# 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**

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

_{0}(set of non-zero rational numbers) defined by a * b= (3ab/5) for all a, b ∈ Q

_{0}. 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 ∈ Q_{0}

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 ∈ 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

⇒ a * (b * c) = (a * b) * c, for all a, b, c ∈ Q0

So we can say that * is associative on Q_{0}.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 ∈ Q_{0}

a * x = a and x * a = a, ∀ a ∈ Q_{0}

3ax/5 = a and 3xa/5 = a, ∀ a ∈ Q_{0}

x = 5/3 ∀ a ∈ Q_{0}[a ≠ 0]

So we can say that, 5/3 is the identity element in Q_{0}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 = R**_{0} × R, where R_{0} 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) ∈ R_{0} × R.

_{0}× R, where R

_{0}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) ∈ R

_{0}× 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 ∈ R_{0}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 ∈ R_{0}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 ∈ R_{0}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 Q**_{0} of all non zero rational numbers defined by a * b = ab/2 for all a, b ∈ Q_{0}

_{0}of all non zero rational numbers defined by a * b = ab/2 for all a, b ∈ Q

_{0}

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

**(ii) Find the identity element in Q _{0 }.**

**(iii) Find the invertible element of Q _{0}.**

**Solution:**

(i)We have to show,‘*’ is commutative.

Let a, b ∈ Q_{0}.

a o b = ab/2 = ba/2

⇒ b o a

⇒ a o b = b o a, ∀ a, b ∈ Q_{0}.

So, o is commutative on Q_{0.}Now, we will show, ‘*’ is Associative.

Let a, b, c ∈ Q_{0}

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 Q_{0 }with respect to * such that

a o x = a x o a ,∀ a ∈ Q_{0}

⇒ ax /2 = a and xa /2 = a, ∀ a ∈ Q_{0}

x = 2 ∈ Q_{0,}∀ a ∈_{ }Q_{0 }

So, we can say that, 2 is the identity element in Q_{0}with respect to o.

(iii)Let us assume that a ∈ Q_{0}and b ∈ Q_{0}be the inverse of a.

⇒ a o b = e = b o a = e

⇒ ab/2 = 2 and ba/2 = 2

⇒ b = 4/a ∈ Q_{0}

So, we can say that, 4/a is the inverse of a∈ Q_{0}.

**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 R**_{0} denote the set of all non zero real number and let A = R_{0} x R_{0} . If ‘*’ is a binary operation on A defined by ( a, b) * (c ,d) = (ac , bd) for all (a , b)(c , d) ∈ A.

_{0}denote the set of all non zero real number and let A = R

_{0}x R

_{0}. 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.