Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Ex 1.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Ex 1.4

Question 1.

Determine whether or not each of the definitions 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) a > b, a * b = a – b > 0 which belongs to Z^{+}.

But if a < b, a * b = a – b < 0 which does not belong to Z^{+}.

⇒ given operation * is not a binary operation.

(ii) For all a and b belonging to Z^{+}, ab also belongs to Z^{+}.

∴ The operation *, defined by a * b = ab is a binary operation.

(iii) For all a and b belonging to R, ab² also belongs to R.

⇒ The operation *, defined by a * b = ab² is a binary operation.

(iv) For all a and b belonging to Z^{+}, |a – b| also belongs to Z^{+}.

∴ The operation a * b = |a-b| is a binary operation.

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

a, b ∈ Z^{+} ⇒ a ∈ Z^{+}.

∴ 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 = \(\frac { a }{ 2 }\)

(iv) On Z_{+} define a * b = 2^{a}

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

(vi) On R – {-1}, define a * b = \(\frac { a }{ b+1}\)

Solution:

(i) On Z, operation * is defined as

(a) a * b = a – b

⇒ b * a = b – a

But a – b + b – a ⇒ a * b # b * a

(b) a – (b – c) * (a-b)-c

Binary operation defined as # is not associative.

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

(a) ab + 1 = ba + 1, a * b = b * a

∴ Defined binary operation is commutative.

(b) a* (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1

and (a * b) * c = (ab + 1) * c = (ab + 1)c + 1

= abc + c + 1

⇒ a * (b * c) # (a * b) * c

∴ Binary operation defined as * is not associative.

(iii) (a) On Q, operation * is defined as a * b = \(\frac { ab }{ 2 }\)

∴ a * b = b * a

∴ Operaion binary defined as * is commutative.

(b) a * (b * c) = a* \(\frac { bc }{ 2 }\) = \(\frac { abc }{ 2 }\).

and (a * b) * c = \(\frac { ab }{ 2 }\) * c = \(\frac { abc }{ 4 }\).

⇒ Defined binary operation * is associative.

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

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

⇒ a * b = b * a

∴ Binary operation defined as * is commutative.

(b) a * (b * c) = a * 2^{ab} = 2^{a.2b}

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

Thus, (a * b) * c ≠ a * (b * c).

∴ Binary operation defined as * is not associative.

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

(a) b * a = ba

∴ a^{b} ≠ b^{a} ⇒ a * b # b * a

* is not commutative.

(b) (a * b) * c = a^{b} * c = (a^{b})^{c} = a^{bc}

a * (b * c) = a * b^{c} = a^{bc}

Thus, (a * b) * c # (a * b * c)

∴ Operation * is not associative.

(vi) On Z^{+}, operation * is defined as

a + b = \(\frac { a }{ b+1 }\), b # 1

∴ b * a = \(\frac { b }{ a+1 }\)

(a) a * b * b * a

Binary operation defined above is not commutative.

⇒ Binary operation defined above is not associative.

Question 3.

Consider the infimum binary operation Λ on the set {1, 2, 3, 4, 5} defined by a Λ b = min of a and b. Write the multiplication table of the operation Λ.

Solution:

Operation A table on the set (1, 2, 3, 4, 5} is as follows:

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).

Solution:

(i) From the table given above, we find:

2 * 3 = 1, 1 * 4 = 1

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

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

(ii) Let a, b ∈ {1, 2, 3, 4, 5}

From the given table, we find:

a * a = a,

a * b = b * a = 1, when a or b or both odd and a≠b.

2 * 4 = 4 * 2 = 2, when a and b are even and a≠b.

Thus, a * b = b * a

∴ Given binary operation * is commuative.

(iii) (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 = HCF of a and b. Is.the operation *’ same as operation * defined in quesion 4 above? Justify your answer.

Solution:

The set is {1, 2, 3, 4, 5} and a *’ b = HCF of a and b.

Let us prepare the table of operaion *’:

This table is the same as given in Question 4.

∴ Operations *’ and * (as defined in Question No. 4) are the same.

Question 6.

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

(i) Find 5 * 7 and 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:

Binary operation * 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) a * b = L.C.M. of a and b

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

= a * b = b * a, since L.C.M. of a, b and b, a are equal.

Binary operation * is commutative.

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

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

⇒ a * (b * c) = (a * b) * c

⇒ Given binary operation * is associative.

(iv) Identity of * in N is 1 because

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

(v) Let * : N x N → N defined as a * b = L.C.M. of (a, b)

For a = 1, b = 1, a * b = 1 = b * a. Otherwise a * b * 1.

∴ Binary operation * is not invertible.

⇒ 1 is invertible for operation *.

Question 7.

Is * defined on the set {1,2,3,4,5} by a * b = L.C.M. of a and b as a binary operation. Justify your answer.

Solution:

The given set = {1, 2, 3, 4, 5}.

Binary operation * is defined as a * b = L.C.M. of a and b.

4 * 5 = 20 which does not belong to the given set {1, 2, 3, 4, 5}. So, it is not a binary operation.

Question 8.

Let * be the binary operation of N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does their exist identity for this operation on N?

Solution:

Binary operation on set N is defined as

a * b = H.C.F. of a and b.

(a) We know that H.C.F. of a, b = H.C.F. of b, a.

∴ a * b = b * a

Binary operation * is commutative.

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

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

= H.C.F. of a, b and c.

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

⇒ (a * b) * c – a * (b * c)

So, binary operation * as defined above is associative.

(c) 1 * a = a * 1 = 1 * a.

There does not exists any identity element.

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 = \(\frac { ab }{ 4 }\)

(vi) a * b = ab²

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

Solution:

Operation * is on the set Q.

(i) Defined as a * b = a – b

(a) Now b * a = b – a. But a – b ≠ b – a

∴ a * b ≠ b * a

∴ Operation * is not commutative.

(b) a * (b * c) = a * (b-c)

= a – (b-c)

= a – b + c

(a * b) * c = (a – b) * c

= a – b – c

Thus a * (b * c) ≠ (a * b) * c

The operation *, as defined above is not associative.

(ii) (a) a * b = a² + b²

b * a = b² + a² = a² + b²

∴ a * b = b * a

∴ This binary operation * is commutative.

(b) a * (b * c) = a * (b² + c²) = a² + (b² + c²)²

(a * b) * c = (a² + b²) * c = (a² + b²) + c²

⇒ a * (b * c) ≠ (a * b) * c

∴ The given operation * is not associaive.

(iii) Operation * is defined as a * b = a + ab

(a) b * a = b + ba

∴ a * b = b * a

∴ This operation * is not commutative.

(b) a * (b * c) = a * (b + bc)

= a + a(b + be)

= a +ab + abc

(a * b) * c = (a + ab) * c

= a + ab + (a + ab).c

= a + ab + ac + abc

⇒ a * (b * c) # (a * b) * c

⇒ The given binary operation * is not associative.

(iv) The binary operation * is defined as

a * b = (a – b)²

(a) b * a = (b – a)² = (a – b)²

⇒ a * b = b * a

This binary operation * is commutative.

(b) a * (b * c) = a * (b – c)²

= [a – (b – c)²]²

(a * b) * c = (a – b)² * c

= [(a-b)²-c]²

⇒ a * (b * c) # (a * b) * c

∴ The operation * is not associative.

(v) Binary operaion * is defined as

a * b = \(\frac { ab }{ 4 }\)

∴ The operation * is commutative.

(b) a * (b * c) = a * \(\frac { ab }{ 4 }\) = \(\frac { a }{ 4 }\)( \(\frac { bc }{ 4 }\)) = \(\frac { abc }{ 16 }\)

(a * b) * c = \(\frac { ab }{ 4 }\) * c = \(\frac { ab }{ 4 }\) * c = \(\frac { ab }{ 4 }\).\(\frac { c }{ 4 }\) = \(\frac { abc }{ 16 }\)

⇒ (a * b) * c = a * (b * c)

Thus, the given operation * is associative.

(vi) Binary operation * is defined as

a * b = ab²

(a) b * a = ba² ≠ ab²

∴ a * b * b * a.

∴ The operation * is not commuative.

(b) a * (b * c) = a * bc² = a(bc²)² = abc²

(a * b) * c = ab² *c = (ab²)c² = ab²c²

∴ a * (b * c) * (a * b) * c.

∴ Given binary operation * is not associative.

Question 10.

Show that none of the operations given above has identity.

Solution:

The binary operation * on set Q is

(i) defined as a * b = a – b.

For idenity element e, a * e = e * a = a.

But a * e = a – e # a and e * a = e – a # a.

∴ There is no identity element for this operation.

(ii) Binary operation * is defined as

a * b = a² + b²

Here, a * e = a² + e² # a

and e * a – e² + a² # a.

So, this operation * has no identity.

(iii) The binary operation * is defined as

a * b = a + ab

We have: a * e = a + ae # a

and e * a = e + ea # a.

∴ There is no identity element.

(iv) The binary operation * is defined as

a * b = (a – b)²

Put b = e. We get a * e = (a – e)² # a

and e * a = (e – a)² # a

for any value of e ∈ Q.

⇒ There is no identity element.

(v) The operation * is

a * b = \(\frac { ab }{ 4 }\).

and e * a = \(\frac { ab }{ 4 }\) # a

for any value of e ∈ Q.

∴ Operation * has no identity element.

(vi) The operation * is a * b = ab².

Put b = e. We get a * e = ae² # a

and e * a = ea² # a

for any value of e ∈Q.

⇒ There is no identity element.

Thus, these operations have no identity.

Question 11.

Let A = N x 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 for * oh A, if any.

Solution:

A = N x N

Binary operation * is defined as

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

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

= (a + c, b + d)

⇒ (a, b) * (c, d) = (c, d) * (a, b)

∴ This operation * is commutative.

(b) Next

(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)]

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

∴ The given binary operation is associative.

(c) Identity element does not exist, because there is no e and e’ in N x N such that (a, b) * (e, e’) = (a + e, b + e’) = (a, b).

Question 12.

State whether the following statements are true or false. Justify.

(i) For any arbitrary binary operation * on a set N,

a * a = a ∀a ∈ N.

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

Solution:

(i) A binary operation on N is defined as

a * a = a ∀a ∈ N.

Here, operation * is not defined.

∴ Given statement is false.

(ii) * is a binary commutative operation on N.

⇒ c * b = b * c [∵ * is commutative]

∴ (c * b) * a = (b * c) * a = a * (b * c)

∴ a * (b * c) = (c * b) * a.

∴ This statement is true.

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:

The binary operation * on set N is defined as

a * b = a³ +b³

Now, b * a = b³ + a³ = a³ +b³.

⇒ a * b = b * a.

∴ The operation * is commutative.

Further, a * (b * c) = a * (b³ + c³) = a³ + (b³ + c³)³

and (a * b) * c = (a³ + b³) * c = (a³ +b³)³ + c³

⇒ a * (b * c) ≠ (a * b) * c

∴ This operation is not associative.

Thus, this operation is commutative but not associative.

Part (B) is the correct answer.