# Exercise 1.4 Solutions

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Exercise 1.4
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}
1.i Determine whether or not * given by {a * b = a - b} gives a binary operation.
Consider two positive integers {a, b}Z+
Now, when {a \gt b}, then {a * b = a - b \gt 0}
a - bZ+
However, when {a \lt b}, then {a * b = a - b \lt 0}
{⇒ a - b}Z+ as {a - b \lt 0}
* is not a binary operation on Z+
1.ii Determine whether or not * given by {a * b = ab} gives a binary operation.
Consider two positive integers {a, b}Z+
As we know, the product of two positive numbers is always positive.
⇒ As aZ+ and bZ+, the product ab also belongs to Z+. i.e., abZ+ (∵ the product of two positive integers is also a positive integer).
* is a binary operation on Z+
1.iii Determine whether or not * given by {a * b = ab^2} gives a binary operation.
Consider two positive integers {a, b}Z+
As bR, b^2R (∵ the square of a real number is also a real number)
As it is given that aR and also as we’ve seen above that b^2R,
{⇒ ab^2}R
* is a binary operation on R
1.iv Determine whether or not * given by {a * b = |a - b|} gives a binary operation.
Consider two positive integers {a, b}Z+
Now when {a - b \le 0} then {|a - b| \gt 0}
And also when {a - b \geq 0} then {|a - b| \gt 0}
As it is given that aZ+ and bZ+, |a - b|Z+ (∵ The modulus of any number, whether it is positive or negative, is always positive)
* is a binary operation on Z+
1.v Determine whether or not * given by {a * b = a} gives a binary operation.
Consider two positive integers {a, b}Z+
As aZ+, the result a * b = aZ+
* is a binary operation on Z+

2. For each 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 = \dfrac{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 = \dfrac{a}{b + 1}}
2.i For the operation * defined on Z as {a * b = a – b} determine whether * is binary, commutative or associative.
To check whether * on Z is binary.
Consider two integers {a, b}Z
As we know, the difference of two integers is always an integer.
So, as aZ and bZ, {a - b}Z
* is a binary operation on Z
To check whether * on Z is commutative.
Consider two integers a, bZ
Now, {a * b = a - b} and {b * a = b - a}
As we know, a - bb - a
⇒ a * bb * a
∴ The binary operation * is not commutative on the set Z.
To check whether * on Z is associative.
Consider three integers {a, b} and cZ
We have {(a * b) * c = (a - b) * c}
{= (a - b) - c}
{= a - b - c}
Now, {a * (b * c) = a * (b - c)}
{= a - (b - c)}
{= a - b + c}
Clearly, a - b - ca - b + c
⇒ (a * b) * ca * (b * c)
∴ The binary operation * is associative on the set Z.
2.ii For the operation * defined on Q as {a * b = ab + 1} determine whether * is binary, commutative or associative.
To check whether * on Q is binary.
Consider two rational numbers {a, b}Q
As we know, the product of two rational numbers is also a rational number.
Thus, when aQ and bQ, abQ
{⇒ ab + 1}Q
* is binary operation on Q
To check whether * on Q is commutative.
Consider two rational numbers {a, b}Q
Now, {a * b = ab + 1}
And {b * a = ba + 1}
We know that the product of two rational numbers is commutative.
{⇒ ab = ba}
{⇒ ab + 1 = ba + 1}
{⇒ a * b = b * a}
∴ The binary operation * is commutative on the set Q
To check whether * on Q is associative.
consider three rational numbers a, b, cQ
Now {(a * b) * c = (ab + 1) * c}
{= (ab + 1) × c + 1}
{= abc + c + 1}
Now, {a * (bc + 1)}
{= a × (bc + 1) + 1}
{= abc + a + 1}
Clearly {abc + c + 1}{abc + a + 1}
{⇒ (a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Q
2.iii For the operation * defined on Q as {a * b = \dfrac{ab}{2}} determine whether * is binary, commutative or associative.
To check whether * on Q is binary.
Consider two rational numbers {a, b}Q
As we know, the product of two rational numbers is also a rational number.
Thus, when aQ and bQ, abQ
{⇒ \dfrac{ab}{2}}Q
* on Q is binary.
To check whether * on Q is commutative.
Consider two elements {a, b}Q
Now {a * b = \dfrac{ab}{2}}
And {b * a = \dfrac{ba}{2}}
As we know, the product of two rational numbers is commutative.
{⇒ ab = ba}
{⇒ \dfrac{ab}{2} = \dfrac{ba}{2}}
{⇒ a * b = b * a}
∴ The binary operation * is commutative on the set Q.
To check whether * on Q is associative
Consider three rational numbers a, b, cQ
Now {(a * b) * c = \dfrac{ab}{2} * c}
{= \dfrac{\dfrac{ab}{2}\times{c}}{2}}
{= \dfrac{abc}{4}}
{a * (b * c) = a * \dfrac{bc}{2}}
{= \dfrac{a\times\dfrac{bc}{2}}{2}}
{= \dfrac{abc}{4}}
{⇒ (a * b) * c = a * (b * c)}
∴ The binary operation * is associative on the set Q.
2.iv For the operation * defined on Z+ as {a * b = 2^{ab}} determine whether * is binary, commutative or associative.
To check whether * on Z+ is binary.
Consider two positive integers {a, b}Z+
As we know, when aZ+ and bZ+, abZ+
⇒ 2^{ab}Z+
* on Z+ is binary.
To check whether * on Z+ is commutative.
Consider any two positive integers a, bZ+
As multiplication on Z+ is commutative,
We have, {ab = ba}
{⇒ 2^{ab} = 2^{ba}}
{⇒ a * b = b * a}
∴ The binary operation * is commutative on the set Z+.
To check whether * on Z+ is associative
Consider any three positive integers a, b, cZ+
Now {(a * b) * c = 2^{ab} * c}
{= 2^{2^{ab} × c}}
Also, {a * (b * c) = a * 2^{bc}}
{= 2^{a × 2^{bc}}}
So, {(a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Z+.
2.v For the operation * defined on Z+ as {a * b = a^b} determine whether * is binary, commutative or associative.
To check whether * on Z+ is binary.
Consider two positive integers a, bZ+
As we know, when a, bZ+, a^bZ+
* on Z+ is binary.
To check whether * on Z+ is commutative.
Consider any two positive integers {a, b}Z+.
We have {a * b = a^b}
and {b * a = b^a}
Clearly {a * b}{b * a}
∴ The binary operation * is not commutative on the set Z+.
To check whether * on Z+ is associative.
Consider any three positive integers, {a, b, c}Z+
We have {(a * b) * c = a^b * c}
{= (a^b)^c}
{= a^{bc}}
Now, {a * (b * c) = a * b^c}
{= a^{b^c}}
So, {(a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Z+.
2.vi For the operation * defined on R – {-1} as {a * b =\dfrac{a}{b + 1}}, determine whether * is binary, commutative or associative.
To check whether * on R – {-1} is binary.
Consider any two real numbers {a, b}R – {-1}.
We have
{a * b = \dfrac{a}{b + 1}}
The above equation is valid for all {a, b}R – {-1}.
* on R – {-1} is binary.
To check whether * on R – {-1} is commutative.
Consider two real numbers {a, b}R – {-1}
Now {a * b = \dfrac{a}{b + 1}}
{b * a = \dfrac{b}{a + 1}}
So, a * bb * a
∴ The binary operation * is not commutative on the set R – {-1}
To check whether * on R – {-1} is associative.
Consider any three real numbers, a, b, cR – {-1}
We have {(a * b) * c = \dfrac{a}{b + 1} * c}
{= \dfrac{\dfrac{a}{b + 1}}{c + 1}}
{= \dfrac{a}{(b + 1)\times(c + 1)}}
Now {a * (b * c) = a * \dfrac{b}{c + 1}}
{= \dfrac{a}{\dfrac{b}{c + 1} + 1}}
{= \dfrac{a}{\dfrac{b + c + 1}{c + 1}}}
{= \dfrac{{a}\times{(c + 1)}}{b + c + 1}}
{⇒ (a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set R – {-1}.

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 ∧
The set is given as {1, 2, 3, 4, 5} and the binary operation ∧ is given as a ∧ b = min {a, b}
The following is the operation table of the operation ∧ on the set {1, 2, 3, 4, 5}
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

4. Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).
(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
*
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
4.(i) To compute {(2 * 3) * 4} and {2 * (3 * 4)}
{(2 * 3) * 4}
{= 1 * 4} {(∵ 2 * 3 = 1} from the table)
{= 1~(∵ 1 * 4 = 1} from the table)
And {2 * (3 * 4)}
{= 2 * 1} {(∵ 3 * 4 = 1} from the table)
{= 1~(∵ 2 * 1 = 1} from the table)
4.(ii) To check whether * is commutative
From the given table it can be noticed that n^{th} row = n^{th} column
{⇒ a * b = b * a}
∴ The binary operation * is commutative on the set {1, 2, 3, 4, 5} {∀ a, b} ∈ {1, 2, 3, 4, 5}
4.(iii) To compute {(2 * 3) * (4 * 5)}
{(2 * 3) * (4 * 5)}
{= 1 * 1} {(∵ 2 * 3 = 1} and {4 * 5 = 1} from the given table)
{= 1~(∵ 1 * 1 = 1} from the given table)

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.
Given that *′ is the binary operation on the set {1, 2, 3, 4, 5} defined by {a *}b = H.C.F. of a and b
To check whether *′ same as * defined in the exercise 4 above, let’s first build the table for *′.
Operation table for *
*′
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
It can be noted that the operation tables for both * and *′ are same.
∴ The operations * and *′ are the same.

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 *?
i. To find {5 * 7}, {20 * 16}
We’ve {5 * 7} = L.C.M. of 5 and 7 = 35
2
20,
16
2
10,
8
5,
4
Now {20 * 16} = L.C.M. of 20 and 16 = 80
ii. To check whether * is commutative on N:
We have {a * b} = L.C.M of a and b
And {b * a} = L.C.M of b and a
As we know, L.C.M. of a and b = L.C.M. of b and a
{⇒ a * b = b * a}
∴ The binary operation * is commutative on the set N
iii. To check whether * is associative on N:
We have {(a * b) * c} = (L.C.M. of a and b) * c
= L.C.M. of (L.C.M. of a and b) and c
= L.C.M. of a, b and c
Similarly {a * (b * c)} = a * (L.C.M. of b and c)
= L.C.M. of a and (L.C.M. of b and c)
= L.C.M. of a, b and c
Thus we see that {(a * b) * c = a * (b * c)}
∴ The binary operation * is associative on the set N
iv. To find the identity of * in N:
An element e will be the identity of * if it satisfies the condition
{a * e = e * a = a}
⇒ L.C.M. of a and e = L.C.M of e and a = a
As we know, in the set of natural numbers N, the L.C.M. of any element a and 1 is a.
Thus, we have L.C.M. of a and 1 = L.C.M. of 1 and a {= a}
{⇒ a * 1} {= 1 * a} {= a~∀~a}N
∴ 1 is the identity element of the binary operation * on the set N.
To find which elements are invertible for the operation *
Any element aN is invertible with respect to the binary operation *, if there exists an element bN such that
{a * b} {= e} {= b * a}
Where e is the identity element under the binary operation * and as we have seen in the previous case, {e = 1}
As {a * b} {= b * a}
⇒ L.C.M of {a} and b = L.C.M. of b and a
This condition is satisfied only when both a and b are equal to 1.
∴ The only invertible element in the binary operation * on the set N is 1.

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.
The binary operation * is defined as
{a * b} = L.C.M of a and b where {a, b} ∈ {1, 2, 3, 4, 5}
The corresponding operation table will be as follows:
L.C.M
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
Considering {a = 3} and {b = 4}, We have
{a * b} = L.C.M. of a and b
= L.C.M. of 3 and 4
= 12
As 12 ∉ {1, 2, 3, 4, 5}, the operation * is not a binary operation on the set {1, 2, 3, 4, 5}

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?
To check whether * is commutative:
For the operation * to be commutative, we should have
{a * b = b * a}
Now, {a * b} = H.C.F of a and b
and {b * a} = H.C.F. of b and a
We know that H.C.F of a and b = H.C.F of b and a
{∴ a * b = b * a}
∴ The binary operation * is commutative.
To check whether * is associative.
Consider there elements a, b and cN
We have {a * (b * c)} {= a * (}H.C.F. of b and c) = H.C.F. of {a, b} and c.
Similarly {(a * b) * c} = (H.C.F. of a and b) {* c} = H.C.F. of {a, b} and c.
This can be well understood by considering the following example.
Consider three numbers 3, 6 and 11 ∈ N. The calculation of their H.C.F’s will be as follows:
Factors of 3 = 1, 3
Factors of 6 = 1, 2, 3, 6
Factors of 11 = 1, 11
Now {(3 * 6) * 11}
= (H.C.F of 3 and 6) * 11
= 3 * 11
= H.C.F. of 3 and 11
= 1
and {3 * (6 * 11)}
= 3 * (H.C.F. of 6 and 11)
= 3 * 1
= H.C.F. of 3 and 1
= 1
Thus we see that {(3 * 6) * 11 = 3 * (6 * 11)}
∴ The binary operation * is associative on the set N.
Identity element of the binary operation *.
For the binary operation *, the identitiy element will be an element eN such that aN, {a * e = e * a = a}
⇒ H.C.F of a and e = H.C.F. of e and a {= a}
However, the set N does not have an element that satisfies the above condition.
∴ The binary operation * does not have an identity element.

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 = \dfrac{ab}{4}}
(vi)
{a * b = ab^2}
Find which of the binary operations are commutative and which are associative.
(i).a To check whether {a * b = a - b} is commutative:
Consider two rational numbers a, bQ.
We have, {a * b = a - b}
and {b * a = b - a}
As {a - b}{b - a}, it is clear that {a * b}{b * a}.
∴ The binary operation * is not commutative on the set Q.
(i).b To check whether {a * b = a - b} is associative.:
Consider three rational numbers a, b, cQ.
We have, {(a * b) * c} {= (a - b) * c} {= a - b - c}
and {a * (b * c))} {= (a * (b - c)} {= a - (b - c)} {= a - b + c}
Clearly {(a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Q.
(ii).a. To check whether {a * b = a^2 + b^2} is commutative.
Consider two rational numbers a, bQ.
We have {a * b = a^2 + b^2}
and {b * a = b^2 + a^2}
So, {a * b} {= a^2 + b^2} {= b^2 + a^2} {= b * a}
∴ The binary operation * is commutative on the set Q.
(ii).b To check whetehr {a * b = a^2 + b^2} is associative.
Consider three rational numbers {a, b, c}Q.
We have {(a * b) * c} {= (a^2 + b^2) * c} {= (a^2 + b^2)^2 + c^2}
and {a * (b * c)} {= a * (b^2 + c^2)} {= a^2 + (b^2 + c^2)^2}
So, it is clear that {(a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Q.
(iii).a To check whetehr a * b = a + ab is commutative.
Consider two rational numbers a, bQ
We have {a * b = a + ab}
and {b * a = b + ba}
So, it is clear that {a * b}{b * a}.
∴ The binary operation * is not commutative on the set Q.
(iii).b To check whether {a * b = a + ab} is associative on the set Q.
Consider three rational numbers a, b, c ∈ Q
We have {(a * b) * c} {= (a + ab) * c} {= a + ab + (a + ab)c} {= a + ab + ac + abc}
and {a * (b * c)} {= a * (b + bc)} {= a + a(b + bc)} {= a + ab + abc}
So, clearly {(a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Q.
(iv).a To check whether {a * b = (a - b)^2} is commutative.
Consider two rational numbers a, bQ
We have {a * b} {= (a - b)^2} {= a^2 - 2ab + b^2}
and {b * a} {= (b - a)^2} {= b^2 - 2ba + a^2} {= a^2 - 2ab + b^2}
So, it is clear that {a * b = b * a}
∴ The binary operation * is commutative on the set Q.
(iv).b To check whether {a * b = (a - b)^2} is associative.
Consider three rational numbers a, b and cQ
We have {(a * b) * c} {= (a - b)^2 * c} {= ((a - b)^2 - c)^2} {= (a^2 - 2ab + b^2 - c)^2}
and {a * (b * c)} {= a * (b - c)^2} {= (a - (b - c)^2)^2} {= (a - (b^2 - 2bc + c^2)^2} {= (a - b^2 + 2bc + c^2)^2}
So, clearly {(a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Q.
(v).a To check whether {a * b = \dfrac{ab}{4}} is commutative.
Consider two rational numbers a, bQ
We have {a * b = \dfrac{ab}{4}}
and {b * a = \dfrac{ba}{4} = \dfrac{ab}{4}}
So, clearly {a * b = b * a}
∴ The binary operation * is commutative on the set Q
(v).b To check whether {a * b = \dfrac{ab}{4}} is associative.
Consider three rational numbers a, b and cQ
We have {(a * b) * c} {= \dfrac{ab}{4} * c} {= \dfrac{\dfrac{ab}{4} × c}{4}} {= \dfrac{abc}{16}}
and {a * (b * c)} {= a * \dfrac{bc}{4}} {= \dfrac{a × \dfrac{bc}{4}}{4}} {= \dfrac{abc}{16}}
So, it is clear that {(a * b) * c = a * (b * c)}
∴ The binary operation * is associative on the set Q.
(vi).a To check whether {a * b = ab^2} is commutative.
Consider two rational numbers {a, b}Q
We have {a * b = ab^2}
and {b * a = ba^2}
Clearly, {a * b}{b * a}
∴ The binary operation * is not commutative on the set Q
(vi).b To check whether {a * b = ab^2} is associative.
Consider three rational numbers {a, b} and cQ
We have {(a * b) * c} {= ab^2 * c} {= ab^2c^2}
and {a * (b * c)} {= a * bc^2} {= a × (bc^2)^2} {= ab^2c^4}
So, it is clear that {(a * b) * c}{a * (b * c)}
∴ The binary operation * is not associative on the set Q.
A check for whether these binary operations are commutative and/or associative is summarised in the following table.
*
{a * b}
Commutative?
Associative?
(i)
{a - b}
(ii)
{a^2 + b^2}
(iii)
{a + ab}
(iv)
{(a - b)^2}
(v)
{\dfrac{ab}{4}}
(vi)
{ab^2}

10. Find which of the operations given above has identity.
Note: If the birnary operation * is not commutative, then we can directly conclude that there is no identity element for the binary operation * on the set Q. However, if the binary operation * is binary, it does not necessarily mean that there is an identity. We have to ensure that it satisfy the condition {a * e = e * a = a}
10.i. Identity element for the binary operation * defined by {a * b = a - b}
An element eQ will be an identity element of the binary operation * on the set Q, if for any arbitrary element aQ, we have
{a * e = e * a = a}
Now, {a * e} {= a - e}
and {e * a} {= e - a}
Obviously there is no element e ∈ Q such that {a - e = e - a = a}
∴ The binary operation * does not have an identity element on the set Q.
10.ii. Identity element for the binary operation * defined by {a * b = a^2 + b^2}
An element eQ will be an identity element of the binary operation * on the set Q, if for any arbitrary element aQ, we have
{a * e = e * a = a}
Now, {a * e} {= a^2 + e^2}
and {e * a} {= e^2 + a^2}
Obviously there is no element eQ such that {a^2 + e^2 = e^2 + a^2 = a}
∴ The binary operation * does not have an identity element on the set Q.
10.iii. Identity element for the binary operation * defined by {a * b = a + ab}
An element eQ will be an identity element of the binary operation * on the set Q, if for any arbitrary element aQ, we have
{a * e = e * a = a}
Now, {a * e} {= a + ae}
and {e * a} {= e + ea}
Obviously there is no element eQ such that {a + ae = e + ea = a}
∴ The binary operation * does not have an identity element on the set Q.
10.iv. Identity element for the binary operation * defined by {a * b = (a - b)^2}
An element eQ will be an identity element of the binary operation * on the set Q, if for any arbitrary element aQ, we have
{a * e = e * a = a}
Now, {a * e} {= (a - e)^2}
and {e * a} {= (e - a)^2}
Obviously there is no element eQ such that {(a - e)^2 = (e - a)^2 = a}
∴ The binary operation * does not have an identity element on the set Q.
10.v. Identity element for the binary operation * defined by {a * b = \dfrac{ab}{4}}
An element eQ will be an identity element of the binary operation * on the set Q, if for any arbitrary element aQ, we have
{a * e = e * a = a}
Now, {a * e} {= \dfrac{ae}{4}}
and {e * a} {= \dfrac{ea}{4}}
Obviously the element 4 ∈ Q exists such that {\dfrac{a × 4}{4} = \dfrac{4 × a}{4} = a}
∴ 4 is the identity element for the binary operation * on the set Q.
10.vi. Identity element for the binary operation * defined by {a * b = ab^2}
An element eQ will be an identity element of the binary operation * on the set Q, if for any arbitrary element aQ, we have
{a * e = e * a = a}
Now, {a * e} {= ae^2}
and {e * a} {= ea^2}
Obviously there is no element eQ such that {ae^2 = ea^2 = a}
∴ The binary operation * does not have an identity element on the set Q.
Whether the binary operation * has the identity element or not is summarized in the following table.
S. No.
{a * b}
Has Identity?
(i)
{a - b}
(ii)
{a^2 + b^2}
(iii)
{a + ab}
(iv)
{(a - b)^2}
(v)
{\dfrac{ab}{4}}
Yes – 4
(vi)
{ab^2}

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.
To check whether the binary operation * defined by {(a, b) * (c, d) = (a + c, b + d)} is commutative.
Consider four elements a, b, c, dN, such that {(a, b), (c, d)}A where A = N × N
Now, {(a, b) * (c, d)} {= (a + c, b + d)}
and {(c, d) * (a, b)} {= (c + a, d + b)} {= (a + c, b + d)} {(∵ c + a = a + c} and {d + b = b + d)}
So, clearly {(a, b) * (c, d)} {= (c, d) * (a, b)} {∀~a, b, c, d}N
∴ The binary operation * is commutative on the set A.
To check whether teh binary operation * defined by {(a, b) * (c, d) = (a + c, b + d)} is associative.
Consider six elements {a, b, c, d, e, f}N, such that {(a, b), (c, d), (e, f)}A
Now {((a, b) * (c, d)) * (e, f)} {= (a + c, b + 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)}
So, obviously {((a, b) * (c, d)) * (e, f)} {= (a, b) * ((c, d) * (e, f))}
∴ The binary operation * is associative on the set A
To find the identity element of the binary operation defined by {(a, b) * (c, d) = (a + c, b + d)}
The identity element of the binary operation * should be an element {(e, e)} such that {(a, b) * (e, e) = (e, e) * (a, b) = (a, b)} where {a, b, e}N such that {(a, b), (e, e)}N
We have {(a, b) * (e, e)} {= (a + e, b + e)}
We’ll have {(a + e, b + e) = (a, b)} only when {e = 0}.
Similary, {(e, e) * (a, b)} {= (e + a, e + b)}
We’ll have {(a + e, b + e) = (a, b)} only when {e = 0}
However, as 0 ∉ N, the identity element {(e, e)} for the binary operation * does not existing on the set A = N × N.
∴ The binary operation * does not have an identity element {(e, e)} on the set A = N × N
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}
To check the truth value of the statement (i)
Let’s assume that the arbitrary binary operation * is defined as {a * a = a × a} for an arbitrary number aN
Now {a * a = a × a = a^2}
As we know, a^2a.
∴ The arbitrary binary operation defined as {a * a = a~∀~a}N is not true.
∴ The statement (i) is False.
To check the truth value of the statement (ii)
It is given in the statement that the binary operation * is commutative.
This implies that for any two elements a, bN, we have {a * b = b * a}
Now, consider any three elements a, b and cN.
We have
{a * (b * c)}
{= (b * c) * a} {(∵ *} is commutative)
{= (c * b) * a} {(∵ *} is commutative, we’ve {b * c = c * b)}
{⇒ a * (b * c) = (c * b) * a}
∴ The given statement (ii) is True.
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?
To check whether the binary operation * defined as {a * b = a^3 + b^3} is commutative.
Consider any natural numbers a, bN
We have {a * b = a^3 + b^3}
and {b * a} {= b^3 + a^3} {= a^3 + b^3}
∴ The binary operation * is commutative on the set N
To check whether the binary operation * defined as {a * b = a^3 + b^3} is associative.
Consider any three natural numbers {a, b, c}N
we have {(a * b) * c} {= (a^3 + b^3) * c} {= (a^3 + b^3)^3 + c^3}
and {a * (b * c)} {= a * (b^3 + c^3)} {= a^3 + (b^3 + c^3)^3}
So, clearly {(a * b) * c)}{a * (b * c)}
∴ The binary operation * is not associative on the set N.
To summarise, the binary operation * is commutative but not associative on the set N.
So, the option B is the correct answer.