**RD Sharma Solutions for Class 12 Maths Exercise 3.4 Chapter 3 Binary Operations** is provided here for students to study and excel in their main exams. This exercise deals with the inverse of an element of binary operations. It mainly consists of a set of examples before the exercise problems to make it possible for students to solve questions effortlessly. Students can access exercise-wise solutions in PDF format to solve problems of **RD Sharma textbook** easily. It improves problem-solving abilities of students, which are important from the exam point of view. RD Sharma Solutions for Class 12 Maths Chapter 3 Binary Operations Exercise 3.4 PDF are available here.

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**1. Let * be a binary operation on Z defined by a * b = a + b â€“ 4 for all a, b âˆˆ 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 have to prove commutativity of *

Let a, b âˆˆ Z. then,

a * b = a + b â€“ 4

= b + a â€“ 4

= b * a

Therefore,

a * b = b * a, âˆ€ a, b âˆˆ Z

Thus, * is commutative on Z.

Now we have to prove associativity of Z.

Let a, b, c âˆˆ Z. then,

a * (b * c) = a * (b + c – 4)

= a + b + c -4 â€“ 4

= a + b + c â€“ 8

(a * b) * c = (a + b â€“ 4) * c

= a + b â€“ 4 + c â€“ 4

= a + b + c â€“ 8

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c âˆˆ Z

Thus, * is associative on Z.

(ii) Let e be the identity element in Z with respect to * such that

a * e = a = e * a âˆ€ a âˆˆ Z

a * e = a and e * a = a, âˆ€ a âˆˆ Z

a + e â€“ 4 = a and e + a â€“ 4 = a, âˆ€ a âˆˆ Z

e = 4, âˆ€ a âˆˆ Z

Thus, 4 is the identity element in Z with respect to *.

(iii) Let a âˆˆ Z and b âˆˆ Z be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b â€“ 4 = 4 and b + a â€“ 4 = 4

b = 8 â€“ a âˆˆ Z

Thus, 8 â€“ a is the inverse of a âˆˆ Z

**2. Let * be a binary operation on Q _{0} (set of non-zero rational numbers) defined by a * b = (3ab/5) for all a, b âˆˆ Q_{0}. Show that * is commutative as well as associative. Also, find its identity element, if it exists.**

**Solution:**

First we have to prove commutativity of *

Let a, b âˆˆ Q_{0}

a * b = (3ab/5)

= (3ba/5)

= b * a

Therefore, a * b = b * a, for all a, b âˆˆ Q_{0}

Now we have to prove associativity of *

Let a, b, c âˆˆ Q_{0}

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

Therefore a * (b * c) = (a * b) * c, for all a, b, c âˆˆ Q_{0}

Thus * is associative on Q_{0}

Now we have to find the identity element

Let e be the identity element in Z with respect to * such that

a * e = a = e * a âˆ€ a âˆˆ Q_{0}

a * e = a and e * a = a, âˆ€ a âˆˆ Q_{0}

3ae/5 = a and 3ea/5 = a, âˆ€ a âˆˆ Q_{0}

e = 5/3 âˆ€ a âˆˆ Q_{0 }[because a is not equal to 0]

Thus, 5/3 is the identity element in Q_{0} with respect to *.

**3. Let * be a binary operation on Q â€“ {-1} defined by a * b = a + b + ab for all a, b âˆˆ Q â€“ {-1}. Then,**

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

**(ii) Find the identity element in Q â€“ {-1}**

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

**Solution:**

(i) First we have to check commutativity of *

Let a, b âˆˆ Q â€“ {-1}

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, âˆ€ a, b âˆˆ Q â€“ {-1}

Now we have to prove associativity of *

Let a, b, c âˆˆ Q â€“ {-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

Therefore,

a * (b * c) = (a * b) * c, âˆ€ a, b, c âˆˆ Q â€“ {-1}

Thus, * is associative on Q â€“ {-1}.

(ii) Let e be the identity element in I^{+} with respect to * such that

a * e = a = e * a, âˆ€ a âˆˆ Q â€“ {-1}

a * e = a and e * a = a, âˆ€ a âˆˆ Q â€“ {-1}

a + e + ae = a and e + a + ea = a, âˆ€ a âˆˆ Q â€“ {-1}

e + ae = 0 and e + ea = 0, âˆ€ a âˆˆ Q â€“ {-1}

e (1 + a) = 0 and e (1 + a) = 0, âˆ€ a âˆˆ Q â€“ {-1}

e = 0, âˆ€ a âˆˆ Q â€“ {-1} [because a not equal to -1]

Thus, 0 is the identity element in Q â€“ {-1} with respect to *.

(iii) Let a âˆˆ Q â€“ {-1} and b âˆˆ Q â€“ {-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 â€“ {-1}

b = -a/1 + a Q â€“ {-1} [because a not equal to -1]

Thus, -a/1 + a is the inverse of a âˆˆ Q â€“ {-1}

**4. LetÂ AÂ =Â R _{0}Â Ã—Â R, whereÂ R_{0}Â denote the set of all non-zero real numbers.Â AÂ binary operation ‘O’ is defined onÂ AÂ as follows: (a,Â b) O (c,Â d) = (ac,Â bcÂ +Â d) for all (a,Â b), (c,Â d) âˆˆÂ R_{0}Â Ã—Â R.**

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

**(ii) Find the identity element in A**

**(iii) Find the invertible element in A.**

**Solution:**

(i) Let X = (a, b) and Y = (c, d) âˆˆ A, âˆ€ a, c âˆˆ R_{0 } and b, d âˆˆ R

Then, X O Y = (ac, bc + d)

And Y O X = (ca, da + b)

Therefore,

X O Y = Y O X, âˆ€ X, Y âˆˆ A

Thus, O is not commutative on A.

Now we have to check associativity of O

Let X = (a, b), Y = (c, d) and Z = (e, f), âˆ€ a, c, e âˆˆ R_{0 } and b, d, f âˆˆ 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)

Therefore, X O (Y O Z) = (X O Y) O Z, âˆ€ X, Y, Z âˆˆ A

(ii) Let E = (x, y) be the identity element in A with respect to O, âˆ€ x âˆˆ R_{0 } and y âˆˆ R

Such that,

X O E = X = E O X, âˆ€ X âˆˆ A

X O E = X and EOX = X

(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b)

Considering (ax, bx + y) = (a, b)

ax = a

x = 1

And bx + y = b

y = 0 [since x = 1]

Considering (xa, ya + b) = (a, b)

xa = a

x = 1

And ya + b = b

y = 0 [since x = 1]

Therefore (1, 0) is the identity element in A with respect to O.

(iii) Let F = (m, n) be the inverse in A âˆ€ m âˆˆ R_{0 } and n âˆˆ R

X O F = E and F O X = E

(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0)

Considering (am, bm + n) = (1, 0)

am = 1

m = 1/a

And bm + n = 0

n = -b/a [since m = 1/a]

Considering (ma, na + b) = (1, 0)

ma = 1

m = 1/a

And na + b = 0

n = -b/a

Therefore the inverse of (a, b) âˆˆ A with respect to O is (1/a, -b/a)