# RD Sharma Class 12 Ex 3.3 Solutions Chapter 3 Binary Operations

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### 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 ∈ Q0.

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 *.

### Question3. 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 Q.

(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 Q0.

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 ∈ Q0.
So, we can say that o is associative on Q0.

(ii) Let x be the identify element in Qwith respect to * such that
a o x = a x o a ,∀ a ∈ Q0
⇒ ax /2 = a and xa /2 = a, ∀ a ∈ Q0
x = 2 ∈ Q0,∀ a ∈ Q
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 *.

### 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.

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