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1800-102-2727Binary Operations,Properties, Practice problems & FAQs
Right from the early school days, we have come across four fundamental operations namely addition, multiplication, subtraction and division.The main feature of these operations is that any two given numbers p and q can be associated with the help of these operations.
Note that we can add or multiply only two numbers at a time. When we need to add three or more numbers, we first add two numbers and the result is then added to the third number. Thus, addition, multiplication, subtraction and division are examples of binary operation, as ‘binary’ means two. Let us try to get an in depth understanding of Binary operations.
Table of Contents
The binary operation can be understood as an operation which is performed on the two elements p & q from the set X. Thus, the binary operation performed on operands p and q is symbolized as p*q. The function is given by *:AAA. The result of the operation on p and q is another element from the same set X.
Note: We can denote a binary operation using any symbol ( !, @ , * , $ etc.)
Example: Show that subtraction and division are not binary operations on N.
Solution:
- : , given by (a, b) a – b, is not binary operation, as the image of (2, 7) under '-' is 2-7 =-5 . Similarly, : , given by (a, b) a ÷ b is not a binary operation, as the image of (2, 7) under '' is 2 ÷ 7 =2/7 .
Example:Show that addition, subtraction and multiplication are binary operations on R , but division is not a binary operation on R . Further, show that division is binary operation on the set R of non-zero real numbers.
Solution
+: R × R → R is given by +(a,b) → a+b
-: R × R → R is given by -(a,b) → a-b
: R × R → R is given by (a,b) → ab
Since +,- and are functions, they are binary operations on R .
But : R × R → R , given by (a,b)a/b is not a function and hence not a binary operation, as for b=0, a/b is not defined.
However, : R × R → R is given by (a,b)a/b is a function and hence a binary operation on set R of non zero real numbers.
Example:Let M be the set of all subsets of a given set A. Show that : M × MM given by (P, Q)PQ and : M × MM given by (P, Q)PQ are binary operations on the set M.
Solution
Proving for Union
Since M is the set of all the subsets of the given set A and P, Q are in set M we can say that P and Q are also the subsets of A.
If we calculate PQ then it will also be a subset of A as the union of subsets is also a subset
Hence, PQ will also be in set M i.e. is a binary operation.
Proving for Intersection
Since M is the set of all the subsets of the given set A and P, Q are in set M we can say that P and Q are also the subsets of A
If we calculate PQ then it will also be a subset of A as the intersection of subsets is also a subset.Hence, PQ will also be in set M i.e. is a binary operation.
Example :Show that the operation : R× R→ R given by (p, q) max {p, q} and the operation
: R× R → R given by (p, q)min {p, q} are binary operations.
Solution:
Since, pq=max of p and q; p,qR
pq will give output as either p or q.
Hence, pq R. Therefore, is a binary operation.
Also, pq=min of p and q, p,qR
pq will give output as either p or q.
Hence, pq R. Therefore, is a binary operation.
Note:
A binary operation * is commutative on the set X, if a*b=b*a for every a,bX.
Example: Addition and multiplication are commutative binary operations but subtraction and division are not.
A binary operation on set X is associative if for every a,b,cX, a*(b*c)=(a*b)*c
Example: Addition and multiplication are associative binary operations on the set of real numbers but subtraction and division are not.
An element eX is called the identity of the operation *: XXX, if
a*e=a=e*a, for every element aX.
Example: 0 and 1 are the identities for addition and multiplication operation on the set of
real numbers. There is no identity for subtraction and division operations on .
For a binary operation * on a non-empty set X, let e be the identity element. If aX, then a is invertible if there exists an element bX such that a*b=b*a=e.
Example: 1/a is the inverse of a0 of the multiplication operation() on but it's not an inverse of X on the set of natural numbers .
Let *:AAA be a binary operation on set A and n(A)=p
Now, n(AA)=n(A)n(A)=pp=p2
Hence, number of functions from AAA will be pp2
(Since, number of functions from AB where n(A)=a, n(B)=b is ba)
The total number of binary operations on a set consisting of n elements is given by pp2.
Example:Number of binary operations on the set {a, b, c, d} is
Solution
Number of binary operations on a set consisting of n elements is given by nn2
Here n=4 so, the number of binary operations =442=416.
Hence option (a) is correct.
Example:Determine whether or not * given below gives a binary operation.
(i) On Z+, * : a*b = a-b
(ii) On Z+, * : a*b = ab2
(iii) On R, * : a*b = a2b
(iv) On Z+, * :a*b=|a-b|
Here, Z+ denotes the set of all non-negative integers and R denotes the set of real numbers.
Solution:
(i) Given: On Z+, * : a*b = a-b
If a = 1 and b = 2 in Z+, then
a * b = a – b=1-2=-1
-1 ∉ Z+ (as Z+ is the set of non-negative integers)
For a=1 and b=2, a*b ∉ Z+
Thus, * is not a binary operation on Z+.
(ii) Given: On Z+, * : a*b = ab2
Let a, b ∈ Z+
⇒ a*b=ab2 ∈ Z+ (as the square of a non-negative integer is a non-negative integer and the product of two integers is also an integer)
Thus, * is a binary operation on R.
(iii) Given: On R, * : a*b = a2b
Let a, b ∈ R
⇒ a2, b ∈ R
⇒ a2b ∈ R
⇒ a * b ∈ R
Thus, * is a binary operation on R.
(iv) Given: On Z+, * :a*b=|a-b|
Let a, b ∈ Z+
⇒ | a – b | ∈ Z+(as modulus function always gives positive output)
⇒ a*b ∈ Z+
Therefore,a*b ∈ Z+, ∀ a, b ∈ Z+
Thus, * is a binary operation on Z+.
Example : Show that the binary operation * on Z defined by a*b=a+b+1 a,bZ satisfies (i)associative law (ii)commutative law (iii)Find the identity element (iv)Inverse of an element a Z
Solution:
(i)For all a,b,cZ, we have
(a*b)*c=(a+b+1)*c
=(a+b+1)+c+1
= a+b+c+2
a*(b*c)=a*(b+c+1)
=a+(b+c+1)+1
= a+b+c+2
Therefore, (a*b)*c=a*(b*c)
(ii)For all a,bZ, we have
a*b=a+b+1
=b+a+1
=b*a
(iii)Let e be te identity element in Z.
Then, a*e=aa+e+1=a e=-1
Thus -1Z is the identity element for *
(iv)Let aZ and it’s inverse be b. Then,
a*b=-1a+b+1=-1
b=-(2+a)
Also, b=-(2+a)Z. Hence the inverse is -(2+a)
Example : Consider a binary operation * on the set {1,2,3,4,5} given by the binary operation table:
1 |
2 |
3 |
4 |
5 |
|
1 |
1 |
1 |
1 |
1 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
3 |
1 |
1 |
3 |
1 |
1 |
4 |
1 |
2 |
1 |
4 |
1 |
5 |
1 |
1 |
1 |
1 |
5 |
(i) Compute (2 *3)*4 and 2*(3* 4)
(ii) Is * commutative
(iii) Compute ( 2*3)*(4*5)
Solution :
(i) 2*3 =1 and 3*4=1
Now, (2*3)*4=1*4=1 and 2*(3*4)=2*1=1
(ii) 2*3=1 and 3*2=1
Hence, 2*3=3*2 and this is true for all other elements also.Hence, the operation * is commutative.
(iii) (2*3)*(4*5)=1*1=1
Question.1 What is closure property?
Answer: An operation * on a non-empty set X satisfies closure property,if aX, bXa*b X .Operations which satisfy closure property are Binary operations.
Question.2 Will the identity element always exist for the binary operation on a given set?
Answer: Let us consider a binary operation * on Z defined by a*b=a-b. Now for identity element ,
a-e=e-a=a. There is no possible value of e which satisfies this relation.
Hence,It is not mandatory that the identity element would always exist.
Question.3 How to make a Binary Operation Table?
AnswerThe steps to make a binary operation table are given as