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.

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HOW TO KNOW THE DIVISIBILITY TEST OF A NUMBER || DIVISIBILITY RULES FOR NUMBERS

How to know whether a given number, however large is divisible by 2, divisible by 3,4,5,6 and 10 .There are fixed divisibility test and divisibility rules for checking the divisibility of any given numbers. For different numbers there are  different rules to divide with . So in this post  we are going to discuss these divisibility rules with examples  one by one. Although there are divisibility test  for fraction also ,but this post will be restricted to natural numbers.

Divisibility rules for 2

To check whether the given number, however large  is divisible by 2 ,we have to check its right most digit/Unit place digit, if it is even number or zero.Then the given number is definitely divisible by 2

For Example

12 is divisible by 2 as its right most digit is 2 which is Even .

3548 is divisible by 2 as its right most digit is 8 which is Even 

999998 is divisible by 2 as its right most digit is 8 which is Even .

65989564 is divisible by 2 as its right most digit is 4 which is Even. 

22222229 is not divisible by 2 as its right most digit is 9 which is not A Even number.

589423100780      is divisible by 2 as its right most digit is 0 .

357913579571536   is divisible by 2 as its right most digit is 6 which is Even although all the remaining digits are odd.

Divisibility rules for 3

To check whether the given number, however large  is divisible by 3,we have to check the SUM of all its digits, if its sum is divisible by 3 then the  given number is divisible by 3, If sum of  all the digits of given number is again a large number then add the result so obtained and apply the rule again which is said earlier.

For Example

15 is divisible by 3 as sum of all its digits 1 + 5 = 6 is divisible by 3 .

833 is not divisible by 3 as sum of all its digits 8 + 3 + 3 = 14 = 1 + 4 = 5 is not divisible by 3 .

2678 is not divisible by 3 as sum of all its digits 2 + 6 + 7 + 8 = 23 = 3 + 3 = 5 is not divisible by3 .

98784552 is not divisible by 3 as sum of all its digits 9 + 8+ 7+ 8+ 4 + 5 + 5 + 2 = 48 = 4 + 8 = 12 = 1 + 2 = 3 is not divisible by 3.

359875269 is divisible by 3 as sum of all its digits 3 + 5 + 9 +8 +7 + 5 + 2 + 6 + 9 = 54 = 5 + 4 = 9 is divisible by 3.

3597841137 is divisible by 3 as sum of all its digits 3 + 5 + 9 + 7 + 8 + 4 + 1 + 1 + 3 + 7 = 48 = 12   is divisible by 3 .

Divisibility rules for 4

If the last two digits of a number is divisible by 4 ,Then the number is divisible by 4 . The number having two or more zeros at the end is also divisible by 4.

For Example

568928 : Here in this number last two digits are 28 ,which are divisible by 4,Hence the given number is divisible by 4.

134826900 : As ther are two zeros at the end,so the given number is divisible by 4.

13444255452 : As the last two digits number (52) is divisible by 4 ,the given number is divisible by 4.

35888875698549 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4

97971349999567776 : As the last two digits number (76) is divisible by 4 ,the given number is divisible by 4.

44444444444444444449 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4.

Divisibility rules for 5

To check whether the given number, however large is divisible by 5 ,we have to check its right most digit/Unit place digit,if it is 5 or zero. Then the given number is definitely divisible by 5.I.e if the given number ends with 0 or 5 then it is divisible by by 5.

For Example

35 is divisible by 5 as its right most digit is 5.

97835 is divisible by 5 as its right most digit is 5.

6854940 is divisible by 5 as its right most digit is 0.

35000000355 is divisible by 5 as its right most digit is 5.

3579515465855 is divisible by 5 as its right most digit is 5.

12345678888880 is divisible by 5 as its right most digit is 0.

568954975311525 is divisible by 5 as its right most digit is 5.


55555555555556 is not divisible by 5 as its right most digit is 6.

66666666666666665 is divisible by 5 as its right most digit is 5.

Divisibility rule for 6

For any number to be divisible by 6 ,it must be divisible by both 2 and 3 ,then the given number is divisible by 6, Therefore
1) The number should ends up with an even digits or zero and
2) The sum of its digit should be divisible by 3.


For Example :

56898 is divisible by 6 as sum of its digits is 5 + 6 + 8 + 9 + 8 = 36 = 3 + 6 = 9 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

3578952  As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 +7 + 8 + 9 + 5 + 2 = 39 = 4 + 2 = 6  which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

25689879798 is divisible by 6 as sum of its digits is 2 + 5+ 6 + 8 + 9 + 8 + 7 + 9 + 7 + 9 + 8 =78 = 15 = 1 + 5 = 6 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

35658999962 is not divisible by 6 as sum of its digits is 3 + 5 + 6 +  5 + 8 + 9 + 9 + 9 + 9 + 6 + 2 = 71 = 7 + 1 = 8 so it is not divisible by 3 and last digit is not even ,so it is not divisible by 6.

35789248956 As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 + 7+ 8+ 9+ 2+ 4 + 8+ 9 + 5 + 6 = 66 =  6 + 6 = 12 = 1 + 2 = 3 which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

Divisibility rule for 10

It is the most easiest number to identify whether it is divisible by 10 , if the given number however large ends up with 0 then it is divisible by 0.

For Example

4546546546540 ,44545454560, 445474456110 5555598959550 are divisible by 10 as all the given numbers ends with 0.

and 445645489,454545,456445555 ,454545577 and 545454 are not divisible by 10 as all these numbers do not ends up with 0.

Conclusion

Thanks for giving your valuable time to read this post of the divisibility rules and divisibility test rules for  divisibility rule for 2 , divisibility rule for 3,    divisibility rule of 4 etc . If  you liked this post  , Then share it with your friends and family members . You can also read my others articles on Mathematics Learning and understanding Maths in easy ways.

Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

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HOW TO MULTIPLY TWO DIFFERENT NUMBERS IN FASTEST AND QUICKEST WAYS

        

                     Whenever we are to multiply two numbers then most of the time we undergo long calculations , and if we have to do easy calculation then we were lucky.

         But what to do when we have to multiply two numbers in very short time. Suppose we have to multiply 32 with 11 then it is easy, because just put right most digit as it is and then increase every digit by one to left and at last put the left most digit as it is and put 352 as answer.

Example

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