NCERT Solutions for Maths Class 12 Exercise 1.4

# NCERT Solutions for Maths Class 12 Exercise 1.4

NCERT Solutions for Maths Class 12 Exercise 1.4

## NCERT Solutions for Maths Class 12 Exercise 1.4

### Maths Class 12 Ex 1.4 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 = ab²
(iv) On Z+, define * by a * b = |a – b|
(v) On Z+, define * by a * b = a

Solution:
(i) For all a, b
Z+ and a > b, a * b = a – b > 0, which belongs to Z+.
But if a < b, then a * b = a – b < 0, which does not belong to Z+.
Thus, the given operation * is not a binary operation.
(ii) For all a, b
Z+, a * b = ab > 0, which belongs to Z+.
Thus, the given operation * is a binary operation.
(iii) For all a, b
R, a * b = ab² also belongs to R.
Thus, the given operation * defined by a * b = ab² is a binary operation.
(iv) For all a, b
Z+, a * b = |a – b| also belongs to Z+.
Thus, the given operation * defined by a * b = |a – b| is a binary operation.
(v) For all a, b
Z+, a * b = a Z+.
Thus, the given operation * defined by a * b = a is a binary operation.

### Maths Class 12 Ex 1.4 Question 2.

For each binary operation * defined below, determine whether * is binary, 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 = 2ab
(v) On Z+, define a * b = ab
(vi) On R – {-1}, define a * b = a/(b + 1)

Solution:
(i) (a) On Z, operation * is defined as
a * b = a – b
b * a = b – a
Since a – b ≠ b – a
a * b ≠ b * a
Hence, the binary operation * is not commutative.

(b) Since a – (b – c) ≠ (a – b) – c a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

(ii) (a) On Q, operation * is defined as a * b = ab + 1
Then, ab + 1 = ba + 1
a * b = b * a
Hence, the binary operation * is commutative.

(b) We have (a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1
Again, a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a+ 1
Therefore, a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

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

Again, b * a = ba/2. Since ab/2 = ba/2 a * b = b * a
Hence, the binary operation * is commutative.

(b) We have a * (b * c) = a * bc/2 = abc/4 and (a * b) * c = ab/2 * c = abc/4

Therefore, (a * b) * c = a * (b * c).
Hence, the binary operation * is associative.

(iv) (a) On Z+, operation * is defined as a * b = 2ab
Then, a * b = 2ab and b * a = 2ba = 2ab
Therefore, a * b = b * a.
Hence, the binary operation * is commutative.

(b) We have a * (b * c) = a * 2bc = 2a.2bc
Again, (a * b) * c = 2ab * c = 22ab.c
Thus, (a * b) * c ≠ a * (b * c).
Hence, the binary operation * is not associative.

(v) (a) On Z+, operation * is defined as a * b = ab
And b * a = ba.
Since, ab ≠ ba  a * b ≠ b * a.
Hence, the binary operation * is not commutative.

(b) We have (a * b) * c = ab * c = (ab)c

And a * (b * c) = a * bc = (a)bc
Thus, (a * b) * c ≠ a * (b * c)
Hence, the binary operation * is not associative.

(vi) (a) On R – {-1}, operation * is defined as a * b = a/(b + 1).

And b * a = b/(a + 1). Thus, a * b ≠ b * a.

Hence, the binary operation * is not commutative.

(b) Similarly, a * (b * c) ≠ (a * b) * c.

Hence, the binary operation * is not associative.

### Maths Class 12 Ex 1.4 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 ^.

Solution:
The table for the binary operation ^ on the set {1, 2, 3, 4, 5} is as follows.

 ^ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5

### Maths Class 12 Ex 1.4 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)

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

Solution:
(i) From the given table, we can see that 2 * 3 = 1 and 1 * 4 = 1
Therefore, (2 * 3) * 4 = 1 * 4 = 1
Again, we have 3 * 4 = 1. Therefore, 2 * (3 * 4) = 2 * 1 = 1
(ii) Let a, b
{1, 2, 3, 4, 5}

From the given table, we can see that a * a = a
Also, a * b = b * a = 1, when a or b or both are odd.
Again, 2 * 4 = 4 * 2 = 2, when a and b are even and a ≠ b.
Thus, a * b = b * a
Hence, the given binary operation * is commutative.
(iii) From the given table, we can see that 2 * 3 = 1 and 4 * 5 = 1.

Therefore, (2 * 3) * (4 * 5) = 1 * 1 = 1.

### Maths Class 12 Ex 1.4 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 question no. 4 above? Justify your answer.

Solution:
The binary operation *’ on the set {1, 2, 3, 4, 5} is defined by a *’ b = H.C.F. of a and b.
Let us prepare a table for the operation *’.

 *’ (HCF of a, b) 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

Since the table obtained above is the same as the table given in question no. 4, therefore, the operation *’ is the same as the operation * in question no. 4.

### Maths Class 12 Ex 1.4 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 *?

Solution:
The binary operation * on the set N is defined by a * b = L.C.M. of a and b.
(i) We have 5 * 7 = L.C.M. of 5 and 7 = 35
Again, 20 * 16 = L.C.M. of 20 and 16 = 80

(ii) We have a * b = L.C.M. of a and b and b * a = L.C.M. of b and a

Since L.C.M. of a and b = L.C.M. of b and a
Therefore, a * b = b * a.

Hence, the binary operation * is commutative.

(iii) We have a * (b * c) = L.C.M. of a, b, c and (a * b) * c = L.C.M. of a, b, c
Therefore, a * (b * c) = (a * b) * c
Hence, the binary operation * is associative.

(iv) Since 1 * a = a * 1 = a = L.C.M. of 1 and a.
Therefore, the identity of * in N is 1.

(v) Let * : N × N → N defined as a * b = L.C.M. of (a, b)
For a = 1, b = 1, a * b = L.C.M. of (a, b) = 1.
Otherwise, a * b ≠ 1
Therefore, the element 1 of N is invertible for the operation *.

### Maths Class 12 Ex 1.4 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.

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

Here, 4 * 5 = L.C.M. of 4 and 5 = 20, which does not belong to the given set {1, 2, 3, 4, 5}.
Therefore, the given operation * is not a binary operation.

### Maths Class 12 Ex 1.4 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 set N is defined as a * b = H.C.F. of a and b
(a) We have a * b = H.C.F. of a and b, and b * a = H.C.F. of b and a

Since H.C.F. of a and b = H.C.F. of b and a
Therefore, a * b = b * a
Hence, the binary operation * is commutative.

(b) We have a * (b * c) = a * (H.C.F. of b, c) = H.C.F. of a and (H.C.F. of b, c) = HCF of a, b and c.
Similarly, (a * b) * c = H.C.F. of a, b and c
Therefore, (a * b) * c = a * (b * c).
Hence, the binary operation * is associative.

(c) We have 1 * a = a * 1 = 1 ≠ a
Therefore, it does not exist any identity for this binary operation on N.

### Maths Class 12 Ex 1.4 Question 9.

Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = ab/4
(vi) a * b = ab²
Find which of the binary operations are commutative and which are associative.

Solution:
(i) The binary operation * on the set Q is defined as
(a) a * b = a – b and b * a = b – a
But a – b ≠ b – a
Therefore, a * b ≠ b * a
Hence, the binary operation * is not commutative.
(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c

And (a * b) * c = (a – b) * c = a – b – c

Since a – b + c ≠ a – b – c
Therefore, a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

(ii) The binary operation * on the set Q is defined as

(a) a * b = a² + b² and b * a = b² + a² = a² + b².
Therefore, a * b = b * a.
Hence, the binary operation * is commutative.
(b) a * (b * c) = a * (b² + c²) = a² + (b² + c²)²
And (a * b) * c = (a² + b²) * c = (a² + b²)² + c²
Thus, a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

(iii) The binary operation * on the set Q is defined as

(a) a * b = a + ab and b * a = b + ba
Therefore, a * b ≠ b * a
Hence, the binary operation * is not commutative.
(b) a * (b * c) = a * (b + bc) = a + a(b + bc) = a + ab + abc

And (a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c = a + ab + ac + abc
Therefore, a * (b * c) ≠ (a * b) * c.
Hence, the binary operation * is not associative.

(iv) The binary operation * on the set Q is defined as

(a) a * b = (a – b)² and b * a = (b – a)² = (a – b)²

Therefore, a * b = b * a
Hence, the binary operation * is commutative.
(b) a * (b * c) = a * (b – c)² = [a – (b – c)²]²

And (a * b) * c = (a – b)² * c = [(a – b)² – c]²
Therefore, a * (b * c) ≠ (a * b) * c.
Hence, the binary operation * is not associative.

(v) The binary operation * on the set Q is defined as

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

Therefore, a * b = b * a
Hence, the binary operation * is commutative.
(b) a * (b * c) = a * bc/4 = abc/16

And (a * b) * c = ab/4 * c = abc/16
Therefore, a * (b * c) = (a * b) * c.
Hence, the binary operation * is associative.

(vi) The binary operation * on the set Q is defined as

(a) a * b = ab² and b * a = ba²

Therefore, a * b ≠ b * a
Hence, the binary operation * is not commutative.
(b) a * (b * c) = a * bc² = ab2c4

And (a * b) * c = ab² * c = ab2
Therefore, a * (b * c) ≠ (a * b) * c.
Hence, the binary operation * is not associative.

### Maths Class 12 Ex 1.4 Question 10.

Find which of the operations given above has identity.

Solution:
(i) The binary operation * on the set Q is defined as a * b = a – b
For identity element e, a * e = e * a = a
Putting b = e, we get a * e = a – e ≠ a and e * a = e – a ≠ a
Thus the given operation * has no identity.

(ii) The binary operation * on the set Q is defined as a * b = a² + b² ≠ a

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = a² + e² ≠ a
Thus, the given operation * has no identity.

(iii) The binary operation * on the set Q is defined as a * b = a + ab

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = a + ae ≠ a
Thus, the given operation * has no identity.

(iv) The binary operation * on the set Q is defined as a * b = (a – b)²

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = (a – e)² ≠ a for any value of e
Q.
Thus, the given operation * has no identity.

(v) The binary operation * on the set Q is defined as a * b = ab/4

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = ae/4 and e * a = ea/4 ≠ a for any value of e
Q.
Thus, the given operation * has no identity.

(vi) The binary operation * on the set Q is defined as a * b = ab²

For identity element e, a * e = e * a = a

Putting b = e, we get a * e = ae² and e * a = ea² ≠ a for any value of e Q.
Thus, the given operation * has no identity.

Thus, the operations * given in question no. 9 have no identity.

### Maths Class 12 Ex 1.4 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.

Solution:
For A = N × N, the binary operation * is defined as (a, b) * (c, d) = (a + c, b + d)
(i) Now, (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
Therefore, (a, b) * (c, d) = (c, d) * (a, b)
Hence, the binary operation * is commutative.
(ii) Again, (a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f) = ((a + c + e), (b + d + f))
and [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c + e, b + d + f))
Therefore, (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e, f)
Hence, the binary operation * is associative.
(iii) For identity element e, a * e = e * a = a

Here, it is not true. Hence, the identity element does not exist.

### Maths Class 12 Ex 1.4 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.

Solution:
(i) For an arbitrary binary operation * on a set N, it is given that
a * a = a
a N
Here, the binary operation * is not defined.
Hence, the given statement is false.
(ii) If * is a commutative binary operation on N, then c * b = b * c
(c * b) * a = (b * c) * a = a * (b * c)
Thus, a * (b * c) = (c * b) * a.
Hence, the given statement is true.

### Maths Class 12 Ex 1.4 Question 13.

Consider a binary operation * on N defined as a * b = a³ + b³. 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?

Solution:
(B) Here, option (B) is correct because it is commutative but not associative.

Since a³ + b³ = b³ + a³ implies that a * b = b * a.

And a * (b * c) ≠ (a * b) * c.