HOW TO UNDERSTAND BINARY OPERATIONS IN RELATIONS AND FUNCTIONS

Hello Friends Welcome 

Today  we shall discuss  Binary operation and how to  understand binary operation with the help of some Examples. Because with the help of Binary operations we can crack many quizzes like .

3                 4                      15

6                 7                      49 
9                 ?                      99
If someone ask us  to solve the problem given below,
            

Binary Operations

Then how will you solve this problem or such types of problems?  so with the of help Binary operations we can solve such problems,

 Commutative Property

If a person leaves for his office at 9 am daily ,which is 5 KM from his home , and comes back  home at 6 pm , then its distance from home to office and back office to home is same 5 KM , then this Property is called commutative Property .

For example 
(1) If  Shayam slaps Ram twice , and in return Ram also slaps Shyam then it is called commutative.
(2) if we add 4  to 5 we shall get 9 and in other case if we add 5  to 4 then  we also  get 9. implies 4 + 5 = 5 + 4 = 9

In Mathematics it written as a ∗ b = b∗ a  for all  values of a ,b  .
but if we subtract 4 from 5 we shall get  1 but if we subtract 5 from 4 we shall get -1, then this relation is not called commutative ,because answer are not same in both the cases.

Associative Property

If we have three numbers 4, 5, and 6 , and we have to add these three numbers in two  ways that 1st we add 4 and 5 and then 6 will be added to the result obtained in last step. and in second ways we shall 5 and 6 then resultant will be added to 4 . And in both the cases result comes out same then this  will be called Associated property. 

( 4 + 5 ) + 6 = 4 + ( 5 + 6 )  

           9 + 6 = 4 + 11 =15

But if we subtract these numbers in place of sum ,then these numbers do not satisfy the property of Associativity .

( 4 - 5 ) -  6  ≠  4 -  ( 5 -  6 )  

      (-1) -  6 ≠   4 -  (-1) 

          -1 - 6   ≠  4 + 1

              -7  ≠ 5
This concept can be better understand with the help of this video

  

Binary Operations

A binary operation ∗ on a set A is a function ∗ : A × A → A. We denote ∗ (a, b) by a ∗ b

Question

Let ∗ be binary operation on the set Q of rational numbers defined below

(1)  a ∗ b = a – b
(2)  a ∗ b = ab + 1
(3)  a ∗ b = a + ab
(4)  a * b = | a - b |


Determine whether ∗ is binary, commutative or associative.

Solution

(1) a ∗ b = a – b

 Commutative Property

 a ∗ b = b∗ a 

Now 

 a ∗ b = a – b   -----------------(1)

b*a = b - a = - (a-b) -----------(2)

Therefore   a ∗ b ≠ b∗ a 

Therefore * is not a commutative operation under Q

Associative Property

(a ∗ b) *c = a∗ (b*c) 

(a ∗ b )*c = (a – b)*c         Replaced a*b with a - b as given

Now assume "a-b" as 1st number(blue) and "c" as 2nd number(red)  Again using eq (1)

(a ∗ b )*c =(a – b) * c    

                 = (a - b ) - c

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

 From (1) and (2)

   a * (b * c)  = a*(b - c)

Now assume "a" as 1st number(blue) and "b - c" as 2nd number(red) 

a * (b * c)  = (a) - (b - c )

                = a - b + c ------------(4)

From (1) and (2)

(a ∗ b) *c  ≠ a ∗ (b * c) 

Therefore * is not a associative  operation under Q

(2) a ∗ b = ab + 1   (Product of two numbers and plus one)

 For commutative Property

 a ∗ b = b∗ a 

Now 

 a ∗ b = ab + 1   ------------------(5)

b ∗ a = ba + 1 = ab +1 --------(6)

Therefore   a ∗ b = b∗a 

Therefore * is a commutative operation under Q

For Associative Property

(a ∗ b) *c = a∗ (b*c) 

(a ∗ b )*c =(ab + 1) * c         Replaced a*b with ab+1 as given

Now assume " ab+1 " as 1st number(blue) and "c" as 2nd number(red)

(a ∗ b )*c =(ab + 1)*c    (Product of two numbers and plus one)

                 = (ab + 1)(c) + 1

                 =  (ab + 1)c +1

                 = abc + c + 1   -------(7)

   a * (b*c)  = a*(bc + 1)   (Product of two numbers and plus one)

Now assume "a" as 1st number(blue) and "bc+1" as 2nd number(red) 

a * (b*c)  = (a)(bc + 1) + 1

                =abc + a + 1  ------------------------(8)

From (7) and (8)

(a ∗ b) *c ≠ a∗ (b*c) 

Therefore * is not a associative  operation under Q

(3)  a ∗ b =a + ab  (Product of two numbers and plus 1st num ) 

so if we put all the values of aand  b according to table then we can solve problem discussed in begining of the post.

 Binary Operations

For commutative Property

 a ∗ b = b ∗ a 

Now 

 a ∗ b = a + ab   -----------------(9)

b ∗ a = b + ba = b + ab -------(10)

Therefore   a ∗ b ≠ b ∗ a 

Therefore * is a  not commutative operation under Q

For Associativeness

(a ∗ b) *c = a∗ (b*c) 

(a ∗ b )*c =(a+ab)*c         Replaced a*b with a+ab as given

Now assume "a + ab" as 1st number(blue) and "c" as 2nd number(red)

(a ∗ b )*c =(a + ab) * c   (Product of two numbers and plus 1st number , here 1st num is a + ab and 2nd num is c)

                 = (a + ab ) + ( a + ab )c

                 =  a + ab + ( a + ab )c

                 = a + ab + ac + abc           ------------------------(11)

   a * (b*c)  = a*( b + bc)   (Product of two numbers and plus 1st number )

Now assume "a" as 1st number(blue) and "b + bc" as 2nd number(red) 

a * (b * c)  = (a) + a( b + bc)

                =a + a( b + bc)     

                =a + ab + abc    ------------------------(12)

From (11) and (12)

(a ∗ b) *c ≠ a∗ (b*c) 

Therefore * is not a associative  operation under Q

(4) 

 For commutative Property

 a ∗ b = b ∗ a 

Now 

 a ∗ b = |a - b | ---------------------(13)

b ∗ a = |b - a |= |a - b | ---------------(14)

Therefore   a ∗ b = b ∗ a 

Therefore * is a  commutative operation under Q

For Associativeness

( a ∗ b ) * c = a ∗ (b * c) 

(a ∗ b )*c  = (| a - b | ) * c         Replaced a*b with | a - b | as given

Now assume "| a - b |" as 1st number and "c" as 2nd number

(a ∗ b )*c =(| a - b | ) * c   ( modulus of difference of two numbers , here 1st num is | a - b | and 2nd num is c)

                 =  | (| a - b |) - c |

                 = | | a - b |  - c |           ------------------------(15)

   a * (b*c)  = a* ( |b-c |)   (Modulus of difference of  two numbers )

Now assume "a" as 1st number(blue) and "|b-c |" as 2nd number(red) 

a * (b * c)  = | (a) - |b-c | |

                 =  

                 = |  a - |b-c | | -------- (16)

From (15) and (16)

(a ∗ b) *c ≠ a∗ (b*c) 

Therefore * is not a associative  operation under Q

Conclusion

Thanks for devoting your precious time to read this post. I hope this post on How to understand     Binary Operations , commutative , Associative has helped you more , If you find this post little bit of your concern then, then follow me on my blog and read my other posts . We shall meet  in next post, till then Bye.