# NCRT Solution Exercise 4

**Question 1**: Determine whether or not each of the definition of given below gives a binary operation.

In the event that * is not a binary operation, give justification for this.

(i) On Z^{+}, define * 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

**Answer 1:**

(i) On Z^{+}, * is defined by a * b = a − b.

It is not a binary operation

as the image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ Z^{+}.

(ii) On Z+, * is defined by a * b = ab.

It is seen that for each a, b ∈ Z^{+}, there is a unique element ab in Z^{+}.

This means that * carries each pair (a, b) to a unique element a * b = ab in Z^{+}.

Therefore, * is a binary operation.

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

It is seen that for each a, b ∈ R, there is a unique element ab^{2 }in R.

This means that * carries each pair (a, b) to a unique element a * b = ab^{2} in R.

Therefore, * is a binary operation.

(iv) On Z^{+}, * is defined by a * b = |a − b|.

It is seen that for each a, b ∈ Z^{+}, there is a unique element |a − b| in Z^{+}.

This means that * carries each pair (a, b) to a unique element a * b = |a − b| in Z^{+}.

Therefore, * is a binary operation.

(v) On Z^{+}, * is defined by a * b = a.

It is seen that for each a, b ∈ Z^{+}, there is a unique element a in Z^{+}.

This means that * carries each pair (a, b) to a unique element a * b = a in Z^{+}.

Therefore, * is a binary operation.

**Question 2: **

For each binary operation * defined below, determine whether * is commutative

or associative.

(i) On Z, define a * b = a − b

(ii) On Q, define a * b = ab + 1

(iii) On Q, define a * b =ab/2

(iv) On Z^{+}, define a * b = 2^{ab}

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

(vi)On R − {−1}, define a ∗ b =(a/b+1)

**Answer 2:**

(i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = −1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1, where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also, we have

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Z

Hence, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1.

It is known that: ab = ba for all a, b ∈ Q

⇒ ab + 1 = ba + 1 for all a, b ∈ Q

⇒ a * b = a * b for all a, b ∈ Q

Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q

Therefore, the operation * is not associative.

(iii) On Q, * is defined by a * b =(ab/2)

It is known that: ab = ba for all a, b ∈ Q

⇒(ab/2)=(ba/2) for all a, b ∈ Q

⇒ a * b = b * a for all a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have

= (a ∗b)*c = (ab/2)*c = ((ab/2)c) / 2 = (abc / 4)

And

= a ∗(b*c) = a*(bc/2) = (a(bc/2)) / 2 = (abc / 4)

∴ (a*b)*c = a*(b*c), where a, b, c ∈ Q

Therefore, the operation * is associative.

(iv) On Z^{+}, * is defined by a * b = 2^{ab}.

It is known that: ab = ba for all a, b ∈ Z^{+}

⇒ 2^{ab}= 2^{ba} for all a, b ∈ Z^{+}

⇒ a * b = b * a for all a, b ∈ Z^{+}

Therefore, the operation * is commutative.

It can be observed that

(1 ∗ 2) ∗ 3 = 2^{1×2}∗ 3 = 4 ∗ 3 = 2^{4×3} = 2^{12} and

1 ∗ (2 ∗ 3) = 1 ∗ 2^{2×3} = 1 ∗ 2^{6} = 1 ∗ 64 = 2^{1×64} = 2^{64}

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Z^{+}

Therefore, the operation * is not associative.

(v) On Z^{+}, * is defined by a * b = a^{b}

It can be observed that

1*2 = 1^{2} = 1 and 2*1 = 2^{1} = 2

∴ 1 * 2 ≠ 2 * 1, where 1, 2 ∈ Z^{+}

Therefore, the operation * is not commutative.

It can also be observed that

(2 ∗ 3) ∗ 4 = 2^{3} ∗ 4 = 8 ∗ 4 = 8^{4} = 2^{12} and

2 ∗ (3 ∗ 4) = 2 ∗ 3^{4} = 2 ∗ 81 = 2^{81}

∴ (2 * 3) * 4 ≠ 2 * (3 * 4), where 2, 3, 4 ∈ Z^{+}

Therefore, the operation * is not associative.

(vi) On R, * − {−1} is defined by a ∗ b = (a / b+1)

It can be observed that

1 ∗ 2 =(1 /(2+1))=1/3 and 2 ∗ 1 =(2 / (1+1))=(2/2)= 1

∴1 * 2 ≠ 2 * 1, where 1, 2 ∈ R − {−1}

Therefore, the operation * is not commutative.

It can also be observed that

(1 ∗ 2) ∗ 3 =(1 / (2 + 1))∗ 3 =(1/3)∗ 3 =(1/3) / (3 + 1)=(1/12)

and

1 ∗ (2 ∗ 3) = 1 ∗(2 / (3 + 1))= 1 ∗(2 / 4)= 1 ∗(1/2) = (1/((1/2)+ 1)=((1/3) /2)=2/3

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ R − {−1}

Therefore, the operation * is not associative.

**Question 3: **Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}.

Write the operation table of the operation ∧.

**Answer 3:**

The binary operation ∧ on the set {1, 2, 3, 4, 5} is defined as a ∧ b = min {a, b}

for all a, b ∈ {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation ∧ can be given as:

∧ |
1 |
2 |
3 |
4 |
5 |

1 |
1 |
1 |
1 |
1 |
1 |

2 |
1 |
2 |
2 |
2 |
2 |

3 |
1 |
3 |
3 |
3 |
3 |

4 |
1 |
2 |
3 |
4 |
4 |

5 |
1 |
2 |
3 |
4 |
5 |

**Question 4: **Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following

multiplication table.

(i) Compute (2 * 3) * 4 and 2 * (3 * 4)

(ii) Is * commutative?

(iii) Compute (2 * 3) * (4 * 5).

(Hint: use the following table)

* |
1 |
2 |
3 |
4 |
5 |

1 |
1 |
1 |
1 |
1 |
1 |

2 |
1 |
2 |
1 |
2 |
1 |

3 |
1 |
1 |
3 |
1 |
1 |

4 |
1 |
2 |
1 |
4 |
1 |

5 |
1 |
1 |
1 |
1 |
5 |

**Answer 4:**

(i) (2 * 3) * 4 = 1 * 4 = 1

2 * (3 * 4) = 2 * 1 = 1

(ii) For every a, b ∈ {1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative.

(iii) (2 * 3) = 1 and (4 * 5) = 1

∴ (2 * 3) * (4 * 5) = 1 * 1 = 1

**Question 5: **

Let *′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *′ b = H.C.F.

of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above?

Justify your answer.

Answer 5:

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b.

The operation table for the operation *′ can be given as:

*’ |
1 |
2 |
3 |
4 |
5 |

1 |
1 |
1 |
1 |
1 |
1 |

2 |
1 |
2 |
1 |
2 |
1 |

3 |
1 |
1 |
3 |
1 |
1 |

4 |
1 |
2 |
1 |
4 |
1 |

5 |
1 |
1 |
1 |
1 |
5 |

We observe that the operation tables for the operations * and *′ are the same.

Thus, the operation *′ is same as the operation*.

**Question 6: **

Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find

(i) 5 * 7, 20 * 16 (ii) Is * commutative?

(iii) Is * associative? (iv) Find the identity of * in N

(v) Which elements of N are invertible for the operation *?

**Answer 6:**

The binary operation * on N is defined as a * b = L.C.M. of a and b.

(i) 5 * 7 = L.C.M. of 5 and 7 = 35

20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that

L.C.M of a and b = L.C.M of b and a for all a, b ∈ N.

∴ a * b = b * a

Thus, the operation * is commutative.

(iii) For a, b, c ∈ N, we have

(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c

a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c

∴ (a * b) * c = a * (b * c)

Thus, the operation * is associative.

(iv)It is known that:

L.C.M. of a and 1 = a = L.C.M. 1 and a for all a ∈ N

⇒ a * 1 = a = 1 * a for all a ∈ N

Thus, 1 is the identity of * in N.

(v) An element a in N is invertible with respect to the operation * if there exists

an element b in N, such that a * b = e = b * a.

Here, e = 1

This means that

L.C.M of a and b = 1 = L.C.M of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.

**Question 7**: Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary

operation? Justify your answer.

**Answer 7:**

The operation * on the set A = {1, 2, 3, 4, 5} is defined as a * b = L.C.M. of a and b.

Then, the operation table for the given operation * can be given as:

* |
1 |
2 |
3 |
4 |
5 |

1 |
1 |
2 |
3 |
4 |
5 |

2 |
2 |
2 |
6 |
4 |
10 |

3 |
3 |
6 |
3 |
12 |
15 |

4 |
4 |
4 |
12 |
4 |
20 |

5 |
5 |
10 |
15 |
20 |
5 |

It can be observed from the obtained table that

3 * 2 = 2 * 3 = 6 ∉ A,

5 * 2 = 2 * 5 = 10 ∉ A,

3 * 4 = 4 * 3 = 12 ∉ A,

3 * 5 = 5 * 3 = 15 ∉ A,

4 * 5 = 5 * 4 = 20 ∉ A

Hence, the given operation * is not a binary operation.

**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?

**Answer 8: **

The binary operation * on N is defined as: a * b = H.C.F. of a and b

It is known that

H.C.F. of a and b = H.C.F. of b and a for all a, b ∈ N.

∴ a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ N, we have

(a * b)* c = (H.C.F. of a and b) * c = H.C.F. of a, b and c

a *(b * c) = a *(H.C.F. of b and c) = H.C.F. of a, b, and c

∴ (a * b) * c = a * (b * c)

Thus, the operation * is associative.

Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a

for all a ∈ N.

But this relation is not true for any a ∈ N.

Thus, the operation * does not have any identity in N.

**Question 9: **Let * be a binary operation on the set Q of rational numbers as follows:

(i) a * b = a − b

(ii) a * b = a^{2}+ b^{2}

(iii) a * b = a + ab

(iv) a * b = (a − b)^{2}

(v) a ∗ b =(ab/4)

(vi) a * b = ab^{2}

Find which of the binary operations are commutative and which are associative.

**Answer 9: **(i) On Q, the operation * is defined as a * b = a − b. It can be observed that:

(1/2)∗(1/3)=(1/2)−(1/3) = ((3 – 2)/6) = (1/6)

and

(1/3)∗(1/2)=(1/3)−(1/2)=((2 – 3)/6)=(−1/6)

∴(1/2)∗(1/3)≠(1/3)∗(1/2), where (1/2),(1/3)∈ Q

Thus, the operation * is not commutative.

It can also be observed that

((1/2)∗(1/3)) ∗ (1/4) = ((1/2)−(1/3)) ∗ (1/4) = ((3 – 2) / 6) ∗ (1/4) = (1/6) ∗(1/4) = ((1/6)−(1/4)) = ((2 – 3) /12) = (−1/12)

And

(1/2)∗((1/3) ∗ (1/4)) = (1/2) ∗ ((1/3) - (1/4)) = (1/2) ∗ ((4 – 3) / 12) = (1/2) ∗ (1/12) = (1/2) - (1/12) = ((6 – 1) /12) = (−1/12)

∴(1/2)∗ (1/3)∗ (1/4) ≠ (1/2) ∗((1/3)∗ (1/4)) , where (1/2), (1/3), (1/4) ∈ Q

Thus, the operation * is not associative.

(ii) On Q, the operation * is defined as a * b = a^{2}+ b^{2}.

For a, b ∈ Q, we have

a * b = a^{2}+ b^{2}= b^{2}+ a^{2} = b * a

∴ a * b = b * a

Thus, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1^{2}+ 2^{2}) * 3 = (1 + 4) * 3 = 5 * 3 = 5^{2}+ 3^{2}= 34 and

1 * (2 * 3) = 1 * (2^{2}+ 3^{2}) = 1 * (4 + 9) = 1 * 13 = 1^{2}+ 13^{2}=170

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

(iii)On Q, the operation * is defined as a * b = a + ab.

It can be observed that

1 * 2 = 1 + 1×2 = 1 + 2 = 3

2 * 1 = 2 + 2×1 = 2 + 2 = 4

∴ 1 * 2 ≠ 2 * 1, where 1, 2 ∈ Q

Thus, the operation * is not commutative.

It can also be observed that

(1 * 2) * 3 = (1+ 1×2 ) * 3 = (1 + 2) * 3 = 3 * 3 = 3 + 3×3 = 3 + 9 = 12 and

1 * (2 * 3) = 1 * (2 + 2×3 ) = 1 * (2 + 6) = 1 * 8 = 1 + 1×8 =1 + 8 = 9

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

(iv)On Q, the operation * is defined by a * b = (a − b)^{2}.

For a, b ∈ Q, we have

a * b = (a − b)^{2}

b * a = (b − a)^{2}= [− (a − b)]^{2}= (a − b)^{2}

∴ a * b = b * a

Thus, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1 – 2)^{2}* 3 = (– 1)^{2}* 3 = 1 * 3 = (1 – 3)^{2}= (– 2)^{2}= 4

And

1 * (2 * 3) = 1 * (2 – 3)^{2}= 1 * (– 1)^{2}= 1 * 1 = (1 – 1)^{2}= 0

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

(v) On Q, the operation * is defined as a*b = (ab/4).

For a, b ∈ Q, we have

a*b = (ab/4) = (ba/4) = b*a

∴ a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ Q, we have

(a*b) *c = (ab/4) *c = (((ab/4).c) / 4) = (abc / 16)

And

a*(b *c) = a*(bc/4) = ((a(bc/4)) / 4) = (abc / 16)

∴ (a * b) * c = a * (b * c), where a, b, c ∈ Q

Thus, the operation * is associative.

(vi) On Q, the operation * is defined as a * b = ab^{2}

It can be observed that

(1/2)∗(1/3) = (1/2). (1/3)^{2} =(1/2).(1/9) = (1/18)

And

(1/3)∗(1/2) = (1/3). (1/2)^{2} =(1/3).(1/4) = (1/12)

∴(1/2)*(1/3) ≠ (1/3) * (1/2), where (1/2) and (1/3) ∈ Q

Thus, the operation * is not commutative.

It can also be observed that

((1/2)*(1/3)) * (1/4) =[(1/2)(1/3)^{2}]∗ (1/4) = (1/18).(1/4)^{2} =(1 / (18 × 16)) = (1/288)

And

(1/2)*((1/3)* (1/4)) =(1/2)*[(1/3)∗ (1/4)^{2}] = (1/2).(1/48) =(1/2) (1 × 48)^{2} = (1/(2 x 2304)) = (1 / 4608)

∴ (1/2)*(1/3) * (1/4) ≠ (1/2) * ((1/3)∗ (1/4)), where (1/2),(1/3), (1/4) ∈ Q

Thus, the operation * is not associative.

Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

**Question 10**: Find which of the operations given above has identity.

**Answer 10:**

An element e ∈ Q will be the identity element for the operation *

if a * e = a = e * a, for all a ∈ Q.

However, there is no such element e ∈ Q with respect to each of the six operations satisfying the above condition.

Thus, none of the six operations has identity.

**Question 11: **Let A = N × N and * be the binary operation on A defined by

(a, b) * (c, d) = (a + c, b + d)

Show that * is commutative and associative. Find the identity element for * on A,

if any.

**Answer 11**:

A = N × N and * is a binary operation on A and is defined by

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)

[Addition is commutative in the set of natural numbers]

∴ (a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈ A

Then, a, b, c, d, e, f ∈ N

We have

[(a, b) * (c,d)] * (e, f) = (a + bc + d) * (e, f) = (a+c+e,b+d+f)

and

(a, b) * [(c,d) * (e, f)] = (a, b) * (c + e,d+f) = (a+c+e,b+d+f)

∴[(a, b) * (c,d)] * (e, f) = (a, b) * [(c,d) * (e, f)]

Therefore, the operation * is associative.

Let an element e = (e_{1}, e_{2}) ∈ A will be an identity element for the operation *

if a * e = a = e * a for all a = (a_{1}, a_{2}) ∈ A

i.e., (a_{1} + e_{1}, a_{2} + e_{2}) = (a_{1}, a_{2}) = (e_{1} + a_{1}, e_{2} + a_{2})

Which is not true for any element in A.

Therefore, the operation * does not have any identity element.

**Question 12: **State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set N, a * a = a ∀ a ∈ N.

(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a

**Answer 12: **

(i) Define an operation * on N as a * b = a + b ∀ a, b ∈ N

Then, in particular, for b = a = 3, we have

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (c * b) * a

= (b * c) * a [* is commutative]

= a * (b * c) [Again, as * is commutative]

= L.H.S.

∴ a * (b * c) = (c * b) * a

Therefore, statement (ii) is true.

**Question 13**: Consider a binary operation * on N defined as a * b = a^{3}+ b^{3} Choose the correct answer.

(A)Is * both associative and commutative?

(B) Is * commutative but not associative?

(C) Is * associative but not commutative?

(D)Is * neither commutative nor associative?

**Answer 13**:

On N, the operation * is defined as a * b = a^{3}+ b^{3}.

For, a, b, ∈ N, we have

a * b = a^{3}+ b^{3}= b^{3}+ a^{3}

= b * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1^{3}+ 2^{3}) * 3 = (1 + 8) * 3 = 9 * 3 = 9^{3}+ 3^{3}= 729 + 27 = 756 and

1* (2 * 3) = 1 * (2^{3}+ 3^{3}) = 1* (8 + 27) = 1*35 = 1^{3}+ 35^{3}= 1 + 42875 = 42876

∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ N

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative.

Thus, the correct answer is B.