In this article, we will explore the solutions to Exercise 3.4 from Chapter 3 of RD Sharma's Class 12 Mathematics textbook which focuses on "Binary Operations". This chapter is essential for understanding how operations can be performed on pairs of elements within a set a fundamental concept in algebra and discrete mathematics. The solutions provided here aim to clarify the methods for solving problems related to binary operations ensuring a solid grasp of the topic.
Binary Operations
Binary operations are mathematical procedures that combine two elements from the set to produce another element from the same set. For example: operations like addition, subtraction, multiplication, and division are binary operations where each pair of numbers yields a single result. In the context of binary operations, we explore properties such as associativity, commutativity, and the existence of identity and inverse elements which help in understanding and solving problems related to algebraic structures.
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 *.
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 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 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 ∈ Q0,∀ 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.
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Summary
Chapter 3 of RD Sharma's Class 12 Mathematics textbook focuses on Binary Operations. Exercise 3.4 specifically deals with the properties of binary operations, including closure, commutativity, associativity, and the existence of identity and inverse elements. Students learn to analyze and prove these properties for various binary operations defined on different sets.