Binary Operations SS1 Further Mathematics Lesson Note

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Topic: Binary Operations

Definition:

A binary operation is any rule of a combination of any two elements of a given non-empty set. The rule of a combination of two elements of a set may give rise to another element which may or not belong to the set under consideration.

It is usually denoted by symbols such as *, Ó¨ e.t.c.

Properties:

  1. Closure property: A non-empty set z is closed under a binary operation * if for all a, b € Z.

Example; A binary operation * is defined on the set S= {0, 1, 2, 3, 4} by 

 X*Y = x + y –xy. Find (a) 2 * 4 (b) 3* 1 (c) 0* 3.  Is the set S closed under the operation *?

Solution

2 * 4, i.e, x= 2,y=4

2+ 4 – (2×4)       = 6-8 = -2.        

3* 1 = 3+1-( 3x 1)    = 4 – 3= 1

0*3  = 0 + 3 –( 0 x3) = 3  

Since -2€ S, therefore the operation * is not closed in S.

Commutative Property: If set S, a non-empty set is closed under the binary operation *, for all a,b€ S. Then the operation * is commutative if a*b= b*a

Therefore, a binary operation is commutative if the order of combination does not affect the result.                                                                                                         

Example; The operation * on the set R of real numbers is defined by: 

p*q= p3 + q3-3pq. Is the operation commutative?

Solution

p*q= p3 + q3 -3pq

Commutative condition p*q= q*p

To obtain q*p, use the same operation q*p, use the same operation p*q but replace p by q and q by p.

Hence, q*p= p3+ q3 -3qp

In conclusion p*q= q*p, the operation is commutative.

Associative Property: If a non–empty set S is closed under a binary operation *, that is a*b €S. Then a binary operation is associative if (a*b) * c= a*(b*c)

Such that C also belongs to S.

Example: The operation Ó¨ on the set Z of integers is defined by; a Ó¨ b = 2a +3b -1. Determine whether or not the operation is associative in Z.

Solution

Introduce another element C

Associative condition: (aӨb) Өc = a Ө (b Өc) 

(aÓ¨b)Ó¨c = (2a+ 3b- 1) Ó¨ C

                = 2(2a +3b -1) + 3c -1

                = 4a + 6b- 2+ 3c- 1

= 4a +6b+3c- 3.

Also, the RHS, a Ó¨ (b Ó¨ c) = a Ó¨ (2b+3c- 1)

= 2a+ 3(2b +3c- 1) – 1

= 2a + 6b +9c -3 -1                                                                                                                                                                     a Ө (b Ө c)  = 2a+ 6b+ 9c -4

Since,   (a Ө b) Ө c ≠ a Ө (b Ө c), the operation is not associative in Z.

  Distributive Property: If a set is closed under two or more binary operations

(* Ө) for all a, b and c € S, such that:

              a*(bÓ¨ c) = (a*b )Ó¨( a*c – Left distributive

              (BÓ¨c) *a = (b*a) Ó¨(c*a) – Right distributive over the operation Ó¨

Example: Given the set R of real numbers under the operations * and Ó¨ defined by:

        a*b = a+ b- 3, aӨb= 5ab for all a, b € R. Does * distribute over Ө.

Solution Let a, b,c € R

a* ( bÓ¨c) = (a*b) Ó¨ (a*c)

a* (bÓ¨c) = a* (5ab)

               = a+ 5ab -3.

(a*b) Ó¨ (a*c) = (a+ b -3) Ó¨ ( a+ c-3)

                      = 5(a +b-3)(a +c -3)

  From the expansion, it’s obvious that a* ( bӨc) ≠ (a*b) Ө (a*c)  therefore * does not distribute over Ө.

IDENTITY AND INVERSE ELEMENTS

Identity Element:

Given a non-empty set S which is closed under a binary operation * and if there exists an element e € S such that a*e = e*a = a for all a € S, then e is called the IDENTITY or NEUTRAL element. The element is unique.

Example: The operation * on the set R of real numbers is defined by 

a*b = 2a-1 

            2         ┼ b for all a, b € R. Determine the identity element.

Solution:

      a*e= e*a = a

      a*b= 2a-1 ┼ b

                 2

     a*e = 2a-1 ┼ e = a

                 2

       2a-1+ 2e = 2a

         2e = 2a-2a +1

         e   = ½.

Example: An operation * is defined on the set of real numbers by x*y = x + y -2xy. If the identity element is 0, find the inverse of the element.

Solution;

      X *y = x+ y- 2xy

      x*x-1 = x-1*x= e, e = 0

      x + x-1- 2xx-1 = 0

      x-1 -2xx-1= -x

      x-1(1-2x) = -x

      x-1 = -x/ (1-2x)

The inverse element x-1 = -x/ (1-2x)

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