Relations and Functions�Binary operations

�Exercise 1.4�Class 12

Vihutuo Paira�Little Star Higher Secondary School�Dimapur Nagaland

Binary operation

Consider the set Z, the set of integers and the addition operator (+).�The + operator takes two elements from Z and relates it to another unique element in Z.��For example it relates the pair of numbers 2 and 3 to 5 (as 2+3=5). We may write (2,3) R 5. Similarly (5,1) is related to 6 and so on.��Thus the operator + relates every pair of elements in Z to another element in Z. The collection of of all pair of elements in Z is the cartesian product Z x Z.

Therefore + is a relation (also a function) from Z x Z to Z.�+ is an example of a binary operator on the set Z.

Binary Operation definition

• Formal Definition : �A binary operation * on a set A is a function * : A x A →A. We denote *(a,b) by a*b.
• In other words, a binary operation on a set is a calculation involving two elements of the set to produce another element of the set.
• Addition(+) is a binary operation on R
• Subtraction(-) is a binary operation on R
• Multiplication is also a binary operation on R

Division on R

• Division is NOT a binary operation on R as (a,b)→a/b is not defined when b=0.
• However if we take R_*, the set of non-zero real numbers, then Division is a binary operation on R_*

Exercise 1.4 Question 1

1. Determine whether the following are binary operations. If not, give reasons.��i) On Z⁺, define * by a * b = a - b, Z⁺ is the set of positive integers�Solution: * is not a binary operation on Z⁺ since (1,3)∈ Z⁺ x Z⁺ but its image given by 1 - 3=-2 ∉ Z⁺��ii) On Z⁺, define * by a * b = |a - b|�Solution: Since * carries each pair (a,b) to a unique element given by |a-b| in Z⁺, * is a binary operation on Z⁺

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Try on your own (Ex 1.4)

• Question 1. (ii), (iii), (v) �
• Show that ∨ : R X R→ R given by (a,b)→ max{a,b} is a binary operation

Ex ∨(5,3) = 5� ∨(7,10) = 10

Commutative binary operation

Definition : A binary operation * on the set X is called commutative, if a*b = b*a, for every a,b ∈ X

Example : Show that +:R x R→R and x : R x R → R are commutative binary operations,but - :R x R→R and ÷: R_* x R_*→R_* are not commutative

Solution : Since a + b = b + a and a x b = b x a ∀a,b ∈ R, ‘+’ and ‘x’ are commutative.�However ‘-’ is not commutative since 3 - 4 ≠ 4 - 3�Similarly 3 ÷ 4 ≠ 4 ÷ 3 therefore ÷ is not commutative

Associative binary operation

Definition : A binary operation on the set X is said to be associative if �(a * b) * c = a * (b * c), for every a,b,c ∈ X

Example : Show that addition and multiplication are associative binary operation on R. But subtraction is not associative on R. Division is not associative on R_*

Solution : Since (a+b)+c = a+(b+c) and (a x b) x c = a x (b x c), ∀a,b,c ∈ R, addition and multiplication are associative binary operations��However subtraction is not associative since (3 - 4) -1 = -2 but 3-(4-1) = 0

Similarly division is not associative as (8 ÷ 4) ÷ 2 = 1 but 8 ÷ (4 ÷ 2) = 4��

Exercise 1.4 Question 2(ii)

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Question 2 : Determine whether * is binary, commutative or associative� (ii) On Q, define a *b = ab +1 (Q is the set of rational numbers)

Solution : If a,b∈ Q, then ab+1 ∈ Q as product and sum of rational number is another rational number�So * carries each pair (a,b) to a unique element given by ab +1 in Q, and hence * is a binary operation on Q�Is * commutative? �a*b = ab +1 �b*a = ba + 1 =ab+1 (since ab=ba)�Therefore a*b=b*a, ∀a,b ∈ Q �Hence * is commutative

...continued

Is * associative?

(a*b)*c = (ab +1) * c = (ab+1)c + 1 =abc +c +1�a*(b*c) = a *(bc+1) = a(bc+1) + 1 =abc + a +1

Clearly (a*b)*c ≠ a*(b*c)�Example (1*2)*3 = (1x2+1)*3=3*3=3x3+1=10� 1*(2*3) =1 * (2x3 +1) = 1 * 7 = 1x7 + 1 = 8�� hence * is not associative

Exercise 1.4 Question 2 (vi)

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Question 2 : Determine whether * is binary, commutative or associative� (vi) On R-{-1}, define a *b = a/(b +1)��Solution : Let’s check whether a/(b+1) maybe equal to -1�If, a/(b+1) = -1 �⇒a=-(b+1) �So say, if b=2 then a = -(2+1) = -3, for these values a/(b+1)=-1��-3*2 = -3/(2+1) = -1 ∉ R- {-1}�So -3,2∈R- {-1} but -3*2 ∉ R- {-1}�Hence, * is not a binary operation�By definition, commutative and associative is only applicable to binary operations�

Try yourself

• Question 2 (i), (iii), (iv), (v)

Operation table

When number of elements in a set A is small, we can express a binary operation * on the set A through a table called the operation table for the operation *. �For example, consider A={1,2,3}. Then the operation ∨ (max) on A can be expressed by the following operation table.

From the table ∨(2,3) = 3 (circled on the table)

3

To read the table: read the first value (2) from the left hand column and the second value (3) from the top row. The answer is the intersection cell.

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Operation table

If the table is symmetric with respect to the diagonal line, the table is commutative.�You can see this table is symmetric wrt to the diagonal hence the operation ∨ is commutative�

Try yourself (Ex 1.4)

• Question 3
• Question 4
• Question 5
• Question 7

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Question 9 (iv)

Question 9: Let * be a binary operation on the set Q of rational numbers as follows�(iv) a* b = (a-b)^2�Find out whether * is commutative or associative

Solution : Is * commutative?�a*b = (a-b)²� b*a = (b-a)² = {-1(a-b)}² = (a-b)² (since (-1)² = 1)

Therefore, a*b=b*a�Hence * is commutative

… continued

Is * associative?

(a*b)*c = (a-b)² * c = ((a-b)² - c)^2�a*(b*c) = a * (b-c)² = (a-(b-c)²)²

Clearly (a*b)*c ≠ a*(b*c)�Example : �(1*2)*3 = (1-2)² * 3 = (-1)² * 3 = 1*3 = (1-3)² = (-2)² = 4�1*(2*3) = 1*(2-3)² = 1 * (-1)² = 1 * 1 = (1-1)² = 0

Hence * is not associative

Try yourself (Ex 1.4)

• Question 9 (i),(ii),(iii),(v),(vi)
• Question 13

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Identity

• Definition : Given a binary operation * :AxA→A, an element e∈A,if it exists, is called identity for the operation *, if a * e = e * a = a , for all a∈A
• Identity element is a unique single element in A and it should be the same element for all elements in A
• For the binary operation ‘+’ on the set R, the identity element e = 0 since a+0 = 0+a = a, for all a ∈R
• For the binary operation ‘x’ on the set R, the identity element e = 1 since ax1 = 1xa = a, for all a ∈R

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Inverse of an element

• Definition : Given a binary operation * :AxA→A with the identity element e in A, an element a∈A is said to be invertible with respect to the operation *, if there exists an element b in A such that a * b = b * a = e and b is called the inverse of a denoted by a^-1
• Inverse of each element is usually different
• For the binary operation ‘+’ on the set R, the inverse of a is -a since�a + (-a) = (-a) + a = 0
• For the binary operation ‘x’ on the set R, the inverse of a is 1/a for a ≠ 0 since�a x (1/a) = (1/a) x a = 1 �

Question 10. Find which of the operations have identity

Let * be a binary operation on the set Q of rational numbers

(i) a * b = a - b

Solution : If possible, let e be the identity element�Therefore, a * e = a and e * a =a�⇒ a - e = a and e - a = a�⇒ e= 0 and e = 2a�No single value of e in Q will satisfy both the equations, hence * does not have an identity element��

Question 10 (v)

v) a* b = ab/4�Solution : If possible, let e be the identity element�Therefore, a * e = a and e * a =a�⇒ ae/4 = a and ea/4 = a�⇒ ae= 4a and ea = 4a�⇒e = 4 and e = 4�Hence * has an identity with e=4

Question 10 (iv)

iv) a* b = (a-b)²�Solution : If possible, let e be the identity element�Therefore, a * e = a and e * a =a�⇒ (a-e)² = a and (e-a)² = a�⇒ a - e= ±√a and e - a = ±√a�⇒e =a±√a and e = a±√a��Since e varies with the value of a and thus is not unique, * does not have an identity�(Remember that e is a single unique element in A, it cannot be different for each a)

Exercise 1.4 Question 6

Question 6 : Let * be the binary operation on N given by a*b=LCM of a and b. Find�(i) 5*7, 20* 16 (ii) Is * commutative (iii) Is * associative (iv) Find the identity �(v) Which elements of N are invertible?��Solution : �(i) 5*7 = LCM of 5 and 7 = 35� 20*16 = LCM of 20 and 16 = 80��(ii) Is * commutative ?�We know that, LCM of a and b = LCM of b and a, hence�a*b = b*a for all a,b∈N. Therefore *is commutative

… continued

(iii) Is * associative?�(a*b)*c = (LCM of a and b) * c = LCM of a,b and c�a*(b*c) = a * (LCM of b and c) = LCM of a,b and c�(a*b)*c = a*(b*c) for all a,b,c ∈N�Hence * is associative

(iv) The identity e = 1, since LCM of a and 1 = a for all a∈N��(v) Let a∈N be invertible with its inverse being a^-1�By definition, a * a^-1 = e and a^-1 * a = e �⇒a * a^-1 = 1 and a^-1 * a = 1 (since e =1)�⇒LCM of a and a^-1 = 1�The only only pair of numbers whose LCM is 1 is 1 and 1. Therefore 1 is the only element that is invertible with 1^-1=1

Question 11

Question 11 : Let A = NxN and * be the binary operation on A defined by �(a,b) * (c,d) =(a+c,b+d)�Show that * is commutative and associative. Find identity element

Solution : Is * commutative ? �(a,b) * (c,d) = (a+c,b+d)�(c,d) * (a,b) = (c+a, b+b) = (a+c, b+d)�Therefore (a,b) * (c,d) = (c,d) * (a,b)

Hence * is commutative

...continued

Is * associative ? �((a,b)*(c,d))*(e,f) = (a+c,b+d) * (e,f) = (a+c +e,b+d+f)�(a,b)*((c,d)*(e,f)) = (a,b) * (c+e,d+f) = (a+c+e,b+d+f)

Therefore ((a,b)*(c,d))*(e,f) = (a,b)*((c,d)*(e,f))�Hence * is associative�Identity element�If possible, let e =(e1,e2) ∈ NxN be the identity element�Therefore, (a,b) * (e1,e2) = (a,b) and (e1,e2) * (a,b) = (a,b) �⇒(a+e1,b+e2) = (a,b) (We will only work with the first condition as * is commutative)�⇒a+e1 = a and b + e2 = b�⇒e1=0 and e2=0�But (0,0) ∉ NxN, hence identity does not exist

Try yourself

• Question 10 (ii), (iii), (vi)
• Question 8
• Question 12�(i) Take the case of + (addition) and check�(ii) Use commutative property