{"id":539893,"date":"2021-09-25T07:59:16","date_gmt":"2021-09-25T07:59:16","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=539893"},"modified":"2022-12-26T09:19:17","modified_gmt":"2022-12-26T09:19:17","slug":"rd-sharma-solutions-for-class-12-maths-chapter-3-binary-operations","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-3-binary-operations\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Chapter 3\u2013Binary Operations"},"content":{"rendered":"\n

Class 12: Maths Chapter 3 solutions. Complete Class 12 Maths Chapter 3 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 3\u2013Binary Operations<\/h2>\n\n\n\n

RD Sharma 12th Maths Chapter 3, Class 12 Maths Chapter 3 solutions<\/p>\n\n\n\n

Exercise 3.1 Page No: 3.4<\/p>\n\n\n\n

1. Determine whether the following operation define a binary operation on the given set or not:<\/strong><\/p>\n\n\n\n

(i) \u2018*\u2019 on N defined by a * b = ab<\/sup> for all a, b \u2208 N.<\/strong><\/p>\n\n\n\n

(ii) \u2018O\u2019 on Z defined by a O b = ab<\/sup> for all a, b \u2208 Z.<\/strong><\/p>\n\n\n\n

(iii)  \u2018*\u2019 on N defined by a * b = a + b \u2013 2 for all a, b \u2208 N<\/strong><\/p>\n\n\n\n

(iv) \u2018\u00d76<\/sub>\u2018 on S = {1, 2, 3, 4, 5} defined by a \u00d76<\/sub> b = Remainder when a b is divided by 6.<\/strong><\/p>\n\n\n\n

(v) \u2018+6<\/sub>\u2019 on S = {0, 1, 2, 3, 4, 5} defined by a +6<\/sub> b<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vi) \u2018\u2299\u2019 on N defined by a \u2299 b= ab<\/sup> + ba<\/sup> for all a, b \u2208 N<\/strong><\/p>\n\n\n\n

(vii) \u2018*\u2019 on Q defined by a * b = (a \u2013 1)\/ (b + 1) for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

(i) Given \u2018*\u2019 on N defined by a * b = ab<\/sup> for all a, b \u2208 N.<\/p>\n\n\n\n

Let a, b \u2208 N. Then,<\/p>\n\n\n\n

ab <\/sup>\u2208 N      [\u2235 ab<\/sup>\u22600 and a, b is positive integer]<\/p>\n\n\n\n

\u21d2 a * b \u2208 N<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b \u2208 N, \u2200 a, b \u2208 N<\/p>\n\n\n\n

Thus, * is a binary operation on N.<\/p>\n\n\n\n

(ii) Given \u2018O\u2019 on Z defined by a O b = ab<\/sup> for all a, b \u2208 Z.<\/p>\n\n\n\n

Both a = 3 and b = -1 belong to Z.<\/p>\n\n\n\n

\u21d2 a * b = 3-1<\/sup><\/p>\n\n\n\n

= 1\/3 \u2209 Z<\/p>\n\n\n\n

Thus, * is not a binary operation on Z.<\/p>\n\n\n\n

(iii)  Given \u2018*\u2019 on N defined by a * b = a + b \u2013 2 for all a, b \u2208 N<\/p>\n\n\n\n

If a = 1 and b = 1,<\/p>\n\n\n\n

a * b = a + b \u2013 2<\/p>\n\n\n\n

= 1 + 1 \u2013 2<\/p>\n\n\n\n

= 0 \u2209 N<\/p>\n\n\n\n

Thus, there exist a = 1 and b = 1 such that a * b \u2209 N<\/p>\n\n\n\n

So, * is not a binary operation on N.<\/p>\n\n\n\n

(iv) Given \u2018\u00d76<\/sub>\u2018 on S = {1, 2, 3, 4, 5} defined by a \u00d76<\/sub> b = Remainder when a b is divided by 6.<\/p>\n\n\n\n

Consider the composition table,<\/p>\n\n\n\n

X6<\/sub><\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
1<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
2<\/td>2<\/td>4<\/td>0<\/td>2<\/td>4<\/td><\/tr>
3<\/td>3<\/td>0<\/td>3<\/td>0<\/td>3<\/td><\/tr>
4<\/td>4<\/td>2<\/td>0<\/td>4<\/td>2<\/td><\/tr>
5<\/td>5<\/td>4<\/td>3<\/td>2<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Here all the elements of the table are not in S.<\/p>\n\n\n\n

\u21d2 For a = 2 and b = 3,<\/p>\n\n\n\n

a \u00d76<\/sub> b = 2 \u00d76<\/sub> 3 = remainder when 6 divided by 6 = 0 \u2260 S<\/p>\n\n\n\n

Thus, \u00d76<\/sub> is not a binary operation on S.<\/p>\n\n\n\n

(v) Given \u2018+6<\/sub>\u2019 on S = {0, 1, 2, 3, 4, 5} defined by a +6<\/sub> b<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Consider the composition table,<\/p>\n\n\n\n

+6<\/sub><\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
0<\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
1<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td>0<\/td><\/tr>
2<\/td>2<\/td>3<\/td>4<\/td>5<\/td>0<\/td>1<\/td><\/tr>
3<\/td>3<\/td>4<\/td>5<\/td>0<\/td>1<\/td>2<\/td><\/tr>
4<\/td>4<\/td>5<\/td>0<\/td>1<\/td>2<\/td>3<\/td><\/tr>
5<\/td>5<\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Here all the elements of the table are not in S.<\/p>\n\n\n\n

\u21d2 For a = 2 and b = 3,<\/p>\n\n\n\n

a \u00d76<\/sub> b = 2 \u00d76<\/sub> 3 = remainder when 6 divided by 6 = 0 \u2260 Thus, \u00d76<\/sub> is not a binary operation on S.<\/p>\n\n\n\n

(vi) Given \u2018\u2299\u2019 on N defined by a \u2299 b= ab<\/sup> + ba<\/sup> for all a, b \u2208 N<\/p>\n\n\n\n

Let a, b \u2208 N. Then,<\/p>\n\n\n\n

ab<\/sup>, ba<\/sup> \u2208 N<\/p>\n\n\n\n

\u21d2 ab<\/sup> + ba<\/sup> \u2208 N      [\u2235Addition is binary operation on N]<\/p>\n\n\n\n

\u21d2 a \u2299 b \u2208 N<\/p>\n\n\n\n

Thus, \u2299 is a binary operation on N.<\/p>\n\n\n\n

(vii) Given \u2018*\u2019 on Q defined by a * b = (a \u2013 1)\/ (b + 1) for all a, b \u2208 Q<\/p>\n\n\n\n

If a = 2 and b = -1 in Q,<\/p>\n\n\n\n

a * b = (a \u2013 1)\/ (b + 1)<\/p>\n\n\n\n

= (2 \u2013 1)\/ (- 1 + 1)<\/p>\n\n\n\n

= 1\/0 [which is not defined]<\/p>\n\n\n\n

For a = 2 and b = -1<\/p>\n\n\n\n

a * b does not belongs to Q<\/p>\n\n\n\n

So, * is not a binary operation in Q.<\/p>\n\n\n\n

2. Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
(i) On Z+<\/sup>, defined * by a * b = a \u2013 b<\/strong><\/p>\n\n\n\n

(ii) On Z+<\/sup>, define * by a*b = ab<\/strong><\/p>\n\n\n\n

(iii) On R, define * by a*b = ab2<\/sup><\/strong><\/p>\n\n\n\n

(iv) On Z+<\/sup> define * by a * b = |a \u2212 b|<\/strong><\/p>\n\n\n\n

(v) On Z+ <\/sup>define * by a * b = a<\/strong><\/p>\n\n\n\n

(vi) On R, define * by a * b = a + 4b2<\/sup><\/strong><\/p>\n\n\n\n

Here, Z<\/em>+<\/sup> denotes the set of all non-negative integers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given On Z<\/em>+<\/sup>, defined * by a * b = a \u2013 b<\/p>\n\n\n\n

If a = 1 and b = 2 in Z+<\/sup>, then<\/p>\n\n\n\n

a * b = a \u2013 b<\/p>\n\n\n\n

= 1 \u2013 2<\/p>\n\n\n\n

= -1 \u2209 Z+ <\/sup>[because Z+<\/sup> is the set of non-negative integers]<\/p>\n\n\n\n

For a = 1 and b = 2,<\/p>\n\n\n\n

a * b \u2209 Z+<\/sup><\/p>\n\n\n\n

Thus, * is not a binary operation on Z+<\/sup>.<\/p>\n\n\n\n

(ii) Given Z+<\/sup>, define * by a*b = a b<\/p>\n\n\n\n

Let a, b \u2208 Z+<\/sup><\/p>\n\n\n\n

\u21d2 a, b \u2208 Z+<\/sup><\/p>\n\n\n\n

\u21d2 a * b \u2208 Z+<\/sup><\/p>\n\n\n\n

Thus, * is a binary operation on R.<\/p>\n\n\n\n

(iii) Given on R, define by a*b = ab2<\/sup><\/p>\n\n\n\n

Let a, b \u2208 R<\/p>\n\n\n\n

\u21d2 a, b2<\/sup> \u2208 R<\/p>\n\n\n\n

\u21d2 ab2<\/sup> \u2208 R<\/p>\n\n\n\n

\u21d2 a * b \u2208 R<\/p>\n\n\n\n

Thus, * is a binary operation on R.<\/p>\n\n\n\n

(iv) Given on Z+<\/sup> define * by a * b = |a \u2212 b|<\/p>\n\n\n\n

Let a, b \u2208 Z+<\/sup><\/p>\n\n\n\n

\u21d2 | a \u2013 b | \u2208 Z+<\/sup><\/p>\n\n\n\n

\u21d2 a * b \u2208 Z+<\/sup><\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b \u2208 Z+<\/sup>, \u2200 a, b \u2208 Z+<\/sup><\/p>\n\n\n\n

Thus, * is a binary operation on Z+<\/sup>.<\/p>\n\n\n\n

(v) Given on Z+ <\/sup>define * by a * b = a<\/p>\n\n\n\n

Let a, b \u2208 Z+<\/sup><\/p>\n\n\n\n

\u21d2 a \u2208 Z+<\/sup><\/p>\n\n\n\n

\u21d2 a * b \u2208 Z+<\/sup><\/p>\n\n\n\n

Therefore, a * b \u2208 Z+<\/sup> \u2200 a, b \u2208 Z+<\/sup><\/p>\n\n\n\n

Thus, * is a binary operation on Z+<\/sup>.<\/p>\n\n\n\n

(vi) Given On R, define * by a * b = a + 4b2<\/sup><\/p>\n\n\n\n

Let a, b \u2208 R<\/p>\n\n\n\n

\u21d2 a, 4b2<\/sup> \u2208 R<\/p>\n\n\n\n

\u21d2 a + 4b2<\/sup> \u2208 R<\/p>\n\n\n\n

\u21d2 a * b \u2208 R<\/p>\n\n\n\n

Therefore, a *b \u2208 R, \u2200 a, b \u2208 R<\/p>\n\n\n\n

Thus, * is a binary operation on R.<\/p>\n\n\n\n

3. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b \u2212 3. Find the value of 3 * 4.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given a<\/em> * b<\/em> = 2a<\/em> + b<\/em> \u2013 3<\/p>\n\n\n\n

3 * 4 = 2 (3) + 4 \u2013 3<\/p>\n\n\n\n

= 6 + 4 \u2013 3<\/p>\n\n\n\n

= 7<\/p>\n\n\n\n

4. Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

LCM<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
1<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
2<\/td>2<\/td>2<\/td>6<\/td>4<\/td>10<\/td><\/tr>
3<\/td>3<\/td>5<\/td>3<\/td>12<\/td>15<\/td><\/tr>
4<\/td>4<\/td>4<\/td>12<\/td>4<\/td>20<\/td><\/tr>
5<\/td>5<\/td>10<\/td>15<\/td>20<\/td>5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.<\/p>\n\n\n\n

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 \u2209 {1, 2, 3, 4, 5}.<\/p>\n\n\n\n

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.<\/p>\n\n\n\n

5. Let S = {a, b, c}. Find the total number of binary operations on S<\/em>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Number of binary operations on a set with n elements is nn2nn2<\/p>\n\n\n\n

Here, S = {a, b, c}<\/p>\n\n\n\n

Number of elements in S = 3<\/p>\n\n\n\n

Number of binary operations on a set with 3 elements is 332332<\/p>\n\n\n\n

\n\n\n\n

Exercise 3.2 Page No: 3.12<\/p>\n\n\n\n

1. Let \u2018*\u2019 be a binary operation on N defined by a * b = l.c.m. (a, b) for all a, b \u2208 N
(i) Find 2 * 4, 3 * 5, 1 * 6.<\/strong><\/p>\n\n\n\n

(ii) Check the commutativity and associativity of \u2018*\u2019 on N.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given a * b = 1.c.m. (a, b)<\/p>\n\n\n\n

2 * 4 = l.c.m. (2, 4)<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

3 * 5 = l.c.m. (3, 5)<\/p>\n\n\n\n

= 15<\/p>\n\n\n\n

1 * 6 = l.c.m. (1, 6)<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

(ii) We have to prove commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 N<\/p>\n\n\n\n

a * b = l.c.m (a, b)<\/p>\n\n\n\n

= l.c.m (b, a)<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

a * b = b * a \u2200 a, b \u2208 N<\/p>\n\n\n\n

Thus * is commutative on N.<\/p>\n\n\n\n

Now we have to prove associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 N<\/p>\n\n\n\n

a * (b * c ) = a * l.c.m. (b, c)<\/p>\n\n\n\n

= l.c.m. (a, (b, c))<\/p>\n\n\n\n

= l.c.m (a, b, c)<\/p>\n\n\n\n

(a * b) * c = l.c.m. (a, b) * c<\/p>\n\n\n\n

= l.c.m. ((a, b), c)<\/p>\n\n\n\n

= l.c.m. (a, b, c)<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

(a * (b * c) = (a * b) * c, \u2200 a, b , c \u2208 N<\/p>\n\n\n\n

Thus, * is associative on N.<\/p>\n\n\n\n

2. Determine which of the following binary operation is associative and which is commutative:<\/strong><\/p>\n\n\n\n

(i) * on N defined by a * b = 1 for all a, b \u2208 N<\/strong><\/p>\n\n\n\n

(ii) * on Q defined by a * b = (a + b)\/2 for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We have to prove commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 N<\/p>\n\n\n\n

a * b = 1<\/p>\n\n\n\n

b * a = 1<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b = b * a, for all a, b \u2208 N<\/p>\n\n\n\n

Thus * is commutative on N.<\/p>\n\n\n\n

Now we have to prove associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 N<\/p>\n\n\n\n

Then a * (b * c) = a * (1)<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

(a * b) *c = (1) * c<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

Therefore a * (b * c) = (a * b) *c for all a, b, c \u2208 N<\/p>\n\n\n\n

Thus, * is associative on N.<\/p>\n\n\n\n

(ii) First we have to prove commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 N<\/p>\n\n\n\n

a * b = (a + b)\/2<\/p>\n\n\n\n

= (b + a)\/2<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore, a * b = b * a, \u2200 a, b \u2208 N<\/p>\n\n\n\n

Thus * is commutative on N.<\/p>\n\n\n\n

Now we have to prove associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 N<\/p>\n\n\n\n

a * (b * c) = a * (b + c)\/2<\/p>\n\n\n\n

= [a + (b + c)]\/2<\/p>\n\n\n\n

= (2a + b + c)\/4<\/p>\n\n\n\n

Now, (a * b) * c = (a + b)\/2 * c<\/p>\n\n\n\n

= [(a + b)\/2 + c] \/2<\/p>\n\n\n\n

= (a + b + 2c)\/4<\/p>\n\n\n\n

Thus, a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

If a = 1, b= 2, c = 3<\/p>\n\n\n\n

1 * (2 * 3) = 1 * (2 + 3)\/2<\/p>\n\n\n\n

= 1 * (5\/2)<\/p>\n\n\n\n

= [1 + (5\/2)]\/2<\/p>\n\n\n\n

= 7\/4<\/p>\n\n\n\n

(1 * 2) * 3 = (1 + 2)\/2 * 3<\/p>\n\n\n\n

= 3\/2 * 3<\/p>\n\n\n\n

= [(3\/2) + 3]\/2<\/p>\n\n\n\n

= 4\/9<\/p>\n\n\n\n

Therefore, there exist a = 1, b = 2, c = 3 \u2208 N such that a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on N.<\/p>\n\n\n\n

3. Let A be any set containing more than one element. Let \u2018*\u2019 be a binary operation on A defined by a * b = b for all a, b \u2208 A Is \u2018*\u2019 commutative or associative on A?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let a, b \u2208 A<\/p>\n\n\n\n

Then, a * b = b<\/p>\n\n\n\n

b * a = a<\/p>\n\n\n\n

Therefore a * b \u2260 b * a<\/p>\n\n\n\n

Thus, * is not commutative on A<\/p>\n\n\n\n

Now we have to check associativity:<\/p>\n\n\n\n

Let a, b, c \u2208 A<\/p>\n\n\n\n

a * (b * c) = a * c<\/p>\n\n\n\n

= c<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

a * (b * c) = (a * b) * c, \u2200 a, b, c \u2208 A<\/p>\n\n\n\n

Thus, * is associative on A<\/p>\n\n\n\n

4. Check the commutativity and associativity of each of the following binary operations:<\/strong><\/p>\n\n\n\n

(i) \u2018*\u2019 on Z defined by a * b = a + b + a b for all a, b \u2208 Z <\/strong><\/p>\n\n\n\n

(ii) \u2018*\u2019 on N defined by a * b = 2ab<\/sup> for all a, b \u2208 N<\/strong><\/p>\n\n\n\n

(iii) \u2018*\u2019 on Q defined by a * b = a \u2013 b for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(iv) \u2018\u2299\u2019 on Q defined by a \u2299 b = a2<\/sup> + b2<\/sup> for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(v) \u2018o\u2019 on Q defined by a o b = (ab\/2) for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(vi) \u2018*\u2019 on Q defined by a * b = ab2<\/sup> for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(vii) \u2018*\u2019 on Q defined by a * b = a + a b for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(viii) \u2018*\u2019 on R defined by a * b = a + b -7 for all a, b \u2208 R<\/strong><\/p>\n\n\n\n

(ix) \u2018*\u2019 on Q defined by a * b = (a \u2013 b)2<\/sup> for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(x) \u2018*\u2019 on Q defined by a * b = a b + 1 for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(xi) \u2018*\u2019 on N defined by a * b = ab<\/sup> for all a, b \u2208 N<\/strong><\/p>\n\n\n\n

(xii) \u2018*\u2019 on Z defined by a * b = a \u2013 b for all a, b \u2208 Z<\/strong><\/p>\n\n\n\n

(xiii) \u2018*\u2019 on Q defined by a * b = (ab\/4) for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

(xiv) \u2018*\u2019 on Z defined by a * b = a + b \u2013 ab for all a, b \u2208 Z<\/strong><\/p>\n\n\n\n

(xv) \u2018*\u2019 on Q defined by a * b = gcd (a, b) for all a, b \u2208 Q<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Z<\/p>\n\n\n\n

Then a * b = a + b + ab<\/p>\n\n\n\n

= b + a + ba<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b = b * a, \u2200 a, b \u2208 Z<\/p>\n\n\n\n

Now we have to prove associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Z, Then,<\/p>\n\n\n\n

a * (b * c) = a * (b + c + b c)<\/p>\n\n\n\n

= a + (b + c + b c) + a (b + c + b c)<\/p>\n\n\n\n

= a + b + c + b c + a b + a c + a b c<\/p>\n\n\n\n

(a * b) * c = (a + b + a b) * c<\/p>\n\n\n\n

= a + b + a b + c + (a + b + a b) c<\/p>\n\n\n\n

= a + b + a b + c + a c + b c + a b c<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c) = (a * b) * c, \u2200 a, b, c \u2208 Z<\/p>\n\n\n\n

Thus, * is associative on Z.<\/p>\n\n\n\n

(ii) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 N<\/p>\n\n\n\n

a * b = 2ab<\/sup><\/p>\n\n\n\n

= 2ba<\/sup><\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore, a * b = b * a, \u2200 a, b \u2208 N<\/p>\n\n\n\n

Thus, * is commutative on N<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 N<\/p>\n\n\n\n

Then, a * (b * c) = a * (2bc<\/sup>)<\/p>\n\n\n\n

=2a\u22172bc2a\u22172bc<\/p>\n\n\n\n

(a * b) * c = (2ab<\/sup>) * c<\/p>\n\n\n\n

=2ab\u22172c2ab\u22172c<\/p>\n\n\n\n

Therefore, a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on N<\/p>\n\n\n\n

(iii) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a * b = a \u2013 b<\/p>\n\n\n\n

b * a = b \u2013 a<\/p>\n\n\n\n

Therefore, a * b \u2260 b * a<\/p>\n\n\n\n

Thus, * is not commutative on Q<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a * (b * c) = a * (b \u2013 c)<\/p>\n\n\n\n

= a \u2013 (b \u2013 c)<\/p>\n\n\n\n

= a \u2013 b + c<\/p>\n\n\n\n

(a * b) * c = (a \u2013 b) * c<\/p>\n\n\n\n

= a \u2013 b \u2013 c<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on Q<\/p>\n\n\n\n

(iv) First we have to check commutativity of \u2299<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a \u2299 b = a2<\/sup> + b2<\/sup><\/p>\n\n\n\n

= b2<\/sup> + a2<\/sup><\/p>\n\n\n\n

= b \u2299 a<\/p>\n\n\n\n

Therefore, a \u2299 b = b \u2299 a, \u2200 a, b \u2208 Q<\/p>\n\n\n\n

Thus, \u2299 on Q<\/p>\n\n\n\n

Now we have to check associativity of \u2299<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a \u2299 (b \u2299 c) = a \u2299 (b2<\/sup> + c2<\/sup>)<\/p>\n\n\n\n

= a2<\/sup> + (b2<\/sup> + c2<\/sup>)2<\/sup><\/p>\n\n\n\n

= a2<\/sup> + b4<\/sup> + c4<\/sup> + 2b2<\/sup>c2<\/sup><\/p>\n\n\n\n

(a \u2299 b) \u2299 c = (a2<\/sup> + b2<\/sup>) \u2299 c<\/p>\n\n\n\n

= (a2<\/sup> + b2<\/sup>)2<\/sup> + c2<\/sup><\/p>\n\n\n\n

= a4<\/sup> + b4<\/sup> + 2a2<\/sup>b2<\/sup> + c2<\/sup><\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

(a \u2299 b) \u2299 c \u2260 a \u2299 (b \u2299 c)<\/p>\n\n\n\n

Thus, \u2299 is not associative on Q.<\/p>\n\n\n\n

(v) First we have to check commutativity of o<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a o b = (ab\/2)<\/p>\n\n\n\n

= (b a\/2)<\/p>\n\n\n\n

= b o a<\/p>\n\n\n\n

Therefore, a o b = b o a, \u2200 a, b \u2208 Q<\/p>\n\n\n\n

Thus, o is commutative on Q<\/p>\n\n\n\n

Now we have to check associativity of o<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a o (b o c) = a o (b c\/2)<\/p>\n\n\n\n

= [a (b c\/2)]\/2<\/p>\n\n\n\n

= [a (b c\/2)]\/2<\/p>\n\n\n\n

= (a b c)\/4<\/p>\n\n\n\n

(a o b) o c = (ab\/2) o c<\/p>\n\n\n\n

= [(ab\/2) c] \/2<\/p>\n\n\n\n

= (a b c)\/4<\/p>\n\n\n\n

Therefore a o (b o c) = (a o b) o c, \u2200 a, b, c \u2208 Q<\/p>\n\n\n\n

Thus, o is associative on Q.<\/p>\n\n\n\n

(vi) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a * b = ab2<\/sup><\/p>\n\n\n\n

b * a = ba2<\/sup><\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b \u2260 b * a<\/p>\n\n\n\n

Thus, * is not commutative on Q<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a * (b * c) = a * (bc2<\/sup>)<\/p>\n\n\n\n

= a (bc2<\/sup>)2<\/sup><\/p>\n\n\n\n

= ab2<\/sup> c4<\/sup><\/p>\n\n\n\n

(a * b) * c = (ab2<\/sup>) * c<\/p>\n\n\n\n

= ab2<\/sup>c2<\/sup><\/p>\n\n\n\n

Therefore a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on Q.<\/p>\n\n\n\n

(vii) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a * b = a + ab<\/p>\n\n\n\n

b * a = b + ba<\/p>\n\n\n\n

= b + ab<\/p>\n\n\n\n

Therefore, a * b \u2260 b * a<\/p>\n\n\n\n

Thus, * is not commutative on Q.<\/p>\n\n\n\n

Now we have to prove associativity on Q.<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a * (b * c) = a * (b + b c)<\/p>\n\n\n\n

= a + a (b + b c)<\/p>\n\n\n\n

= a + ab + a b c<\/p>\n\n\n\n

(a * b) * c = (a + a b) * c<\/p>\n\n\n\n

= (a + a b) + (a + a b) c<\/p>\n\n\n\n

= a + a b + a c + a b c<\/p>\n\n\n\n

Therefore a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on Q.<\/p>\n\n\n\n

(viii) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 R, then<\/p>\n\n\n\n

a * b = a + b \u2013 7<\/p>\n\n\n\n

= b + a \u2013 7<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b = b * a, for all a, b \u2208 R<\/p>\n\n\n\n

Thus, * is commutative on R<\/p>\n\n\n\n

Now we have to prove associativity of * on R.<\/p>\n\n\n\n

Let a, b, c \u2208 R, then<\/p>\n\n\n\n

a * (b * c) = a * (b + c \u2013 7)<\/p>\n\n\n\n

= a + b + c -7 -7<\/p>\n\n\n\n

= a + b + c \u2013 14<\/p>\n\n\n\n

(a * b) * c = (a + b \u2013 7) * c<\/p>\n\n\n\n

= a + b \u2013 7 + c \u2013 7<\/p>\n\n\n\n

= a + b + c \u2013 14<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c ) = (a * b) * c, for all a, b, c \u2208 R<\/p>\n\n\n\n

Thus, * is associative on R.<\/p>\n\n\n\n

(ix) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a * b = (a \u2013 b)2<\/sup><\/p>\n\n\n\n

= (b \u2013 a)2<\/sup><\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b = b * a, for all a, b \u2208 Q<\/p>\n\n\n\n

Thus, * is commutative on Q<\/p>\n\n\n\n

Now we have to prove associativity of * on Q<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a * (b * c) = a * (b \u2013 c)2<\/sup><\/p>\n\n\n\n

= a * (b2<\/sup> + c2<\/sup> \u2013 2 b c)<\/p>\n\n\n\n

= (a \u2013 b2<\/sup> \u2013 c2<\/sup> + 2bc)2<\/sup><\/p>\n\n\n\n

(a * b) * c = (a \u2013 b)2<\/sup> * c<\/p>\n\n\n\n

= (a2<\/sup> + b2<\/sup> \u2013 2ab) * c<\/p>\n\n\n\n

= (a2<\/sup> + b2<\/sup> \u2013 2ab \u2013 c)2<\/sup><\/p>\n\n\n\n

Therefore, a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on Q.<\/p>\n\n\n\n

(x) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a * b = ab + 1<\/p>\n\n\n\n

= ba + 1<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

a * b = b * a, for all a, b \u2208 Q<\/p>\n\n\n\n

Thus, * is commutative on Q<\/p>\n\n\n\n

Now we have to prove associativity of * on Q<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a * (b * c) = a * (bc + 1)<\/p>\n\n\n\n

= a (b c + 1) + 1<\/p>\n\n\n\n

= a b c + a + 1<\/p>\n\n\n\n

(a * b) * c = (ab + 1) * c<\/p>\n\n\n\n

= (ab + 1) c + 1<\/p>\n\n\n\n

= a b c + c + 1<\/p>\n\n\n\n

Therefore, a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on Q.<\/p>\n\n\n\n

(xi) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 N, then<\/p>\n\n\n\n

a * b = ab<\/sup><\/p>\n\n\n\n

b * a = ba<\/sup><\/p>\n\n\n\n

Therefore, a * b \u2260 b * a<\/p>\n\n\n\n

Thus, * is not commutative on N.<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

a * (b * c) = a * (bc<\/sup>)<\/p>\n\n\n\n

=
\"RD<\/p>\n\n\n\n

(a * b) * c = (ab<\/sup>) * c<\/p>\n\n\n\n

= (ab<\/sup>)c<\/sup><\/p>\n\n\n\n

= abc<\/sup><\/p>\n\n\n\n

Therefore, a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on N<\/p>\n\n\n\n

(xii) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Z, then<\/p>\n\n\n\n

a * b = a \u2013 b<\/p>\n\n\n\n

b * a = b \u2013 a<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b \u2260 b * a<\/p>\n\n\n\n

Thus, * is not commutative on Z.<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Z, then<\/p>\n\n\n\n

a * (b * c) = a * (b \u2013 c)<\/p>\n\n\n\n

= a \u2013 (b \u2013 c)<\/p>\n\n\n\n

= a \u2013 (b + c)<\/p>\n\n\n\n

(a * b) * c = (a \u2013 b) \u2013 c<\/p>\n\n\n\n

= a \u2013 b \u2013 c<\/p>\n\n\n\n

Therefore, a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on Z<\/p>\n\n\n\n

(xiii) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q, then<\/p>\n\n\n\n

a * b = (ab\/4)<\/p>\n\n\n\n

= (ba\/4)<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore, a * b = b * a, for all a, b \u2208 Q<\/p>\n\n\n\n

Thus, * is commutative on Q<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Q, then<\/p>\n\n\n\n

a * (b * c) = a * (b c\/4)<\/p>\n\n\n\n

= [a (b c\/4)]\/4<\/p>\n\n\n\n

= (a b c\/16)<\/p>\n\n\n\n

(a * b) * c = (ab\/4) * c<\/p>\n\n\n\n

= [(ab\/4) c]\/4<\/p>\n\n\n\n

= a b c\/16<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c) = (a * b) * c for all a, b, c \u2208 Q<\/p>\n\n\n\n

Thus, * is associative on Q.<\/p>\n\n\n\n

(xiv) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Z, then<\/p>\n\n\n\n

a * b = a + b \u2013 ab<\/p>\n\n\n\n

= b + a \u2013 ba<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore, a * b = b * a, for all a, b \u2208 Z<\/p>\n\n\n\n

Thus, * is commutative on Z.<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Z<\/p>\n\n\n\n

a * (b * c) = a * (b + c \u2013 b c)<\/p>\n\n\n\n

= a + b + c- b c \u2013 ab \u2013 ac + a b c<\/p>\n\n\n\n

(a * b) * c = (a + b \u2013 a b) c<\/p>\n\n\n\n

= a + b \u2013 ab + c \u2013 (a + b \u2013 ab) c<\/p>\n\n\n\n

= a + b + c \u2013 ab \u2013 ac \u2013 bc + a b c<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c) = (a * b) * c, for all a, b, c \u2208 Z<\/p>\n\n\n\n

Thus, * is associative on Z.<\/p>\n\n\n\n

(xv) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 N, then<\/p>\n\n\n\n

a * b = gcd (a, b)<\/p>\n\n\n\n

= gcd (b, a)<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore, a * b = b * a, for all a, b \u2208 N<\/p>\n\n\n\n

Thus, * is commutative on N.<\/p>\n\n\n\n

Now we have to check associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 N<\/p>\n\n\n\n

a * (b * c) = a * [gcd (a, b)]<\/p>\n\n\n\n

= gcd (a, b, c)<\/p>\n\n\n\n

(a * b) * c = [gcd (a, b)] * c<\/p>\n\n\n\n

= gcd (a, b, c)<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c) = (a * b) * c, for all a, b, c \u2208 N<\/p>\n\n\n\n

Thus, * is associative on N.<\/p>\n\n\n\n

5. If the binary operation o is defined by a0b = a + b \u2013 ab on the set Q \u2013 {-1} of all rational numbers other than 1, show that o is commutative on Q \u2013 [1].<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let a, b \u2208 Q \u2013 {-1}.<\/p>\n\n\n\n

Then aob = a + b \u2013 ab<\/p>\n\n\n\n

= b+ a \u2013 ba<\/p>\n\n\n\n

= boa<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

aob = boa for all a, b \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

Thus, o is commutative on Q \u2013 {-1}<\/p>\n\n\n\n

6. Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let a, b \u2208 Z<\/p>\n\n\n\n

a * b = 3a + 7b<\/p>\n\n\n\n

b * a = 3b + 7a<\/p>\n\n\n\n

Thus, a * b \u2260 b * a<\/p>\n\n\n\n

Let a = 1 and b = 2<\/p>\n\n\n\n

1 * 2 = 3 \u00d7 1 + 7 \u00d7 2<\/p>\n\n\n\n

= 3 + 14<\/p>\n\n\n\n

= 17<\/p>\n\n\n\n

2 * 1 = 3 \u00d7 2 + 7 \u00d7 1<\/p>\n\n\n\n

= 6 + 7<\/p>\n\n\n\n

= 13<\/p>\n\n\n\n

Therefore, there exist a = 1, b = 2 \u2208 Z such that a * b \u2260 b * a<\/p>\n\n\n\n

Thus, * is not commutative on Z.<\/p>\n\n\n\n

7. On the set Z of integers a binary operation * is defined by a 8 b = ab + 1 for all a, b \u2208 Z. Prove that * is not associative on Z.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let a, b, c \u2208 Z<\/p>\n\n\n\n

a * (b * c) = a * (bc + 1)<\/p>\n\n\n\n

= a (bc + 1) + 1<\/p>\n\n\n\n

= a b c + a + 1<\/p>\n\n\n\n

(a * b) * c = (ab+ 1) * c<\/p>\n\n\n\n

= (ab + 1) c + 1<\/p>\n\n\n\n

= a b c + c + 1<\/p>\n\n\n\n

Thus, a * (b * c) \u2260 (a * b) * c<\/p>\n\n\n\n

Thus, * is not associative on Z.<\/p>\n\n\n\n

\n\n\n\n

Exercise 3.3 Page No: 3.15<\/p>\n\n\n\n

1. Find the identity element in the set I+<\/sup> of all positive integers defined by a * b = a + b for all a, b \u2208 I+<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let e be the identity element in I+<\/sup> with respect to * such that<\/p>\n\n\n\n

a * e = a = e * a, \u2200 a \u2208 I+<\/sup><\/p>\n\n\n\n

a * e = a and e * a = a, \u2200 a \u2208 I+<\/sup><\/p>\n\n\n\n

a + e = a and e + a = a, \u2200 a \u2208 I+<\/sup><\/p>\n\n\n\n

e = 0, \u2200 a \u2208 I+<\/sup><\/p>\n\n\n\n

Thus, 0 is the identity element in I+<\/sup> with respect to *.<\/p>\n\n\n\n

2. Find the identity element in the set of all rational numbers except \u2013 1 with respect to * defined by a * b = a + b + ab<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let e be the identity element in I+<\/sup> with respect to * such that<\/p>\n\n\n\n

a * e = a = e * a, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

a * e = a and e * a = a, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

a + e + ae = a and e + a + ea = a, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

e + ae = 0 and e + ea = 0, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

e (1 + a) = 0 and e (1 + a) = 0, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

e = 0, \u2200 a \u2208 Q \u2013 {-1} [because a not equal to -1]<\/p>\n\n\n\n

Thus, 0 is the identity element in Q \u2013 {-1} with respect to *.<\/p>\n\n\n\n

\n\n\n\n

Exercise 3.4 Page No: 3.25<\/p>\n\n\n\n

1. Let * be a binary operation on Z defined by a * b = a + b \u2013 4 for all a, b \u2208 Z.<\/strong><\/p>\n\n\n\n

(i) Show that * is both commutative and associative.<\/strong><\/p>\n\n\n\n

(ii) Find the identity element in Z<\/strong><\/p>\n\n\n\n

(iii) Find the invertible element in Z.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) First we have to prove commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Z. then,<\/p>\n\n\n\n

a * b = a + b \u2013 4<\/p>\n\n\n\n

= b + a \u2013 4<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b = b * a, \u2200 a, b \u2208 Z<\/p>\n\n\n\n

Thus, * is commutative on Z.<\/p>\n\n\n\n

Now we have to prove associativity of Z.<\/p>\n\n\n\n

Let a, b, c \u2208 Z. then,<\/p>\n\n\n\n

a * (b * c) = a * (b + c \u2013 4)<\/p>\n\n\n\n

= a + b + c -4 \u2013 4<\/p>\n\n\n\n

= a + b + c \u2013 8<\/p>\n\n\n\n

(a * b) * c = (a + b \u2013 4) * c<\/p>\n\n\n\n

= a + b \u2013 4 + c \u2013 4<\/p>\n\n\n\n

= a + b + c \u2013 8<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c) = (a * b) * c, for all a, b, c \u2208 Z<\/p>\n\n\n\n

Thus, * is associative on Z.<\/p>\n\n\n\n

(ii) Let e be the identity element in Z with respect to * such that<\/p>\n\n\n\n

a * e = a = e * a \u2200 a \u2208 Z<\/p>\n\n\n\n

a * e = a and e * a = a, \u2200 a \u2208 Z<\/p>\n\n\n\n

a + e \u2013 4 = a and e + a \u2013 4 = a, \u2200 a \u2208 Z<\/p>\n\n\n\n

e = 4, \u2200 a \u2208 Z<\/p>\n\n\n\n

Thus, 4 is the identity element in Z with respect to *.<\/p>\n\n\n\n

(iii) Let a \u2208 Z and b \u2208 Z be the inverse of a. Then,<\/p>\n\n\n\n

a * b = e = b * a<\/p>\n\n\n\n

a * b = e and b * a = e<\/p>\n\n\n\n

a + b \u2013 4 = 4 and b + a \u2013 4 = 4<\/p>\n\n\n\n

b = 8 \u2013 a \u2208 Z<\/p>\n\n\n\n

Thus, 8 \u2013 a is the inverse of a \u2208 Z<\/p>\n\n\n\n

2. Let * be a binary operation on Q0<\/sub> (set of non-zero rational numbers) defined by a * b = (3ab\/5) for all a, b \u2208 Q0<\/sub>. Show that * is commutative as well as associative. Also, find its identity element, if it exists.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

First we have to prove commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q0<\/sub><\/p>\n\n\n\n

a * b = (3ab\/5)<\/p>\n\n\n\n

= (3ba\/5)<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore, a * b = b * a, for all a, b \u2208 Q0<\/sub><\/p>\n\n\n\n

Now we have to prove associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Q0<\/sub><\/p>\n\n\n\n

a * (b * c) = a * (3bc\/5)<\/p>\n\n\n\n

= [a (3 bc\/5)] \/5<\/p>\n\n\n\n

= 3 abc\/25<\/p>\n\n\n\n

(a * b) * c = (3 ab\/5) * c<\/p>\n\n\n\n

= [(3 ab\/5) c]\/ 5<\/p>\n\n\n\n

= 3 abc \/25<\/p>\n\n\n\n

Therefore a * (b * c) = (a * b) * c, for all a, b, c \u2208 Q0<\/sub><\/p>\n\n\n\n

Thus * is associative on Q0<\/sub><\/p>\n\n\n\n

Now we have to find the identity element<\/p>\n\n\n\n

Let e be the identity element in Z with respect to * such that<\/p>\n\n\n\n

a * e = a = e * a \u2200 a \u2208 Q0<\/sub><\/p>\n\n\n\n

a * e = a and e * a = a, \u2200 a \u2208 Q0<\/sub><\/p>\n\n\n\n

3ae\/5 = a and 3ea\/5 = a, \u2200 a \u2208 Q0<\/sub><\/p>\n\n\n\n

e = 5\/3 \u2200 a \u2208 Q0 <\/sub>[because a is not equal to 0]<\/p>\n\n\n\n

Thus, 5\/3 is the identity element in Q0<\/sub> with respect to *.<\/p>\n\n\n\n

3. Let * be a binary operation on Q \u2013 {-1} defined by a * b = a + b + ab for all a, b \u2208 Q \u2013 {-1}. Then,<\/strong><\/p>\n\n\n\n

(i) Show that * is both commutative and associative on Q \u2013 {-1}<\/strong><\/p>\n\n\n\n

(ii) Find the identity element in Q \u2013 {-1}<\/strong><\/p>\n\n\n\n

(iii) Show that every element of Q \u2013 {-1} is invertible. Also, find inverse of an arbitrary element.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) First we have to check commutativity of *<\/p>\n\n\n\n

Let a, b \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

Then a * b = a + b + ab<\/p>\n\n\n\n

= b + a + ba<\/p>\n\n\n\n

= b * a<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * b = b * a, \u2200 a, b \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

Now we have to prove associativity of *<\/p>\n\n\n\n

Let a, b, c \u2208 Q \u2013 {-1}, Then,<\/p>\n\n\n\n

a * (b * c) = a * (b + c + b c)<\/p>\n\n\n\n

= a + (b + c + b c) + a (b + c + b c)<\/p>\n\n\n\n

= a + b + c + b c + a b + a c + a b c<\/p>\n\n\n\n

(a * b) * c = (a + b + a b) * c<\/p>\n\n\n\n

= a + b + a b + c + (a + b + a b) c<\/p>\n\n\n\n

= a + b + a b + c + a c + b c + a b c<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

a * (b * c) = (a * b) * c, \u2200 a, b, c \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

Thus, * is associative on Q \u2013 {-1}.<\/p>\n\n\n\n

(ii) Let e be the identity element in I+<\/sup> with respect to * such that<\/p>\n\n\n\n

a * e = a = e * a, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

a * e = a and e * a = a, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

a + e + ae = a and e + a + ea = a, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

e + ae = 0 and e + ea = 0, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

e (1 + a) = 0 and e (1 + a) = 0, \u2200 a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

e = 0, \u2200 a \u2208 Q \u2013 {-1} [because a not equal to -1]<\/p>\n\n\n\n

Thus, 0 is the identity element in Q \u2013 {-1} with respect to *.<\/p>\n\n\n\n

(iii) Let a \u2208 Q \u2013 {-1} and b \u2208 Q \u2013 {-1} be the inverse of a. Then,<\/p>\n\n\n\n

a * b = e = b * a<\/p>\n\n\n\n

a * b = e and b * a = e<\/p>\n\n\n\n

a + b + ab = 0 and b + a + ba = 0<\/p>\n\n\n\n

b (1 + a) = \u2013 a Q \u2013 {-1}<\/p>\n\n\n\n

b = -a\/1 + a Q \u2013 {-1} [because a not equal to -1]<\/p>\n\n\n\n

Thus, -a\/1 + a is the inverse of a \u2208 Q \u2013 {-1}<\/p>\n\n\n\n

4. Let A = R0<\/sub> \u00d7 R, where R0<\/sub> denote the set of all non-zero real numbers. A binary operation \u2018O\u2019 is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) \u2208 R0<\/sub> \u00d7 R.<\/strong><\/p>\n\n\n\n

(i) Show that \u2018O\u2019 is commutative and associative on A<\/strong><\/p>\n\n\n\n

(ii) Find the identity element in A<\/strong><\/p>\n\n\n\n

(iii) Find the invertible element in A.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Let X = (a, b) and Y = (c, d) \u2208 A, \u2200 a, c \u2208 R0 <\/sub>and b, d \u2208 R<\/p>\n\n\n\n

Then, X O Y = (ac, bc + d)<\/p>\n\n\n\n

And Y O X = (ca, da + b)<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

X O Y = Y O X, \u2200 X, Y \u2208 A<\/p>\n\n\n\n

Thus, O is not commutative on A.<\/p>\n\n\n\n

Now we have to check associativity of O<\/p>\n\n\n\n

Let X = (a, b), Y = (c, d) and Z = (e, f), \u2200 a, c, e \u2208 R0 <\/sub>and b, d, f \u2208 R<\/p>\n\n\n\n

X O (Y O Z) = (a, b) O (ce, de + f)<\/p>\n\n\n\n

= (ace, bce + de + f)<\/p>\n\n\n\n

(X O Y) O Z = (ac, bc + d) O (e, f)<\/p>\n\n\n\n

= (ace, (bc + d) e + f)<\/p>\n\n\n\n

= (ace, bce + de + f)<\/p>\n\n\n\n

Therefore, X O (Y O Z) = (X O Y) O Z, \u2200 X, Y, Z \u2208 A<\/p>\n\n\n\n

(ii) Let E = (x, y) be the identity element in A with respect to O, \u2200 x \u2208 R0 <\/sub>and y \u2208 R<\/p>\n\n\n\n

Such that,<\/p>\n\n\n\n

X O E = X = E O X, \u2200 X \u2208 A<\/p>\n\n\n\n

X O E = X and EOX = X<\/p>\n\n\n\n

(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b)<\/p>\n\n\n\n

Considering (ax, bx + y) = (a, b)<\/p>\n\n\n\n

ax = a<\/p>\n\n\n\n

x = 1<\/p>\n\n\n\n

And bx + y = b<\/p>\n\n\n\n

y = 0 [since x = 1]<\/p>\n\n\n\n

Considering (xa, ya + b) = (a, b)<\/p>\n\n\n\n

xa = a<\/p>\n\n\n\n

x = 1<\/p>\n\n\n\n

And ya + b = b<\/p>\n\n\n\n

y = 0 [since x = 1]<\/p>\n\n\n\n

Therefore (1, 0) is the identity element in A with respect to O.<\/p>\n\n\n\n

(iii) Let F = (m, n) be the inverse in A \u2200 m \u2208 R0 <\/sub>and n \u2208 R<\/p>\n\n\n\n

X O F = E and F O X = E<\/p>\n\n\n\n

(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0)<\/p>\n\n\n\n

Considering (am, bm + n) = (1, 0)<\/p>\n\n\n\n

am = 1<\/p>\n\n\n\n

m = 1\/a<\/p>\n\n\n\n

And bm + n = 0<\/p>\n\n\n\n

n = -b\/a [since m = 1\/a]<\/p>\n\n\n\n

Considering (ma, na + b) = (1, 0)<\/p>\n\n\n\n

ma = 1<\/p>\n\n\n\n

m = 1\/a<\/p>\n\n\n\n

And na + b = 0<\/p>\n\n\n\n

n = -b\/a<\/p>\n\n\n\n

Therefore the inverse of (a, b) \u2208 A with respect to O is (1\/a, -b\/a)<\/p>\n\n\n\n

\n\n\n\n

Exercise 3.5 Page No: 3.33<\/p>\n\n\n\n

1. Construct the composition table for \u00d74<\/sub> on set S = {0, 1, 2, 3}.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that \u00d74<\/sub> on set S<\/em> = {0, 1, 2, 3}<\/p>\n\n\n\n

Here,<\/p>\n\n\n\n

1 \u00d74<\/sub> 1 = remainder obtained by dividing 1 \u00d7 1 by 4<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

0 \u00d74<\/sub> 1 = remainder obtained by dividing 0 \u00d7 1 by 4<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

2 \u00d74<\/sub> 3 = remainder obtained by dividing 2 \u00d7 3 by 4<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

3 \u00d74<\/sub> 3 = remainder obtained by dividing 3 \u00d7 3 by 4<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

So, the composition table is as follows:<\/p>\n\n\n\n

\u00d74<\/sub><\/td>0<\/td>1<\/td>2<\/td>3<\/td><\/tr>
0<\/td>0<\/td>0<\/td>0<\/td>0<\/td><\/tr>
1<\/td>0<\/td>1<\/td>2<\/td>3<\/td><\/tr>
2<\/td>0<\/td>2<\/td>0<\/td>2<\/td><\/tr>
3<\/td>0<\/td>3<\/td>2<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

2. Construct the composition table for +5<\/sub> on set S = {0, 1, 2, 3, 4}<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

1 +5 <\/sub>1 = remainder obtained by dividing 1 + 1 by 5<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

3 +5 <\/sub>1 = remainder obtained by dividing 3 + 1 by 5<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

4 +5 <\/sub>1 = remainder obtained by dividing 4 + 1 by 5<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

So, the composition table is as follows:<\/p>\n\n\n\n

+5<\/sub><\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td><\/tr>
0<\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td><\/tr>
1<\/td>1<\/td>2<\/td>3<\/td>4<\/td>0<\/td><\/tr>
2<\/td>2<\/td>3<\/td>4<\/td>0<\/td>1<\/td><\/tr>
3<\/td>3<\/td>4<\/td>0<\/td>1<\/td>2<\/td><\/tr>
4<\/td>4<\/td>0<\/td>1<\/td>2<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

3. Construct the composition table for \u00d76<\/sub> on set S = {0, 1, 2, 3, 4, 5}.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Here,<\/p>\n\n\n\n

1 \u00d76 <\/sub>1 = remainder obtained by dividing 1 \u00d7 1 by 6<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

3 \u00d76 <\/sub>4 = remainder obtained by dividing 3 \u00d7 4 by 6<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

4 \u00d76 <\/sub>5 = remainder obtained by dividing 4 \u00d7 5 by 6<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

So, the composition table is as follows:<\/p>\n\n\n\n

\u00d76<\/sub><\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
0<\/td>0<\/td>0<\/td>0<\/td>0<\/td>0<\/td>0<\/td><\/tr>
1<\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td>5<\/td><\/tr>
2<\/td>0<\/td>2<\/td>4<\/td>0<\/td>2<\/td>4<\/td><\/tr>
3<\/td>0<\/td>3<\/td>0<\/td>3<\/td>0<\/td>3<\/td><\/tr>
4<\/td>0<\/td>4<\/td>2<\/td>0<\/td>4<\/td>2<\/td><\/tr>
5<\/td>0<\/td>5<\/td>4<\/td>3<\/td>2<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

4. Construct the composition table for \u00d75<\/sub> on set Z5<\/sub> = {0, 1, 2, 3, 4}<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Here,<\/p>\n\n\n\n

1 \u00d75 <\/sub>1 = remainder obtained by dividing 1 \u00d7 1 by 5<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

3 \u00d75 <\/sub>4 = remainder obtained by dividing 3 \u00d7 4 by 5<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

4 \u00d75 <\/sub>4 = remainder obtained by dividing 4 \u00d7 4 by 5<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

So, the composition table is as follows:<\/p>\n\n\n\n

\u00d75<\/sub><\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td><\/tr>
0<\/td>0<\/td>0<\/td>0<\/td>0<\/td>0<\/td><\/tr>
1<\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td><\/tr>
2<\/td>0<\/td>2<\/td>4<\/td>1<\/td>3<\/td><\/tr>
3<\/td>0<\/td>3<\/td>1<\/td>4<\/td>2<\/td><\/tr>
4<\/td>0<\/td>4<\/td>3<\/td>2<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

5. For the binary operation \u00d710 <\/sub>set S = {1, 3, 7, 9}, find the inverse of 3.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Here,<\/p>\n\n\n\n

1 \u00d710 <\/sub>1 = remainder obtained by dividing 1 \u00d7 1 by 10<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

3 \u00d710 <\/sub>7 = remainder obtained by dividing 3 \u00d7 7 by 10<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

7 \u00d710 <\/sub>9 = remainder obtained by dividing 7 \u00d7 9 by 10<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

So, the composition table is as follows:<\/p>\n\n\n\n

\u00d710<\/sub><\/td>1<\/td>3<\/td>7<\/td>9<\/td><\/tr>
1<\/td>1<\/td>3<\/td>7<\/td>9<\/td><\/tr>
3<\/td>3<\/td>9<\/td>1<\/td>7<\/td><\/tr>
7<\/td>7<\/td>1<\/td>9<\/td>3<\/td><\/tr>
9<\/td>9<\/td>7<\/td>3<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

From the table we can observe that elements of first row as same as the top-most row.<\/p>\n\n\n\n

So, 1 \u2208 S is the identity element with respect to \u00d710<\/sub><\/p>\n\n\n\n

Now we have to find inverse of 3<\/p>\n\n\n\n

3 \u00d710<\/sub> 7 = 1<\/p>\n\n\n\n

So the inverse of 3 is 7.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 3: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 3\u2013Binary Operations<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 12 Maths Chapter 3\u2013Binary Operations PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 12 Maths :<\/strong><\/h2>\n\n\n\n