Ncert Solutions For Class 12 Maths Ex 1.4

Ncert Solutions For Class 12 Maths Chapter 1 Ex 1.4

Q-1: Check whether each of the following definitions gives a binary operation or not.

In each of the event, that * will not be a binary operation, give explanation for this condition.

(i) On Z+, define * by x * y = x – y

(ii) On Z+, define * by x * y = xy

(iii) On R, define * by x * y = xy2

(iv) On Z+, define * by x * y = |x – y|

(v) On Z +, define * by x * y = x

 

Sol:

(i) On Z+, * will be defined by x * y = x – y

It’s not a binary operation,

Such that the image of (2, 3) under * is 2 * 3 = 2 – 3 = -1 \(\notin \; Z^{+}\)

 

(ii) On Z+, * will be defined by x * y = xy.

We can see that for every x, y ɛ \(\; Z^{+}\), there will be a unique element xy in Z+ .

It means that * carries every pair (x, y) from the unique element x * y = xy in Z+.

Hence, * is the binary operation.

 

(iii) On R, * will be defined as x * y = xy2.

We can see that for every x, y ɛ R, there will be a unique element xy2 in R.

It means that * will carry each pair (x, y) to a unique element x * y = xy2 in R.

Hence, * is a binary operation.

 

(iv) On Z+, * will be defined as x * y =|x – y|.

We can see that for every x, y ɛ Z+, there will be a unique element |x – y| in Z+.

It means that * will carry each pair (x, y) to a unique element x * y = |x – y| in Z+.

Hence, * is a binary operation.

 

(v) On Z+, * will be defined as x * y = x.

We can see that for every x, y ɛ Z+, there will be a unique element x in Z+.

It means that * will carry each pair (x, y) to a unique element x * y = x in Z+.

Hence, * is a binary operation.

 

 

Q-2: For every binary operations * which is defined below, check whether * is associative or commutative in the following cases.

(i) On Z, explain x * y = x – y

(ii) On Q, explain x * y = xy + 1

(iii) On Q, explain x * y = \(\frac{xy}{2}\)

(iv) On Z+, explain x * y = 2xy

(v) On Z+, explain x * y = xy

(vi) On R – {-1}, explain x * y = \(\frac{x}{y + 1}\)

Sol:

(i) On Z, * will be defined as x * y = x – y

(2 * 3) * 4 = (2 – 3)* 4 = -1 * 4 = -1 – 4 = -5

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

Thus,

(2 * 3) * 4 ≠ 2*(3 * 4), where 2, 3, 4 ɛ Z.

Therefore, the given operation * is not associative.

Now, also

We can observe that 2 * 3 = 2 – 3 = -1 and 3 * 2 = 3 – 2 = 1

Since, 2 * 3 ≠ 3 * 2, where 2, 3 ɛ Z

Therefore, the given operation * is not associative.

(ii) On Q, * will be defined as x * y = xy + 1

We can observe here that,

(2 * 3)* 4 = (2 × 3 + 1) * 3 = 7 * 3 = 7 × 3 + 1 = 22

2 *(3 * 4) = 2 * (3 × 4 + 1) = 2 * 13 = 2 × 13 + 1 = 27

Thus,

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ Q.

Therefore, the given operation * is not associative.

Now, also

We know that: xy = ya for every x, y ɛ Q.

\(\boldsymbol{\Rightarrow }\) xy + 1 = yx + 1 for every x, y ɛ Q

\(\boldsymbol{\Rightarrow }\) x * y = x * y for every x, y ɛ Q

Therefore, the given operation * is commutative.

 

(iii) On Q, * will be defined by x * y = \(\frac{xy}{2}\)

For every x, y, z ɛ Q, we have

(x * y)* z = \(\left (\frac{xy}{2} \right) * z = \frac{\left (\frac{xy}{2} \right)z}{2} = \frac{xyz}{4}\)

And,

x *(y * z) = \( x * \left (\frac{yz}{2} \right) = \frac{x \left (\frac{yz}{2} \right)}{2} = \frac{xyz}{4}\)

Thus,

(x * y)* z = x *(y * z), where x, y, z ɛ Q

Hence, the given operation * is associative.

We know that, xy = yx for every x, y ɛ Q

\(\boldsymbol{\Rightarrow }\) \(\frac{xy}{2} = \frac{yx}{2}\) for every x, y ɛ Q.

\(\boldsymbol{\Rightarrow }\) x * y = y * x for every x, y ɛ Q.

Hence, the given operation * is commutative.

 

(iv) On Z+, * will be defined by x * y = 2xy

For every x, y, z ɛ Q, we have

(1 * 3)* 4 = 21 × 3 * 4 = 8 * 4 = 28 × 4

And,

1 *(3 * 4) = 1 * 23 × 4 = 1 * 4096 = 21 × 4096

Thus,

(1 * 3)* 4 ≠ 1 *(3 * 4), where x, y, z ɛ Q

Hence, the given operation * is not associative.

We know that, xy = yx for every x, y ɛ Q

\(\boldsymbol{\Rightarrow }\) 2xy = 2yx for every x, y ɛ Q.

\(\boldsymbol{\Rightarrow }\) x * y = y * x for every x, y ɛ Q.

Hence, the given operation * is commutative.

 

(v) On Z+, * will be defined by x * y = xy

We can see that,

For every x, y, z ɛ Z+, we have

(1 * 3)* 4 = 13 * 4 = 1 * 4 = 1 4 = 1

And,

1 *(3 * 4) = 1 * 3 4 = 1 * 81 = 181 = 1

Thus,

(1 * 3)* 4 = 1 *(3 * 4), where x, y, z ɛ Z+

Hence, the given operation * is associative.

We can also see that,

2 * 3 = 23 = 8 and 3 * 2 = 32 = 9

Hence, 2 * 3 ≠ 3 * 2, where x, y, z ɛ Z+

Therefore, the given operation * is not commutative.

 

(vi) On R, * – {-1} which is defined as x * y = \(\frac{x}{y + 1}\)

We can see that

(2 * 3)* 4 = \(\left (\frac{2}{3 + 1} \right) * 4 = \left (\frac{2}{4} \right) * 4 = \frac{\frac{1}{2}}{4 + 1} = \frac{1}{2 \times 4} = \frac{1}{8}\)

And,

2 *(3 * 4) = \(2 * \left (\frac{3}{4 + 1} \right) = 2 * \left (\frac{3}{5} \right) = \frac{2}{\frac{3}{5} + 1} = \frac{1}{\frac{3 + 5}{5}} = \frac{5}{8}\)

Thus,

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ R – {-1}

Hence, the given operation * is not associative.

 

 

Q-3: Let us consider a binary operation ^ for the set {2, 3, 4, 5, 6} which will be defined as x ^ y = min {x, y}.

Draw the operational table for the operation ^.

Sol:

As per the data given in the question, we have

A binary operation ^ for the set {2, 3, 4, 5, 6} which will be defined as x ^ y = min {x, y}.

Then,

x, y ɛ {2, 3, 4, 5, 6}

Hence, the operational table for the operation ^ given in the question will be given as below:

^ 2 3 4 5 6
2 2 2 2 2 2
3 2 3 3 3 3
4 2 3 4 4 4
5 2 3 4 5 5
6 2 3 4 5 6

 

Q-4: Let us consider a binary operation * for the set {2, 3, 4, 5, 6} which is given by the multiplication table given below:

(i) Find the value of (3 * 4) * 5 and 3 *(4 * 5).

(ii) Check whether * is commutative.

(iii) Find the value of (3 * 4)*(5 * 6)

* 2 3 4 5 6
2 2 2 2 2 2
3 2 3 2 2 3
4 2 2 4 2 2
5 2 2 2 5 5
6 2 3 2 5 6

 

Sol:

(i) (3 * 4)* 5 = 2 * 5 = 2

3 *(4 * 5) = 3 * 2 = 2

(ii) For each x, y ɛ {2, 3, 4, 5, 6}, we have x * y = y * x

Hence, the given operation is commutative.

(iii) (3 * 4)*(5 * 6) = 2 * 5 = 2

Q-5: Let us consider *’ be a binary operation for the set {2, 3, 4, 5, 6} which will be defined by x *’ y = HCF of x * y. Check whether the operation *’ is the same as the operation * which is defined and used previously? Also, do justify your answer.

Sol:

A binary operation for the set {2, 3, 4, 5, 6} which will be defined by x * y = HCF of x and y.

The operational table for the operation *’ is given as the table below:

*’ 2 3 4 5 6
2 2 2 2 2 2
3 2 3 2 2 3
4 2 2 4 2 2
5 2 2 2 5 5
6 2 3 2 5 6

 

Here, we will observe that the operational table for the operation * and *’ is the same.

Hence, the operation *’ is the same as * operation.

 

 

Q-6: Let us consider a binary operation * on N which is given by x * y = LCM of x and y. Find the following:

(i) 6 * 7, 21 * 18 (ii) Is the operation * commutative?

(iii) Is the operation * associative? (iv) Obtain the identity of * in N.

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

Sol:

As per the data given in the question, we have

A binary operation * on N which is given by x * y = LCM of x and y.

(i) 6 * 7 = LCM of 6 and 7 = 42

21 * 18 = LCM of 21 and 18 = 126

 

(ii) We know that,

LCM of x and y = LCM of y and x for every x, y ɛ N.

Hence, x * y = y * x

Therefore, the given operation * is commutative.

(iii) For x, y, z ɛ N, we have

(x * y) * z = (LCM of x and y) * z = LCM of x, y and z

x * (y * z) = x * (LCM of y and z) = LCM of x, y and z

Hence,

(x * y) * z = x * (y * z)

Therefore, the given operation * is associative.

 

(iv) We know that:

LCM of x and 1 = LCM of 1 and x = x for all x ɛ N

\(\boldsymbol{\Rightarrow }\) x * 1 = x = 1 * x for all x ɛ N

Therefore, 1 is an identity of * in N.

 

(v) The element x in N will be invertible with respect to the operation *, if there will exist an element y in N, so that x * y = e = y * x

Here, e = 1

This means,

LCM of x and y = 1 = LCM of y and x

This case will only be possible when x and y is equal to 1.

Therefore, 1 is the only element which is invertible to N with respect to the operation *

 

 

Q-7: Check whether the operation * is defined for the set {1, 2, 3, 4, 5} by x * y = LCM of x and y, is a binary operation? Also, do justify your answer.

Sol:

The operation * for the set R = {1, 2, 3, 4, 5} which is defines as x * y = LCM of x and y.

Thus,

The operational table for the given operation * will be given by:

* 1 2 3 4 5
1 1 2 3 4 5
2 2 4 6 8 10
3 3 6 9 12 15
4 4 8 12 16 20
5 5 10 15 20 25

 

We can observe from the table defined above that,

2 * 3 = 3 * 2 = 6 \(\notin \) R,

2 * 5 = 5 * 2 = 10 \(\notin \) R,

3 * 4 = 4 * 3 = 12 \(\notin \) R,

3 * 5 = 5 * 3 = 15 \(\notin \) R,

4 * 5 = 5 * 4 = 20 \(\notin \) R

Therefore, the operation * is not a binary operation.

Q- 8: Let us consider a binary operation * on N which is defined by x * y = HCF of x and y. Check whether * is commutative? Check whether * is associative? Check whether there exists any identity for the binary operation * on N?

Sol:

A binary operation * on N which is defined by x * y = HCF of x and y

We know that,

HCF of x and y = HCF of y and x for every x, y ɛ N

Thus,

x * y = y * x

Hence, the binary operation * is commutative.

For x, y, z \(\epsilon \), we have

(x * y)* z = (HCF of x and y)* z = HCF of x, y and z

x *(y * z) = x *(HCF of y and z) = HCF of x, y and z.

Hence, (x * y)* z = x *(y * z)

Thus, the binary operation * is associative.

Then,

An element e ɛ N is the identity for the binary operation * if, x * e = x = e * x for every x ɛ N.

But, this is not true for any x ɛ N.

Therefore, the given binary operation * don’t have any identity in N.

 

 

Q-9: Let us consider a binary operation * for the set Q of the rational numbers for the following:

(i) x * y = x – y (ii) x * y = x2 + y2

(iii) x * y = x + xy (iv) x * y = (x – y)2

(v) x * y = \(\frac{xy}{4}\) (vi) x * y = xy2

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

Sol:

(i) On Q, the binary operation * is defined by x * y = x – y. It will be observed that:

\(\frac{1}{3} * \frac{1}{4} = \frac{1}{3} – \frac{1}{4} = \frac{4 – 3}{12} = \frac{1}{12}\)

And,

\(\frac{1}{4} * \frac{1}{3} = \frac{1}{4} – \frac{1}{3} = \frac{3 – 4}{12} = -\frac{1}{12}\)

Therefore, \(\frac{1}{3} * \frac{1}{4} \neq \frac{1}{4} * \frac{1}{3}\), where, \(\frac{1}{3}, \; \frac{1}{4}\) ɛ Q.

Hence, the binary operation * is not commutative.

It will be observed that,

\(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \left (\frac{1}{3} – \frac{1}{4} \right) * \frac{1}{5} = \left (\frac{4 – 3}{12} \right) * \frac{1}{5} = \frac{1}{12} * \frac{1}{5} = \frac{1}{12} – \frac{1}{5} = \frac{5 – 12}{60} = \frac{-7}{60}\)

And,

\(\frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right) = \frac{1}{3} * \left (\frac{1}{4} – \frac{1}{5} \right) = \frac{1}{3} * \left (\frac{5 – 4}{20} \right) = \frac{1}{3} * \frac{1}{20} = \frac{20 – 3}{60} = \frac{17}{20}\)

Thus,

\(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right)\), where \(\frac{1}{3}, \frac{1}{4} \; and \; \frac{1}{4} \; \epsilon \; Q\)

Therefore, the binary operation * is not associative.

 

(ii) On Q, the binary operation * is defined as x * y = x2 + y2.

For x, y ɛ Q, we have

x * y = x2 + y2 = y2 + x2 = y * x

\(\boldsymbol{\Rightarrow }\) x * y = y * x

Hence, the binary operation * is commutative.

We can observe that,

(2 * 3)* 4 = (22 + 32) * 4 = (4 + 9)* 4 = 13 * 4 = 132 + 42 = 185,

And,

2 * (3 * 4) = 2 * (32 + 42) = 2 * (3 + 4) = 2 * 7 = 22 + 72 = 53

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

 

(iii) On Q, the binary operation * which is defined by x * y = x + xy

It will be observed that,

2 * 3 = 2 + 2 × 3 = 2 + 6 = 8

3 * 2 = 3 + 3 × 2 = 3 + 6 = 9

Thus, the binary operation * is not commutative.

It will be observed that,

(2 * 3)* 4 = (2 + 2 × 3)* 4 = (2 + 6)* 4 = 8 * 4 = 8 + 8 × 4 = 40

And,

2 * (3 * 4) = 2 * (3 + 3 × 4) = 2 * (3 + 12) = 2 * 15 = 2 + 2 × 15 = 32

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

Therefore, the binary operation * is not associative.

 

(iv) On Q, the binary operation * will be defined by x * y = (x – y)2

For x, y ɛ Q, we have

x * y = (x – y)2

y * x = (y – a)2 = [-(x – y)]2 = (x – y)2

Hence, the binary operation * is commutative.

We can also observe that,

(2 * 3)* 4 = (2 – 3)2 * 4 = 1 * 4 = (1 – 4)2 = (-3)2 = 9

And,

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

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

Therefore, the binary operation * is not associative.

 

(v) On Q, the binary operation * will be defined as x * y = \(\frac{xy}{4}\)

For x, y ɛ Q, we will get

x * y = \(\frac{xy}{4}\) = \(\frac{yx}{4}\) = y * x

Hence, x * y = y * x

Therefore, the binary operation * is commutative.

For x, y, z \(\epsilon \)Q, we will get

(x * y) * z = \( \left (\frac{xy}{4} \right) * z = \frac{\left (\frac{xy}{4} \right). z}{4} = \frac{xyz}{16}\)

And,

x * (y * z) = \( x * \left (\frac{yz}{4} \right) = \frac{x * \left (\frac{yz}{4} \right)}{4} = \frac{xyz}{16}\)

Hence, (x * y) * z = x * (y * z), where x, y, z \(\epsilon \)

Therefore, the binary operation * is associative.

(vi) On Q, the binary operation * will be defined as x * y = xy2.

We can observe that,

\(\frac{1}{3} * \frac{1}{4} = \frac{1}{3}.\left (\frac{1}{4} \right)^{2} = \frac{1}{3} . \frac{1}{16} = \frac{1}{48}\)

And,

\(\frac{1}{4} * \frac{1}{3} = \frac{1}{4}.\left (\frac{1}{3} \right)^{2} = \frac{1}{4} . \frac{1}{9} = \frac{1}{36}\)

Hence, \(\frac{1}{3} * \frac{1}{4}\) ≠ \(\frac{1}{4} * \frac{1}{3}\), where \(\frac{1}{3} \; and \; \frac{1}{4} \) ɛ Q.

Therefore, the binary operation * is not commutative.

We can also observe that:

\(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \left [\frac{1}{3}\left (\frac{1}{4} \right)^{2} \right] * \frac{1}{5} = \frac{1}{48} * \frac{1}{5} = \frac{1}{48} * \left (\frac{1}{5} \right)^{2} = \frac{1}{48 \times 25} = \frac{1}{1200}\)

And,

\(\frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right) = \frac{1}{3}\left [\frac{1}{4}\left (\frac{1}{5} \right)^{2} \right] = \frac{1}{3} * \frac{1}{100} = \frac{1}{3} * \left (\frac{1}{100} \right)^{2} = \frac{1}{3 \times 10000} = \frac{1}{30000}\)

Hence, \(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} ≠ \frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right)\), where \(\frac{1}{3}, \frac{1}{4}, \frac{1}{5} \; \epsilon \; Q \)

Therefore, the binary operation * is not associative.

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

Q-10: Check the identity of the operations which are equated above.

 

Sol:

Let us consider an element e ɛ Q which can be the identity element for the operation * if,

x * e = x = e * x, for every x ɛ Q.

Here, among each of the six operations, there is no such element e ɛ Q, which satisfies the above condition.

Hence, none of the six operations evaluated above have identity.

 

 

Q-11: Let us consider M = N × N and * be the binary operation on M which can be defined by

(p, q)* (r, s) = (p + r, q + s).

Prove that the operation * is commutative and associative. If there is any possibility of having an identity element, then what will be the identity element for the operation * on M?

 

Sol:

M = N × N and * be the binary operation on M which can be defined by

(p, q)* (r, s) = (p + r, q + s)

Let, (p, q), (r, s) ɛ M

Then,

p, q, r, s ɛ N

Here, we have

(p, q)*(r, s) = (p + r, q + s)

(r, s)*(p, q) = (r + p, s + q) = (p + r, q + s)

[Addition will be commutative in the set of the natural numbers]

Thus, (p, q)*(r, s) = (r, s)*(p, q)

Hence, the operation * is commutative here.

Let, (p, q), (r, s), (t, u) ɛ M

Then, p, q, r, s, t, u ɛ N

Now,

[(p, q)*(r, s)] * (t, u) = (p + r, q + s)*(t, u) = (p + r + t, q + s + u)

And,

(p, q) * [(r, s) * (t, u)] = (p, q)*(r + t, s + u) = (p + r + t, q + s + u)

Thus, [(p, q)*(r, s)] * (t, u) = (p, q) * [(r, s) * (t, u)]

Hence, the operation * is associative.

Let us consider an element e = (e1, e2) ɛ M which can be an identity element for * operation, if

x * e = x = e * x, for every x = (x1, x2) ɛ M

Then,

(x1 + e1, x2 + e2) = (x1, x2) = (e1 + x1, e2 + x2),

This cannot be true for every element in M.

Hence, the operation * will not have any identity element.

 

 

Q-12: Check whether the statements given below are true or false. Also, justify your answer.

(i) For any arbitrary binary operation * for the set N, x * x = x for every x ɛ N.

(ii) If the operation * is a commutative binary operation for N, then x *(y * z) = (z * y)* x.

Sol:

(i) Explain a binary operation * on N as x * y = x + y for every x ɛ N.

Thus, for y = x = 4, we have

4 * 4 = 4 + 4 = 8 ≠ 3.

Hence, the statement (i) is false.

(ii) RHS = (z * y)* x

= (y * z)* x [* is commutative]

= x * (y * z) [Again, as * is commutative]

= LHS

Therefore, x *(y * z) = (z * y)* x

Hence, statement (ii) is true.

 

 

Q-13: Let us consider a binary operation * for N defined as x * y = x3 + y3. Select the correct answer.

(a) Is * commutative but, not associative?

(b) Is * both commutative and associative?

(c) Is * neither commutative nor associative?

(d) Is * associative but, not commutative?

Sol:

On N, the binary operation * will be defined as x * y = x3 + y3.

For x, y ɛ N, we have

x * y = x3 + y3 = y3 + x3 = y * x [Addition is commutative in N]

Hence, the binary operation * is commutative.

We can observe that,

(2 * 3)* 4 = (23 + 33)* 4 = (8 + 27)* 4 = 35 * 4 = 353 + 44 = 42875 + 16 = 42891

And,

2 *(3 * 4) = 2 *(33 + 43) = 2 *(27 + 64) = 2 * 91 = 23 + 913 = 8 + 753571 = 753579

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ N

Thus, the operation * is not associative.

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

Hence, the correct statement is a.

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