is 0 and
the additive inverse of any integer m is m .
Example 12.3
Verify
the (i) closure property, (ii) commutative property, (iii) associative property
(iv) existence of identity and (v) existence of inverse for the arithmetic
operation – on ℤ.
Solution
(i) Though
− is
not binary on ℕ; it is binary on ℤ. To check the validity of any more
properties satisfied by – on ℤ,
it is better to check them for some particular simple values.
(ii) Take
m = 4 , n = 5 and ( m −
n) = ( 4 − 5) = −1and ( n − m)
=
(5 −
4) =
1.
Hence (m − n)
≠
(n − m)
. So the operation is not commutative on ℤ.
(iii) In
order to check the associative property, let us put m = 4, n =
5 and p = 7 in both
(m  n)

p and m  ( n 
p) .
(m − n
) −
p = (4 −5) – 7 =(−1−7) =−8
…(1)
m − ( n− p)
= 4 −(5− 7) =(4+2)=6.
…(2)
From (1)
and (2), it follows that ( m
–
n) –p ≠ m
–
(n – p
).
Hence –
is not associative on ℤ.
(iv) Identity
does not exist (why?).
(v) Inverse
does not exist (why?).
Verify
the (i) closure property, (ii) commutative property, (iii) associative property (iv) existence of identity and (v) existence
of inverse for the arithmetic operation + on ℤ_{e} = the set of all even integers.
Consider
the set of all even
integers ℤ_{e }= {2
k  k ∈ ℤ} = {..., −6, −4, −2, 0, 2, 4, 6,...}.
Let us
verify the properties satisfied by +
on ℤ_{e
}.
(i) The
sum of any two even integers is also an even integer.
Because x , y
∈ ℤ_{e} ⇒ x
=
2m and y = 2n , m, n
∈ ℤ.
So (
x + y)
=
2m + 2n = 2( m
+
n)∈ ℤ_{e} . Hence +
is a binary operation on ℤ_{e}.
(ii) ∀x , y ∈ ℤ_{e} ,
( x + y) = 2( m + n) = 2( n + m) = ( 2 n + 2m) = (
y + x) .
So +
has commutative property.
(iii) Similarly
it can be seen that ∀x , y, z ∈ ℤ_{e} , ( x + y) +
z = x
+
( y + z)
.
Hence
the associative property is true.
(iv) Now
take x = 2k , then 2 k +
e = e
+
2 k = 2 k ⇒ e = 0.
Thus ∀ ∈ ℤ_{e} , ∃0 ∈ ℤ_{e }⋺ x
+ 0 = 0 + x = x .
So, 0 is
the identity element.
(v) Taking x = 2k and x ′ as its inverse,
we have 2k + x' = 0 = x' + 2k ⇒ x'
= − 2k . i.e., x
' = −x .
Thus ∀x ∈ ℤ_{e} , ∃ − x ∈
ℤ_{e} ⋺ x + ( −x) = ( −x ) + x = 0
Hence x
is the inverse of x ∈ ℤ_{e}.
Verify the
(i) closure property, (ii) commutative property, (iii) associative property
(iv) existence of identity and (v) existence of inverse for the arithmetic operation
+ on ℤ_{o}= the set of all odd integers.
Solution
Consider
the set ℤ_{o} of all odd integers ℤ_{o} = {2 k +1: k ∈ ℤ} = {..., −5, −3, −1, 1, 3, 5,...}
. +
is not a
binary operation on ℤ_{o} because when x = 2 m +1, y = 2 n +1, x + y = 2( m + n) + 2 is even for all m and n. For instance,
consider the two odd numbers 3, 7 ∈ ℤ_{o} . Their sum 3 + 7 = 10 is an even number. In general, if x, y ∈ ℤ_{0} , then ( x + y )∉ ℤ_{0} . Other properties need
not be checked as it is not a binary operation.
Verify
(i) closure property (ii) commutative property, and (iii) associative property
of the following operation on the given set.
(a ∗ b) = a^{b} ; ∀a,
b ∈ ℕ (exponentiation property)
Solution
(i) It
is true that a ∗ b
=
a ^{b}
∈ ℕ; ∀a, b ∈ ℕ. So ∗
is a binary
operation on ℕ.
(ii) a ∗ b = a^{b} and b ∗ a = b^{a} . Put, a = 2
and b = 3 . Then a ∗ b = 2^{3} = 8 but b ∗ a = 3^{2} = 9
So a ∗b need not be equal to b
∗ a
. Hence ∗ does not have commutative property.
(iii) Next
consider a ∗ ( b
∗ c
)
=
a ∗ ( b
^{c} )
=
a( ^{bc}) . Take a =
2, b = 3 and c = 4 .
Then a ∗ ( b
∗ c
)
=
2 ∗ ( 3∗ 4 ) = 23^{4} =
2^{81}
But (
a ∗ b)∗ c
=
(
a ^{b}
)∗ c
=
(
a^{b} )^{c} = a
( ^{bc} ) =
a^{bc} =
2^{12}
Hence a ∗ ( b
∗ c)
≠
(
a ∗ b)∗ c
. So ∗ does not have associative property
on ℕ
Note: This binary operation has no
identity and no inverse. (Justify).
Verify
(i) closure property, (ii) commutative property, (iii) associative property,
(iv) existence of identity, and (v) existence of inverse for following
operation on the given set.
m ∗ n = m + n − mn;
m, n ∈ ℤ
(i) The
output m +
n  m
n is clearly an integer and hence ∗ is a binary operation on ℤ.
(ii) m ∗ n = m + n − m n = n + m − nm = n ∗ m,
∀m,
n ∈ ℤ. So ∗ has commutative
property.
(iii) Consider
( m ∗ n)
∗ p
=
( m + n
−
mn) ∗ p
=
( m + n
−
mn) + p
−
( m + n
−
mn ) p
= m + n + p − mn − mp − np + mnp ... (1)
Similarly
m ∗ ( n ∗ p) = m
∗ ( n + p −
np) = m + ( n +
p − np
) −
m ( n + p −
np)
= m + n + p − np − mn − mp + mnp ... (2)
From (1)
and (2), we see that m ∗ ( n ∗ p ) = ( m ∗ n ) ∗ p
. Hence ∗ has associative property.
(iv) An
integer e is to be found such that
m ∗ e = e ∗ m = m, ∀m ∈ ℤ ⇒
m + e − m e = m
⇒ e (1 − m) = 0 ⇒ e = 0 or m = 1. But m is an
arbitrary integer and hence need not be equal
to 1. So the only possibility is e =
0 . Also m ∗0 = 0 ∗ m
=
m, ∀m ∈ ℤ. Hence 0 is the identity element and hence the existence of
identity is assured.
(v) An
element m′
∈ ℤ is to be found such that m ∗ m′ = m′
∗ m
=
e = 0, ∀m ∈ ℤ.
m ∗ m′ = 0 ⇒ m + m′ − m
m′ _{=} 0 ⇒ m′ = m
/ m1. When m=1, m′
is not defined.
When m =2,
m′ is
an integer. But except m=2, m′ need
not be an integer for all values of m.
Hence inverse does not exist in ℤ.
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