# Class 12 RD Sharma Solutions – Chapter 3 Binary Operations – Exercise 3.1

**Question 1. Determine whether the following operation define a binary operation on the given set or not:**

**(i) ‘*’ on N defined by a * b = ab for all a, b ∈ N.**

**(ii) ‘O’ on Z defined by a O b = ab for all a, b ∈ Z.**

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**(iii) ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N**

**(iv) ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a × 6 b = Remainder when a b is divided by 6.**

**(v) ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b**

**(vi) ‘⊙’ on N defined by a ⊙ b= ab + ba for all a, b ∈ N**

**(vii) ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q**

**Solution:**

(i)Given ‘*’ on N defined by a * b = a^{b}for all a, b ∈ N.Let a, b ∈ N. Then,

a

^{b}∈ N [∵ ab≠0 and a, b is positive integer]⇒ a * b ∈ N

Therefore,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

(ii)Given ‘O’ on Z defined by a O b = a^{b}for all a, b ∈ Z.Both a = 3 and b = -1 belong to Z.

⇒ a * b = 3

^{-1}= ∉ Z

Thus, * is not a binary operation on Z.

(iii)Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ NIf a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 ∉ N

Thus, there exist a = 1 and b = 1 such that a * b ∉ N

So, * is not a binary operation on N.

(iv)Given ‘×_{6}‘ on S = {1, 2, 3, 4, 5} defined by a ×_{6}b = Remainder when a b is divided by 6.Consider the composition table,

X _{6}1 2 3 4 5 1 1 2 3 4 5 2 2 4 0 2 4 3 3 0 3 0 3 4 4 2 0 4 2 5 5 4 3 2 1 Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×

_{6}b = 2 ×_{6}3 = remainder when 6 divided by 6 = 0 ≠ SThus, ×

_{6}is not a binary operation on S.

(v)Given ‘+_{6}’ on S = {0, 1, 2, 3, 4, 5} defined by a +_{6}bConsider the composition table,

+ _{6}0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×

_{6}b = 2 ×_{6}3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×_{6}is not a binary operation on S.

(vi)Given ‘⊙’ on N defined by a ⊙ b= a^{b}+ b^{a}for all a, b ∈ NLet a, b ∈ N. Then,

ab, ba ∈ N

⇒ a

^{b}+ b^{a}∈ N [∵Addition is binary operation on N]⇒ a ⊙ b ∈ N

Thus, ⊙ is a binary operation on N.

(vii)Given ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ QIf a = 2 and b = -1 in Q,

a * b =

=

= [which is not defined]

For a = 2 and b = -1

a * b does not belongs to Q

So, * is not a binary operation in Q.

**Question 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 ^{+}, defined * by a * b = a – b**

**(ii) On Z ^{+}, define * by a*b = ab**

**(iii) On R, define * by a*b = ab ^{2}**

**(iv) On Z ^{+} define * by a * b = |a − b|**

**(v) On Z ^{+} define * by a * b = a**

**(vi) On R, define * by a * b = a + 4b ^{2}**

**Here, Z ^{+} denotes the set of all non-negative integers.**

**Solution:**

(i)Given On Z^{+}, defined * by a * b = a – bIf a = 1 and b = 2 in Z

^{+}, thena * b = a – b

= 1 – 2

= -1 ∉ Z

^{+}[because Z^{+}is the set of non-negative integers]For a = 1 and b = 2,

a * b ∉ Z

^{+}Thus, * is not a binary operation on Z

^{+}.

(ii)Given Z^{+}, define * by a*b = a bLet a, b ∈ Z

^{+}⇒ a, b ∈ Z

^{+}⇒ a * b ∈ Z

^{+}Thus, * is a binary operation on R.

(iii)Given on R, define by a*b = ab^{2}Let a, b ∈ R

⇒ a, b

^{2}∈ R⇒ ab

^{2}∈ R⇒ a * b ∈ R

Thus, * is a binary operation on R.

(iv)Given on Z^{+}define * by a * b = |a − b|Let a, b ∈ Z

^{+}⇒ | a – b | ∈ Z

^{+}⇒ a * b ∈ Z

^{+}Therefore,

a * b ∈ Z

^{+}, ∀ a, b ∈ Z^{+}Thus, * is a binary operation on Z

^{+}.

(v)Given on Z^{+}define * by a * b = aLet a, b ∈ Z

^{+}⇒ a ∈ Z

^{+}⇒ a * b ∈ Z

^{+}Therefore, a * b ∈ Z

^{+}∀ a, b ∈ Z^{+}Thus, * is a binary operation on Z

^{+}.

(vi)Given On R, define * by a * b = a + 4b^{2}Let a, b ∈ R

⇒ a, 4b

^{2}∈ R⇒ a + 4b

^{2}∈ R⇒ a * b ∈ R

Therefore, a *b ∈ R, ∀ a, b ∈ R

Thus, * is a binary operation on R.

**Question 3. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.**

**Solution:**

Given:

a * b = 2a + b – 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 3

= 7

**Question 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.**

**Solution:**

LCM1234511 2 3 4 5 22 2 6 4 10 33 5 3 12 15 44 4 12 4 20 55 10 15 20 5 In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 ∉ {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

**Question 5. Let S = {a, b, c}. Find the total number of binary operations on S.**

**Solution:**

Number of binary operations on a set with n elements is

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is

**Question 6. Find the total number of binary operations on {a, b}.**

**Solution:**

We have,

S = {a, b}

The total number of binary operation on S = {a, b} in

**Question 7. Prove that the operation * on the set**

**M=**** defined by A + B = AB is a binary operation.**

**Solution: **

We have,

and

A + B = AB for all A, B ∈ M

Let A =\ and B =

Now, AB =

Therefore, a ∈ R, b ∈ R, c ∈ R and d ∈ R

⇒ ac ∈ R and bd ∈ R

⇒

⇒ A * B ∈ M

Hence, the operator * defines a binary operation on M

**Question 8. Let S be the set of all rational numbers of the form **** where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation**

**Solution: **

S = set of rational numbers of the form where m ∈ Z and n = 1, 2, 3

Also, a * b = ab

Let a ∈ S and b ∈ S

⇒ ab =

Therefore, a * b ∉ S

Hence, the operator * does not defines a binary operation on S

**Question 9. The binary operation & : R × R → R is defined as a*b = 2a + b**

**Solution:**

It is given that, a*b = 2a + b

Now,

(2*3) = 2 × 2 + 3

= 4 + 3

(2*3)*4 = 7*4 = 2 × 7 + 4

= 14 + 4

= 18

**Question 10. Let * be a binary operation on N given by a*b = LCM(a, b) for all a, b ∈ N. Find 5*7.**

**Solution:**

It is given that a*b = LCM (a, b)

Now,

5*7 = LCM (5, 7)

= 35