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# Some more properties of a binary operation

Commutative property, Associative property, Existence of identity property, Existence of inverse property

Some more properties of a binary operation

### Commutative property

Any binary operation defined on a nonempty set S is said to satisfy the commutative

property, if

a b = b a a , b S .

### Associative property

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 .

### Existence of identity property

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 .

### Existence of inverse property

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 = a1 .

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 and it is the one and only one element with the property n 1 = 1 n = n, n .

(ii) The multiplicative inverse of any element, say 2 in is ½ and no other nonzero rational number x has the property that 2 x = x 2 = 1.

### Note

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.

### Theorem 12.1: (Uniqueness of Identity)

In an algebraic structure the identity element (if exists) must be unique.

### Proof

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 e1 and e2 be any two identity elements of S .

First treat e1 as the identity and e2 as an arbitrary element of S .

Then by the existence of identity property, e2 e1 = e1 e2 = e2 .    ... (1)

Interchanging the role of e1 and e2 , e1 e2 = e2 e1 = e1 .         …(2)

From (1) and (2), e1 = e2 . Hence the identity element is unique which completes the proof.

### Theorem 12.2 (Uniqueness of Inverse)

In an algebraic structure the inverse of an element (if exists) must be unique.

### Proof

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, a1 , a2 .

Treating a1 as an inverse of a gives a a1 = a1 a = e        …(1)

Next treating a2 as the inverse of a gives a a2 = a2 a = e       …(2)

a1 = a1 e = a1 ( a a2 ) = ( a1 a) a2 = e a2 = a2 (by (1) and (2)).

So, a1 = a2 . 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 .

Solution

(i) m + n , m, n . Hence + is a binary operation on .

(ii) Also m + n = n + m, m, n . So the commutative property is satisfied

(iii) m, n, p , m + ( n + p) = ( m + n) + p . Hence the associative property is satisfied.

(iv) m + e = e + m = m e = 0. Thus 0 ( m + 0) = ( 0 + m) = m . Hence the existence of identity is assured.

(v) m + m ' = m '+ m = 0 m ' = − m. Thus m , m m + ( m) = ( m ) + m = 0 . Hence, the existence of inverse property is also assured.

Thus we see that the usual addition + on 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 .

Solution

(i) Though is not binary on ; it is binary on . To check the validity of any more properties satisfied by – on , 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 .

(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 =(17) =−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 .

(iv) Identity does not exist (why?).

(v) Inverse does not exist (why?).

### Example 12.4

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 e = the set of all even integers.

### Solution

Consider the set of all even integers e = {2 k | k } = {..., 6, 4, 2, 0, 2, 4, 6,...}.

Let us verify the properties satisfied by + on e .

(i) The sum of any two even integers is also an even integer.

Because x , y e x = 2m and y = 2n , m, n .

So ( x + y) = 2m + 2n = 2( m + n) e . Hence + is a binary operation on e.

(ii) ∀x , y 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 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 e , ∃0 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 e , x e x + ( −x) = ( −x ) + x = 0

Hence -x is the inverse of x e.

### Example 12.5

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 o= the set of all odd integers.

Solution

Consider the set o of all odd integers o = {2 k +1: k } = {..., 5, 3, 1, 1, 3, 5,...} . + is not a binary operation on 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 o . Their sum 3 + 7 = 10 is an even number. In general, if x, y 0 , then ( x + y ) 0 . Other properties need not be checked as it is not a binary operation.

### Example 12.6

Verify (i) closure property (ii) commutative property, and (iii) associative property of the following operation on the given set.

(a b) = ab ; a, b (exponentiation property)

Solution

(i) It is true that a b = a b ; a, b . So is a binary operation on .

(ii) a b = ab and b a = ba . Put, a = 2 and b = 3 . Then a b = 23 = 8 but b a = 32 = 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 ) = 234 = 281

But ( a b) c = ( a b ) c = ( ab )c = a ( bc ) = abc = 212

Hence a ( b c) ( a b) c . So does not have associative property on

Note: This binary operation has no identity and no inverse. (Justify).

### Example 12.7

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

### Solution

(i) The output m + n - m n is clearly an integer and hence is a binary operation on .

(ii) m n = m + n m n = n + m nm = n m, m, n . So has commutative property.

(iii) Consider ( m n) p = ( m + n mn) p = ( m + n mn) + p ( m + n mn ) p

= m + n + pmnmpnp + mnp            ... (1)

Similarly m ( n p) = m ( n + p np) = m + ( n + p np ) m ( n + p np)

= m + n + pnpmnmp + 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  ⇒ 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 . Hence 0 is the identity element and hence the existence of identity is assured.

(v) An element m is to be found such that m m = m m = e = 0, m .

m m = 0 m + m m m = 0 m = m / m-1. When m=1, m is not defined.

When m =2, mis an integer. But except m=2, mneed not be an integer for all values of m. Hence inverse does not exist in .

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