Class 12 RD Sharma Solutions – Chapter 3 Binary Operations – Exercise 3.4

Question 1. Let * be a binary operation on Z defined by a * b = a + b – 4 for all a, b ∈ Z.

(i) Show that * is both commutative and associative.

(ii) Find the identity element in Z

(iii) Find the invertible element in Z. 

Solution:

(i) First we will prove commutativity of * 
Let a, b ∈ Z. 
a * b = a + b – 4 
= b + a – 4 
= b * a 

⇒ a * b = b * a, ∀ a, b ∈ Z 
So we can say that, * is commutative on Z. 

Now we will prove associativity of Z. 
Let a, b, c ∈ Z. 
a * (b * c) = a * (b + c – 4) 
= a + b + c -4 – 4 
= a + b + c – 8 
⇒ (a * b) * c = (a + b – 4) * c 
= a + b – 4 + c – 4 
= a + b + c – 8 

⇒ a * (b * c) = (a * b) * c, for all a, b, c ∈ Z 
So we can say that, * is associative on Z.

(ii) We have to find identity element in Z. 
Let x be the identity element in Z with respect to * such that 
a * x = a = x * a ∀ a ∈ Z 
a * x = a and x * a = a, ∀ a ∈ Z 
a + x – 4 = a and x + a – 4 = a, ∀ a ∈ Z 
x = 4, ∀ a ∈ Z 
So we can say that, 4 is the identity element in Z with respect to *.

(iii) We have to find the invertible element in Z. 
Let a ∈ Z and b ∈ Z be the inverse of a. So, 
a * b = x = b * a 
a * b = x and b * a = x 
a + b – 4 = 4 and b + a – 4 = 4 
b = 8 – a ∈ Z 
So we can say that, 8 – a is the inverse of a ∈ Z

Question 2. Let * be a binary operation on Q₀ (set of non-zero rational numbers) defined by a * b= (3ab/5) for all a, b ∈ Q₀. Show that * is commutative as well as associative. Also, find its identity element, if it exists.

Solution:

Firstly we will prove commutativity of * 
Let a, b ∈ Q₀
a * b = (3ab/5) 
= (3ba/5) 
= b * a 
⇒ a * b = b * a, for all a, b ∈ Q_0.

Now we will prove associativity of * 
Let a, b, c ∈ Q₀
a * (b * c) = a * (3bc/5) 
= [a (3 bc/5)] /5 
= 3 abc/25 
(a * b) * c = (3 ab/5) * c 
= [(3 ab/5) c]/ 5 
= 3 abc /25 
⇒ a * (b * c) = (a * b) * c, for all a, b, c ∈ Q0 
So we can say that * is associative on Q₀. 

Now we will find the identity element. 
Let x be the identity element in Z with respect to * such that 
a * x = a = x * a ∀ a ∈ Q₀
a * x = a and x * a = a, ∀ a ∈ Q₀
3ax/5 = a and 3xa/5 = a, ∀ a ∈ Q₀
x = 5/3 ∀ a ∈ Q₀ [a ≠ 0] 
So we can say that, 5/3 is the identity element in Q₀ with respect to *.

Question 3. Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab for all a, b ∈ Q – {-1}. Then,

(i) Show that * is both commutative and associative on Q – {-1}

(ii) Find the identity element in Q – {-1}

(iii) Show that every element of Q – {-1} is invertible. Also, find inverse of an arbitrary element.

Solution:

(i) First we will check commutativity of * 
Let us assume that a, b ∈ Q – {-1} 
a * b = a + b + ab 
= b + a + ba 
= b * a 
⇒ 
a * b = b * a, ∀ a, b ∈ Q – {-1} 

Now we will prove associativity of * 
Let us assume that a, b, c ∈ Q – {-1}, Then, 
a * (b * c) = a * (b + c + b c) 
= a + (b + c + b c) + a (b + c + b c) 
= a + b + c + b c + a b + a c + a b c 
= (a * b) * c = (a + b + a b) * c 
= a + b + a b + c + (a + b + a b) c 
= a + b + a b + c + a c + b c + a b c 
⇒ a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Q – {-1} 
So we can say that, * is associative on Q – {-1} 

(ii) Let us assume that x be the identity element in I+ with respect to * such that 
a * x = a = x * a, ∀ a ∈ Q – {-1} 
a * x = a and x * a = a, ∀ a ∈ Q – {-1} 
a + x + ax = a and x + a + xa = a, ∀ a ∈ Q – {-1} 
x + ax = 0 and x + xa = 0, ∀ a ∈ Q – {-1} 
x (1 + a) = 0 and x (1 + a) = 0, ∀ a ∈ Q – {-1} 
x = 0, ∀ a ∈ Q – {-1} [a ≠ -1] 
so we can say that , 0 is the identity element in Q – {-1} with respect to *. 

(iii) Let us assume that a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. Then, 
a * b = e = b * a 
a * b = e and b * a = e 
a + b + ab = 0 and b + a + ba = 0 
b (1 + a) = – a Q – {-1} 
b = -a/1 + a Q – {-1} [a ≠ -1] 
So we can say that, -a/1 + a is the inverse of a ∈ Q – {-1}.

Question 4. Let A = R₀ × R, where R₀ denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R₀ × R.

(i) Show that ‘O’ is commutative and associative on A

(ii) Find the identity element in A

(iii) Find the invertible element in A

Solution:

(i) Let us assume that X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ R₀ and b, d ∈ R 
X O Y = (ac, bc + d) 
Y O X = (ca, da + b) 
⇒ X O Y = Y O X, ∀ X, Y ∈ A 
⇒ O commutative on A. 

Now we have to check associativity of O 
Let X = (a, b), Y = (c, d) and Z = (e, f), ∀ a, c, e ∈ R0 and b, d, f ∈ R 
⇒X O (Y O Z) = (a, b) O (ce, de + f) 
= (ace, bce + de + f) 
⇒ (X O Y) O Z = (ac, bc + d) O (e, f) 
= (ace, (bc + d) e + f) 
= (ace, bce + de + f) 
⇒ X O (Y O Z) = (X O Y) O Z, ∀ X, Y, Z ∈ A 

(ii) Let us assume that E = (x, y) be the identity element in A with respect to O, ∀ x ∈ R₀ and y ∈ R 
X O E = X = E O X, ∀ X ∈ A 
X O E = X and EOX = X 
⇒(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b) 
We know that , (ax, bx + y) = (a, b) 
ax = a 
x = 1 
bx + y = b 
y = 0 [x = 1] 
we know that, (xa, ya + b) = (a, b) 
xa = a 
x = 1 
ya + b = b 
y = 0 [since x = 1] 
So we can say that (1, 0) is the identity element in A with respect to O. 

(iii) Let us assume that F = (m, n) be the inverse in A ∀ m ∈ R₀ and n ∈ R 
X O F = E and F O X = E 
(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0) 
As we know that (am, bm + n) = (1, 0) 
am = 1 
m = 1/a 
bm + n = 0 
n = -b/a [m = 1/a] 
We know that (ma, na + b) = (1, 0) 
ma = 1 
m = 1/a 
na + b = 0 
n = -b/a 
So we can say that, the inverse of (a, b) ∈ A with respect to O is (1/a, -1/a).

Question 5. Let ‘*’ be a binary operation on the set of Q₀ of all non zero rational numbers defined by a * b = ab/2 for all a, b ∈ Q₀

(i) show that ‘*’ is both commutative and associative.

(ii) Find the identity element in Q₀ .

(iii) Find the invertible element of Q₀.

Solution:

(i) We have to show, ‘*’ is commutative. 
Let a, b ∈ Q₀. 
a o b = ab/2 = ba/2 
⇒ b o a 
⇒ a o b = b o a, ∀ a, b ∈ Q₀. 
So, o is commutative on Q_0.

Now, we will show, ‘*’ is Associative. 
Let a, b, c ∈ Q₀
a o (b 0 c) = a o (bc/2) 
= (a(bc/2))/2 
= abc /4 
⇒ (a o b) o c = (ab/2) o c 
= abc/4 
⇒ a o (b o c) = (a o b) o c ∀ a, b, c ∈ Q_0.
So, we can say that o is associative on Q_0.

(ii) Let x be the identify element in Q₀ with respect to * such that 
a o x = a x o a ,∀ a ∈ Q₀
⇒ ax /2 = a and xa /2 = a, ∀ a ∈ Q₀
x = 2 ∈ Q_0,∀ a ∈ Q₀ 
So, we can say that, 2 is the identity element in Q₀ with respect to o. 

(iii) Let us assume that a ∈ Q₀ and b ∈ Q₀ be the inverse of a. 
⇒ a o b = e = b o a = e 
⇒ ab/2 = 2 and ba/2 = 2 
⇒ b = 4/a ∈ Q₀
So, we can say that, 4/a is the inverse of a∈ Q₀.

Question 6. On R -{1}, a binary operation * is defined by a*b = a+b-ab . Prove that * is commutative and associative. Find the identity element for * on R-{1}. Also, prove that every element of R-{1} is invertible.

Solution:

Firstly we will find commutative. 
Let us assume that a, b ∈ R -{1} 
a * b = a + b – ab 
= b + a -ba 
= b*a 
⇒ a * b = b + a ,∀ a , b ∈ R – {1} 
So , we can say that * is commutative on R-{1} 

Now , we will find Associative. 
Let assume that a , b , c ∈ R – {1} 
a * (b * c ) = a * (b + c – bc) 
=a + b + c – bc -a(b + c – bc) 
=a + b + c – bc – ab – ac + abc 
(a * b) * c = (a + b – ab ) * c 
= a + b – ab + c – (a + b – ab)c 
= a + b + c – ab – ac – bc + abc 
⇒ a * (b * c) = (a * c )* c , ∀ a , b , c ∈ R – {1} 
So we can say that , * is associative on R-{1} 

Now we will find identity element. 
Let assume that x be the identity element in R-{1} with respect to * 
a * x = a = x * a , ∀ a ∈ R-{1} 
a * x = a and x * a = a, ∀ a ∈ R-{1} 
⇒ a + x – ax = a and x + a – xa = a , ∀ a ∈ R-{1} 
x(1 – a) = 0 , ∀ a ∈ R-{1} 
⇒ x = 0 [ a ≠ 1 ⇒ 1 – a ≠ 0 ] 
So we can say that , x = 0 will be the identity element with respect to * . 

Now lets find inverse element. 
Let’s assume that b ∈ R-{1} be the inverse element of a ∈ R-{1} 
a * b = b * a = x 
⇒ a + b -ab = 0 [e=0] 
⇒b(1 – a) = -a 
⇒ b = -a /(1 – a) ≠ 1 [ if -a/(1-a) = 1 ⇒ -a = 1 – a ⇒ 1≠ 0] 
So we can say that , b = -a/(1 – a) is the inverse of a ∈ R-{1} with respect to *.

Question 7.Let R₀ denote the set of all non zero real number and let A = R₀ x R₀ . If  ‘*’ is a binary operation on A defined by ( a, b) * (c ,d) = (ac , bd) for all (a , b)(c , d)  ∈  A.

(i) Show that ‘*’ is both commutative and associative on A.

(ii) Find the identity element in A.

(iii) Find the invertible element in A.

Solution:

In the question we have given (a, b) * (c ,d) = (ac , bd) for all (a,b)(c,d) ∈ A. 
(i) Let us assume that , (a,b)(c,d) ∈ A. So, 
(a, b) * (c ,d) = (ac , bd) 
=(ca , bd) [ ac = ca and bd = db ] 
=(c , d)*(a , b) 
⇒ (a, b) * (c,d) = (ac,bd) 
So we can say that , ‘*’ is commutative on A. 

⇒ Now we will find associativity on A. 
Let us assume that , (a,b),(c,d),(e,f) ∈ A. 
⇒ ((a,b)*(c,d))*(e,f) = (ac , bd)*(e,f) 
=(ace , bdf) –(i) 
Now (a,b)*((c,d)*(e,f)) =(a,b)*(ce,df) 
=(ace , bdf) –(ii) 
From equation (i) and (ii). 
((a,b)*(c,d))*(e,f) = (a,b)*((c,d)*(e,f)) 
So we can say that , ‘*’ is associative on A. 

(ii) Let find identity element in A. 
Let assume that (x,y) ∈ A be the identity element with respect to *. 
(a,b) * (x,y) = (x,y)*(a,b) = (a,b) for all (a,b) ∈ A. 
⇒ (ax , by) = (a,b) 
⇒ ax = a & by = b 
⇒ x = 1 & y = 1 
So we can say that (1,1) will be identity element. 

(iii) Now we will find invertible element in A. 
Let assume that (c,d) ∈ A be the inverse of (a,b) ∈ A 
(a,b)*(c,d) = (c,d)*(a,b) = x 
(ac , bd) = (1,1) [e = (1,1) ] 
ac = 1 & bd = 1 
c = 1/a & d = 1/b 
So we can say that (1/a ,1/b) will be the inverse of (a,b) with respect to *.

Question 8. Let * be the binary operation on N defined by a*b = H.C.F of a and b. 

Is  * commutative? Is * associative? Does there exist identity for this binary operation on N?

Solution:

The binary operation * on N can be defined as: 
a*b = H.C.F of a and b 
And we also know that , HCF(a,b) = HCF(b,a) . a,b ∈ N. 
So we can say that , a * b = b * a 
So , the operation * is commutative. 

For a,b,c ∈ N. So we have. 
(a * b) * c = (HCF(a,b))*c = HCF(a,b,c) 
a * (b * c) = a * (HCF(a,b)) = HCF(a,b,c) 
So it can be said that (a * b) * c = a * (b * c) 
So we can say that , the operation * is associative. 

Now , an element e ∈ N will be the identity for the operation. 
* if a * e = a = e * a ,∀ a ∈ N. 
But we can say that , this relation is not true for any a ∈ N. 
So we can say that , the operation * does not have any identity in N.