GSEB Class 12 Maths Solutions Chapter 1 સંબંધ અને વિધેય Exercise 1.4

GSEB Solutions Class 12 Maths Chapter 1 संघ अने विधेय Ex 1.4

 

Question 1. नीचे प्रमाणे व्याख्यायित करेला प्रत्येक क्रिया * ए द्विक्रिया छे के नहि ते नक्की करो. जे प्रश्नमां * द्विक्रिया न होय, तेना माटे कारण आपो.
(i) * ए \( Z^+\) पर, a * b = a – b द्वारा व्याख्यायित छे.
Answer:
\(Z^+ = \{1, 2, 3, 4, ......\}\)
a * b = a - b
\{ धन होय क्यारे a > b होय.
ऋण होय क्यारे a < b होय.
If a = 2, b = 5, then a, b are in \(Z^+\).
So, 2 * 5 = 2 - 5 = -3. This result -3 is not in \(Z^+\).
Therefore, * is not a binary operation on \(Z^+\).
In simple words: A binary operation means that combining two numbers from a set always gives a result that is also in that set. Here, for positive whole numbers, subtracting two numbers might result in a negative number, which is not part of the positive whole number set.

🎯 Exam Tip: For binary operations, always check if the result of the operation on any two elements from the given set remains within that same set.

 

Question. (ii) * ए \( Z^+\) पर, a * b = ab द्वारा व्याख्यायित छे.
Answer:
a, b ∈ \(Z^+\)
a * b = ab ∈ \(Z^+\), ∀ a > b माटे
∀ a < b माटे
Thus, * is a binary operation on \(Z^+\).
In simple words: When you multiply any two positive whole numbers, the answer is always another positive whole number. So, this operation works for the set of positive whole numbers.

🎯 Exam Tip: Multiplication of positive integers always yields a positive integer, making it a straightforward example of a binary operation on \(Z^+\).

 

Question. (iii) * ए R पर, a * b = \(ab^2\) द्वारा व्याख्यायित छे.
Answer:
Let a ∈ R and b ∈ R.
Then, a * b = \(ab^2\) ∈ R.
For example, if a = 4.5 and b = 10, then a, b ∈ R.
So, a * b = 4.5 * 10 = (4.5)\((10)^2\) = 450 ∈ R.
Thus, * is a binary operation on R.
In simple words: For any real numbers 'a' and 'b', if you calculate 'a' multiplied by 'b' squared, the result will always be another real number. This means the operation works for all real numbers.

🎯 Exam Tip: When proving a binary operation on real numbers, show that for any two real numbers, the result of the operation is also a real number, possibly with an example.

 

Question. (iv) * ए \( Z^+\) पर, a * b = |a - b| द्वारा व्याख्यायित छे.
Answer:
\(Z^+ = \{1, 2, 3, 4, ...........\}\)
If a, b ∈ \(Z^+\), then
a * b = |a - b| > 0 \( \implies \) |a - b| ∈ \(Z^+\)
For example, if a = 3 and b = 5, then 3 * 5 = |3 - 5| = |-2| = 2 ∈ \(Z^+\).
Thus, * is a binary operation on \(Z^+\).
In simple words: For any two positive whole numbers, finding the absolute difference between them will always result in a positive whole number. So, this operation works on the set of positive whole numbers.

🎯 Exam Tip: The absolute value of the difference between two positive integers will always be a non-negative integer. Since the set is \(Z^+\) (positive integers), the result will always be a positive integer, confirming it as a binary operation.

 

Question. (v) * ए \( Z^+\) पर, a * b = a द्वारा व्याख्यायित छे.
Answer:
a, b ∈ \(Z^+\)
\(Z^+ = \{1, 2, 3, 4, .......\}\)
a * b = a ∈ \(Z^+\)
For example, 5 * 3 = 5 ∈ \(Z^+\) for 5, 3 ∈ \(Z^+\).
Thus, * is a binary operation on \(Z^+\).
In simple words: When you operate on two positive whole numbers by simply taking the first number, the result will always be a positive whole number. This means the operation works on the set of positive whole numbers.

🎯 Exam Tip: A simple projection operation like this always produces an element from the set, so it's inherently a binary operation.

 

Question 2. नीचे दर्शाव्या प्रमाणे व्याख्यायित प्रत्येक क्रिया * माटे नक्की करो के ते द्विक्रिया छे के नहि. जो ते द्विक्रिया होय, तो ए समक्रमी छे के नहि अथवा जूथना नियमनुं पालन करे छे के नहि :
(i) Z पर व्याख्यायित a * b = a - b
Answer:
* is defined on Z as a * b = a - b.
Z = \{.......-3, -2, -1, 0, 1, 2, 3,.......\}
If a > b, then a * b = a - b is positive.
If a < b, then a * b = a - b is negative.
If a = b, then a * b = a - b = 0.
Thus, * is a binary operation on Z.
Now, let's check for commutativity:
For example, 5 * 3 = 5 - 3 = 2.
And 3 * 5 = 3 - 5 = -2.
Since 5 * 3 ≠ 3 * 5,
it means a * b ≠ b * a for a, b ∈ Z.
Therefore, the binary operation * is not commutative.
Now, let's check for associativity:
Let 2, 6, 8 ∈ Z.
(2 * 6) * 8 = (2 - 6) * 8 = -4 * 8 = -4 - 8 = -12.
2 * (6 * 8) = 2 * (6 - 8) = 2 * (-2) = 2 - (-2) = 4.
Since (2 * 6) * 8 ≠ 2 * (6 * 8),
it means (a * b) * c ≠ a * (b * c) for a, b, c ∈ Z.
Therefore, the binary operation * does not satisfy the associative law.
In simple words: The subtraction operation on whole numbers is a binary operation because the result is always a whole number. However, it's not commutative because changing the order changes the answer (like 5-3 is not 3-5). It's also not associative because grouping the numbers differently changes the answer (like (2-6)-8 is not 2-(6-8)).

🎯 Exam Tip: When proving non-commutativity or non-associativity, a single counterexample is sufficient. For commutativity, check if a * b = b * a. For associativity, check if (a * b) * c = a * (b * c).

 

Question. (ii) Q पर व्याख्यायित a * b = ab + 1
Answer:
* is defined on Q as a * b = ab + 1.
If a, b ∈ Q, then a * b = ab + 1 ∈ Q.
For example, if 5, 3 ∈ Q, then 5 * 3 = (5)(3) + 1 = 15 + 1 = 16 ∈ Q.
Thus, * is a binary operation on Q.
Now, let's check for commutativity:
a * b = ab + 1
b * a = ba + 1 = ab + 1

\( \implies \) a * b = b * a
Therefore, the binary operation * is commutative.
Now, let's check for associativity:
Let a, b, c ∈ Q.
a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1.
(a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1.
Since a * (b * c) ≠ (a * b) * c (because a + 1 ≠ c + 1 in general),
Therefore, the binary operation * does not satisfy the associative law.
In simple words: The operation a*b = ab+1 on rational numbers is binary because the result is always a rational number. It is commutative because a*b gives the same result as b*a. But it is not associative because changing how you group the numbers (a*(b*c) versus (a*b)*c) gives different results.

🎯 Exam Tip: For operations involving multiplication and addition, commutativity often holds, but associativity requires careful verification, as the order of operations can significantly change the outcome.

 

Question. (iii) Q पर व्याख्यायित a * b = \( \frac{ab}{2} \)

Answer:
* is defined on Q as a * b = \( \frac{ab}{2} \).
If a, b ∈ Q, then a * b = \( \frac{ab}{2} \) ∈ Q.
For example, if a = 4 and b = \( \frac{1}{2} \), then 4 * \( \frac{1}{2} \) = \( \frac{(4)(\frac{1}{2})}{2} \) = \( \frac{2}{2} \) = 1 ∈ Q.
Thus, * is a binary operation on Q.
Now, let's check for commutativity:
a * b = \( \frac{ab}{2} \)
b * a = \( \frac{ba}{2} \) = \( \frac{ab}{2} \)
Therefore, a * b = b * a, so the binary operation * is commutative.
Now, let's check for associativity:
Let a, b, c ∈ Q.
a * (b * c) = a * \( \frac{bc}{2} \) = \( \frac{a \left(\frac{bc}{2}\right)}{2} \) = \( \frac{abc}{4} \).
(a * b) * c = \( \frac{ab}{2} \) * c = \( \frac{\left(\frac{ab}{2}\right)c}{2} \) = \( \frac{abc}{4} \).
Since a * (b * c) = (a * b) * c,
Therefore, the binary operation * satisfies the associative law.
In simple words: The operation a*b = ab/2 on rational numbers is binary because the result is always a rational number. It is commutative because a*b is the same as b*a. It is also associative because the way you group the numbers (a*(b*c) or (a*b)*c) does not change the final result.

🎯 Exam Tip: Operations involving division might be tricky for associativity. Always perform the calculations step-by-step for both sides of the associative property to ensure accuracy.

 

Question. (iv) \( Z^+\) पर व्याख्यायित a * b = \( 2^{ab} \)
Answer:
* is defined on \( Z^+\) as a * b = \( 2^{ab} \).
\( Z^+ = \{1, 2, 3,...\} \).
For example, if a = 3, b = 2 ∈ \( Z^+\), then 3 * 2 = \( 2^{(3)(2)} \) = \( 2^6 \) = 64 ∈ \( Z^+\).
So, if a, b ∈ \( Z^+\), then a * b = \( 2^{ab} \) ∈ \( Z^+\).
Thus, * is a binary operation on \( Z^+\).
Now, let's check for commutativity:
a * b = \( 2^{ab} \)
b * a = \( 2^{ba} \) = \( 2^{ab} \)
Therefore, a * b = b * a, so the binary operation * is commutative.
Now, let's check for associativity:
Let a, b, c ∈ \( Z^+\).
a * (b * c) = a * \( 2^{bc} \) = \( 2^{a \cdot 2^{bc}} \).
(a * b) * c = \( 2^{ab} \) * c = \( 2^{2^{ab} \cdot c} \).
Since \( 2^{a \cdot 2^{bc}} \) ≠ \( 2^{2^{ab} \cdot c} \) (in general),
For example, let a = 1, b = 2, c = 3.
1 * (2 * 3) = 1 * \( 2^{(2)(3)} \) = 1 * \( 2^6 \) = 1 * 64 = \( 2^{(1)(64)} \) = \( 2^{64} \).
(1 * 2) * 3 = \( 2^{(1)(2)} \) * 3 = \( 2^2 \) * 3 = 4 * 3 = \( 2^{(4)(3)} \) = \( 2^{12} \).
Since \( 2^{64} \) ≠ \( 2^{12} \), the operation is not associative.
Therefore, the binary operation * does not satisfy the associative law.
In simple words: The operation where a*b is 2 raised to the power of (a times b) on positive whole numbers is a binary operation because the result is always a positive whole number. It is commutative because a times b is the same as b times a. However, it is not associative because grouping the numbers differently (like 1*(2*3) versus (1*2)*3) gives different results.

🎯 Exam Tip: Operations involving exponents often look commutative but fail the associativity test due to the layered nature of exponentiation. Always test with simple values if unsure.

 

Question. (v) \( Z^+\) पर व्याख्यायित a * b = \(a^b\).
Answer:
* is defined on \( Z^+\) as a * b = \(a^b\).
\( Z^+ = \{1, 2, 3,...\} \).
If a, b ∈ \( Z^+\), then a * b = \(a^b\) ∈ \( Z^+\).
For example, if a = 5, b = 3 ∈ \( Z^+\), then 5 * 3 = \( 5^3 \) = 125 ∈ \( Z^+\).
Thus, * is a binary operation on \( Z^+\).
Now, let's check for commutativity:
a * b = \(a^b\).
b * a = \(b^a\).
For example, 5 * 3 = \( 5^3 \) = 125.
3 * 5 = \( 3^5 \) = 243.
Since 5 * 3 ≠ 3 * 5,
Therefore, the binary operation * is not commutative.
Now, let's check for associativity:
Let a, b, c ∈ \( Z^+\).
a * (b * c) = a * \(b^c\) = \(a^{(b^c)}\).
(a * b) * c = \(a^b\) * c = \((a^b)^c\).
Since \(a^{(b^c)}\) ≠ \((a^b)^c\) (in general, because \(a^{(b^c)}\) = \(a^{b^c}\) and \((a^b)^c\) = \(a^{bc}\) and \(b^c \neq bc\) generally).
For example, let a = 2, b = 3, c = 2.
2 * (3 * 2) = 2 * \( 3^2 \) = 2 * 9 = \( 2^9 \) = 512.
(2 * 3) * 2 = \( 2^3 \) * 2 = 8 * 2 = \( 8^2 \) = 64.
Since 512 ≠ 64, the operation is not associative.
Therefore, the binary operation * does not satisfy the associative law.
In simple words: The operation where a*b means 'a' raised to the power of 'b' on positive whole numbers is a binary operation because the result is always a positive whole number. It is not commutative because changing the order (like \(5^3\) versus \(3^5\)) gives different answers. It is also not associative because grouping the numbers differently (like \(a^{(b^c)}\) versus \((a^b)^c\)) changes the answer.

🎯 Exam Tip: Exponentiation is a common example to demonstrate non-commutativity and non-associativity. Clearly showing the difference between \(a^{(b^c)}\) and \((a^b)^c\) is key for associativity.

 

Question. (vi) R- {-1} पर व्याख्यायित a * b = \( \frac{a}{b+1} \)
Answer:
* is defined on R - \{-1\} as a * b = \( \frac{a}{b+1} \).
If a, b ∈ R - \{-1\}, then a * b = \( \frac{a}{b+1} \) ∈ R - \{-1\} (since b ≠ -1, b+1 ≠ 0).
For example, if a = 9, b = 2 ∈ R - \{-1\}, then 9 * 2 = \( \frac{9}{2+1} \) = \( \frac{9}{3} \) = 3 ∈ R - \{-1\}.
Thus, * is a binary operation on R - \{-1\}.
Now, let's check for commutativity:
a * b = \( \frac{a}{b+1} \)
b * a = \( \frac{b}{a+1} \)
For example, if a = 3, b = 5 ∈ R - \{-1\}, then 3 * 5 = \( \frac{3}{5+1} \) = \( \frac{3}{6} \) = \( \frac{1}{2} \).
And 5 * 3 = \( \frac{5}{3+1} \) = \( \frac{5}{4} \).
Since 3 * 5 ≠ 5 * 3,
Therefore, the binary operation * is not commutative.
Now, let's check for associativity:
Let a = 2, b = 3, c = 5 ∈ R - \{-1\}.
(2 * 3) * 5 = \( \left(\frac{2}{3+1}\right) \) * 5 = \( \frac{2}{4} \) * 5 = \( \frac{1}{2} \) * 5 = \( \frac{\frac{1}{2}}{5+1} \) = \( \frac{\frac{1}{2}}{6} \) = \( \frac{1}{12} \).
2 * (3 * 5) = 2 * \( \left(\frac{3}{5+1}\right) \) = 2 * \( \frac{3}{6} \) = 2 * \( \frac{1}{2} \) = \( \frac{2}{\frac{1}{2}+1} \) = \( \frac{2}{\frac{3}{2}} \) = \( \frac{4}{3} \).
Since (2 * 3) * 5 ≠ 2 * (3 * 5),
Therefore, the binary operation * does not satisfy the associative law.
In simple words: The operation a*b = a/(b+1) on real numbers (excluding -1) is a binary operation because the result is always a real number (and the denominator is never zero). It is not commutative because changing the order (like 3*5 versus 5*3) gives different answers. It is also not associative because grouping the numbers differently (like (2*3)*5 versus 2*(3*5)) changes the answer.

🎯 Exam Tip: Operations involving division or fractions require careful calculation of both sides of the commutative and associative properties. Pay close attention to the order of operations and the denominator's value.

 

Question 3. धारो के गण {1, 2, 3, 4, 5} पर द्विक्रिया ^, a ^ b = min {a, b} (अथवा न्यूनतम {a, b}) द्वारा व्याख्यायित छे. क्रिया ^ माटे द्विक्रिया कोष्टक लखो.
Answer:
The operation ^ is defined on the set {1, 2, 3, 4, 5} as a ^ b = min{a, b} (the minimum of a and b).
The binary operation table for ^ 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

In simple words: This table shows the result of the 'minimum' operation for every pair of numbers from 1 to 5. For example, if you pick 3 and 4, the minimum is 3, so 3 ^ 4 = 3.

🎯 Exam Tip: When constructing a binary operation table, ensure that each cell correctly reflects the result of the operation for the corresponding row and column elements. For the 'minimum' operation, the result is always the smaller of the two numbers.

 

Question 4. धारो के गण {1, 2, 3, 4, 5} पर द्विक्रिया *, नीचे आपेला गुणाकार कोष्टक द्वारा व्याख्यायित करेला छे :
(i) (2 * 3) * 4 अने 2 * (3 * 4) नी गणतरी करी.
(ii) * समक्रमी छे ?
(iii) (2 * 3) * (4 * 3) नी गणतरी करो.
(सूचन : नीचे आपेल कोष्टकनो उपयोग करो.)

*	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:
(i) (2 * 3) * 4 and 2 * (3 * 4) are calculated as follows:
From the table, 2 * 3 = 1.
So, (2 * 3) * 4 = 1 * 4 = 1.
Also, from the table, 3 * 4 = 1.
So, 2 * (3 * 4) = 2 * 1 = 1.
(ii) Check if * is commutative:
From the table, 3 * 4 = 1.
And 4 * 3 = 1.
Since 3 * 4 = 4 * 3, the operation * is commutative.
(iii) Calculate (2 * 3) * (4 * 5):
From the table, 2 * 3 = 1.
From the table, 4 * 5 = 1.
So, (2 * 3) * (4 * 5) = 1 * 1 = 1.
In simple words: (i) Using the table, (2 * 3) is 1, so (2 * 3) * 4 becomes 1 * 4, which is 1. Also, (3 * 4) is 1, so 2 * (3 * 4) becomes 2 * 1, which is 1. (ii) Yes, the operation is commutative because changing the order of numbers, like 3*4 and 4*3, gives the same result (1). (iii) (2 * 3) is 1 and (4 * 5) is 1, so (2 * 3) * (4 * 5) becomes 1 * 1, which is 1.

🎯 Exam Tip: For problems involving operation tables, accurately read the values from the table. Commutativity can be checked by comparing values across the main diagonal (e.g., cell(i,j) == cell(j,i)).

 

Question 5. गण {1, 2, 3, 4, 5} पर द्विक्रिया *' ए a *' b = a अने b नो गु.सा.अ. द्वारा व्याख्यायित छे. क्रिया *' ए उपरना प्रश्न 4 मां व्याख्यायित द्विकक्रिया * जेवी ज छे ? तमारा जवाबनी यथार्थता चकासो.
Answer:
The operation *' is defined on the set {1, 2, 3, 4, 5} as a *' b = HCF of a and b.
Let's create the operation table for this *' operation:
1 *' 1 = HCF (1, 1) = 1
1 *' 2 = HCF (1, 2) = 1
2 *' 4 = HCF (2, 4) = 2
The table for this operation is:

*'	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

Comparing this table with the table from Question 4, both tables are identical.
Therefore, the operation *' (HCF) is the same as the operation * defined in Question 4.
In simple words: The new operation 'a *' b' means finding the greatest common factor (HCF) of 'a' and 'b'. When we make a table for this operation, it looks exactly like the table from Question 4. This means both operations are actually the same.

🎯 Exam Tip: To verify if two operations are the same, compare their respective operation tables. If all corresponding entries are identical, then the operations are indeed equivalent.

 

Question 6. N पर a * b = a अने b नो ल.सा.अ. द्वारा व्याख्यायित द्विक्रिया * आपेल छे.
(i) 5 * 7, 20 * 16 मेळवो.
(ii) * समक्रमी छे ?
(iii) * जूथना नियमनुं पालन करे छे ?
(iv) N मां * माटे तटस्थ घटक शोघो.
(v) द्विक्रिया * माटे N ना कया घटको व्यस्तसंपन्न छे ?
Answer:
The operation * is defined on N (natural numbers) as a * b = LCM of a and b.
(i) Calculate 5 * 7 and 20 * 16.
5 * 7 = LCM(5, 7) = 35.
20 * 16 = LCM(20, 16) = 80.

Explanation for LCM(20, 16):

2	20	16
2	10	8
2	5	4
2	5	2
5	5	1
	1	1

LCM = 2 * 2 * 2 * 2 * 5 = 16 * 5 = 80.
(ii) Check if * is commutative:
a * b = LCM(a, b).
LCM(a, b) = LCM(b, a).
= b * a.
Thus, the operation * is commutative.
(iii) Check if * satisfies the associative law:
(a * b) * c = LCM(a, b) * c = LCM(LCM(a, b), c) = LCM(a, b, c).
a * (b * c) = a * LCM(b, c) = LCM(a, LCM(b, c)) = LCM(a, b, c).
Since (a * b) * c = a * (b * c),
Thus, the operation * satisfies the associative law.
(iv) Find the identity element for * in N.
Let e be the identity element for *.
Then, a * e = a = e * a for all a ∈ N.
LCM(a, e) = a.
This means that 'e' must be a factor of 'a' for all 'a'. The only natural number that is a factor of every natural number is 1.
Clearly, e = 1 ∈ N.
For example, 5 * 1 = LCM(5, 1) = 5.
Thus, the identity element for * is e = 1.
(v) Find the invertible elements for * in N.
Let b be the inverse of element a.
Then, a * b = e, where e = 1 is the identity element.
LCM(a, b) = 1.
For the LCM of two natural numbers to be 1, both numbers must be 1.
So, LCM(a, b) = 1 implies a = 1 and b = 1.
Therefore, only the element 1 in N has an inverse under the operation *, and the inverse of 1 is 1 itself.
In simple words: (i) The operation means finding the Least Common Multiple (LCM). So, 5*7 is LCM(5,7)=35, and 20*16 is LCM(20,16)=80. (ii) Yes, it's commutative because LCM(a,b) is always the same as LCM(b,a). (iii) Yes, it's associative because LCM(LCM(a,b),c) is the same as LCM(a,LCM(b,c)), which is just LCM(a,b,c). (iv) The identity element is 1, because LCM(a,1) is always 'a'. (v) Only the number 1 has an inverse, which is also 1, because LCM(a,b) can only be 1 if both 'a' and 'b' are 1.

🎯 Exam Tip: For LCM/HCF operations, commutativity and associativity are generally true. The identity element is crucial: for LCM, it's 1; for HCF, it depends on the set. An element is invertible only if its operation with its inverse yields the identity element.

 

Question 7. गण {1, 2, 3, 4, 5} पर a * b = a अने b नो ल.सा.अ. द्वारा व्याख्यायित * ए द्विक्रिया छे ? तमारा जवाबनी यथार्थता चकासो.
Answer:
The operation * is defined on the set {1, 2, 3, 4, 5} as a * b = LCM of a and b.
Let's check if it is a binary operation on this set.
Let a = 3 and b = 4 from the set {1, 2, 3, 4, 5}.
Then, a * b = 3 * 4 = LCM(3, 4) = 12.
However, 12 is not an element of the set {1, 2, 3, 4, 5}.
Therefore, the operation * is not a binary operation on the set {1, 2, 3, 4, 5}.
In simple words: For an operation to be binary on a set, combining any two numbers from that set must result in a number that is also in the same set. Here, if we take 3 and 4 from the set {1, 2, 3, 4, 5} and find their LCM, we get 12. Since 12 is not in the set, this operation is not a binary operation on this specific set.

🎯 Exam Tip: Always check the closure property for a binary operation. If even one pair of elements from the set produces a result outside the set, it is not a binary operation on that set.

 

Question 8. गण N पर a * b = a अने b नो गु.सा.अ. द्वारा व्याख्यायित द्विक्रिया * समक्रमी छे ? शुं * जूथना नियमनुं पालन करे छे ? N परनी आ द्विक्रिया माटे तटस्थ घटकनुं अस्तित्व छे ?
Answer:
The binary operation * is defined on N as a * b = HCF of a and b.
Let a, b ∈ N.
a * b = HCF(a, b).
Since HCF(a, b) = HCF(b, a),
a * b = b * a.
Thus, the operation * is commutative.
Let a, b, c ∈ N.
(a * b) * c = HCF(a, b) * c = HCF(HCF(a, b), c) = HCF(a, b, c).
a * (b * c) = a * HCF(b, c) = HCF(a, HCF(b, c)) = HCF(a, b, c).
Since (a * b) * c = a * (b * c),
Thus, the operation * satisfies the associative law.
Now, let's find the identity element for * in N.
Let e be the identity element for *.
Then, a * e = a = e * a for all a ∈ N.
HCF(a, e) = a.
For HCF(a, e) to be 'a', 'e' must be a multiple of 'a'. This must hold for all 'a' in N. However, no single natural number 'e' can be a multiple of every natural number 'a' (unless 'a' itself is 1, in which case e can be 1). If 'a' = 2, HCF(2, e) = 2, so e must be a multiple of 2. If 'a' = 3, HCF(3, e) = 3, so e must be a multiple of 3. There is no single 'e' that is a multiple of every 'a' in N.
For example, if we take e=1, then HCF(a,1)=1, which is not 'a' for a>1.
If we take a * e = a, it means HCF(a, e) = a. This implies that 'a' must divide 'e'. For this to be true for all 'a' in N, 'e' would have to be an infinitely large multiple, which is not in N. So, no identity element exists in N for the HCF operation.
Thus, an identity element for this binary operation does not exist in N.
In simple words: The HCF (Highest Common Factor) operation on natural numbers is commutative because HCF(a,b) is the same as HCF(b,a). It is also associative because HCF(HCF(a,b),c) is the same as HCF(a,HCF(b,c)). However, there is no identity element for this operation in natural numbers. An identity element 'e' would mean HCF(a,e) = a for all 'a'. This would require 'e' to be a multiple of every natural number 'a', which is impossible.

🎯 Exam Tip: While HCF is typically commutative and associative, the existence of an identity element is not guaranteed for every binary operation. For HCF on N, an identity element does not exist as no single natural number 'e' can satisfy HCF(a, e) = a for all 'a' ∈ N.

 

Question 9. संमेय संख्याओना गण Q पर द्विक्रिया * नीचे दर्शाव्या प्रमाणे व्याख्यायित छे :
(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 = \( \frac{ab}{4} \)
(vi) a * b = \(ab^2\)
कइ द्विक्रियाओ समक्रमी छे अने कइ क्रियाओ जूथना नियमनुं पालन करे छे ते शोघो.
Answer:
(i) Operation on Q: a * b = a - b
Let a, b ∈ Q.
Commutativity: a * b = a - b. Also, b * a = b - a.
Since a - b ≠ b - a (unless a=b), for example 5 - 3 = 2, but 3 - 5 = -2.
Therefore, a * b ≠ b * a. Thus, * is not commutative.
Associativity: Let a, b, c ∈ Q.
(a * b) * c = (a - b) * c = (a - b) - c = a - b - c.
a * (b * c) = a * (b - c) = a - (b - c) = a - b + c.
Since a - b - c ≠ a - b + c (unless c = 0 or c = -c i.e. c=0), for example (5-3)-2 = 2-2 = 0, but 5-(3-2) = 5-1 = 4.
Therefore, (a * b) * c ≠ a * (b * c). Thus, * is not associative.
In simple words: For rational numbers, subtraction (a*b = a-b) is not commutative because changing the order gives a different result. It is also not associative because grouping the numbers differently changes the answer.

🎯 Exam Tip: Subtraction is a classic operation that is neither commutative nor associative. Use clear examples to demonstrate this.

 

Question. (ii) Operation on Q: a * b = \(a^2 + b^2\)

Answer:
Let a, b ∈ Q.
Commutativity: a * b = \(a^2 + b^2\). Also, b * a = \(b^2 + a^2\).
Since \(a^2 + b^2 = b^2 + a^2\),
Therefore, a * b = b * a. Thus, * is commutative.
Associativity: Let a, b, c ∈ Q.
(a * b) * c = \((a^2 + b^2)\) * c = \((a^2 + b^2)^2 + c^2\).
a * (b * c) = a * \((b^2 + c^2)\) = \(a^2 + (b^2 + c^2)^2\).
Since \((a^2 + b^2)^2 + c^2 \neq a^2 + (b^2 + c^2)^2\) (in general), for example (1*2)*3 = \((1^2+2^2)\)*3 = 5*3 = \(5^2+3^2\) = 25+9=34. And 1*(2*3) = 1*\((2^2+3^2)\) = 1*13 = \(1^2+13^2\) = 1+169=170.
Therefore, (a * b) * c ≠ a * (b * c). Thus, * is not associative.
In simple words: The operation a*b = \(a^2 + b^2\) on rational numbers is commutative because \(a^2 + b^2\) is always the same as \(b^2 + a^2\). However, it is not associative because grouping the numbers differently (like (a*b)*c versus a*(b*c)) gives different results.

🎯 Exam Tip: Sums of squares are often commutative. For associativity, expanding and comparing the expressions is crucial, as the nested squaring can lead to different results.

 

Question. (iii) Operation on Q: a * b = a + ab

Answer:
Let a, b ∈ Q.
Commutativity: a * b = a + ab. Also, b * a = b + ba.
Since a + ab ≠ b + ba (in general), for example 1*2 = 1 + 1(2) = 3, but 2*1 = 2 + 2(1) = 4.
Therefore, a * b ≠ b * a. Thus, * is not commutative.
Associativity: Let a, b, c ∈ Q.
(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c = a + ab + ac + abc.
a * (b * c) = a * (b + bc) = a + a(b + bc) = a + ab + abc.
Since a + ab + ac + abc ≠ a + ab + abc (unless ac = 0, i.e., a=0 or c=0), for example (1*2)*3 = 3*3 = 3 + 3(3) = 12. And 1*(2*3) = 1*(2 + 2(3)) = 1*8 = 1 + 1(8) = 9.
Therefore, (a * b) * c ≠ a * (b * c). Thus, * is not associative.
In simple words: The operation a*b = a + ab on rational numbers is not commutative because changing the order (a*b vs b*a) gives different answers. It is also not associative because grouping numbers differently changes the final result.

🎯 Exam Tip: When an operation involves terms like 'ab' and 'a', direct expansion is usually the best approach to verify commutativity and associativity. Use counterexamples if properties do not hold.

 

Question. (iv) Operation on Q: a * b = \((a - b)^2\)

Answer:
Let a, b ∈ Q.
Commutativity: a * b = \((a - b)^2\). Also, b * a = \((b - a)^2\).
Since \((a - b)^2 = (-(b - a))^2 = (b - a)^2\),
Therefore, a * b = b * a. Thus, * is commutative.
Associativity: Let a, b, c ∈ Q.
(a * b) * c = \((a - b)^2\) * c = \((\,(a - b)^2 - c\,)^2\).
a * (b * c) = a * \((b - c)^2\) = \((\,a - (b - c)^2\,)^2\).
Since \((\,(a - b)^2 - c\,)^2 \neq (\,a - (b - c)^2\,)^2\) (in general), for example (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 = \((1-1)^2\) = \(0^2\) = 0.
Therefore, (a * b) * c ≠ a * (b * c). Thus, * is not associative.
In simple words: The operation a*b = \((a-b)^2\) on rational numbers is commutative because squaring the difference \((a-b)\) gives the same result as squaring \((b-a)\). But it is not associative because grouping the numbers differently (like (a*b)*c versus a*(b*c)) gives different answers.

🎯 Exam Tip: The square of a difference \((a-b)^2\) is equal to \((b-a)^2\), which makes this operation commutative. However, nesting this operation, as in associativity, generally leads to different outcomes due to the nature of squaring differences.

 

Question. (v) Operation on Q: a * b = \( \frac{ab}{4} \)

Answer:
Let a, b ∈ Q.
Commutativity: a * b = \( \frac{ab}{4} \). Also, b * a = \( \frac{ba}{4} \).
Since \( \frac{ab}{4} = \frac{ba}{4} \),
Therefore, a * b = b * a. Thus, * is commutative.
Associativity: Let a, b, c ∈ Q.
(a * b) * c = \( \frac{ab}{4} \) * c = \( \frac{(\frac{ab}{4})c}{4} \) = \( \frac{abc}{16} \).
a * (b * c) = a * \( \frac{bc}{4} \) = \( \frac{a(\frac{bc}{4})}{4} \) = \( \frac{abc}{16} \).
Since \( \frac{abc}{16} = \frac{abc}{16} \),
Therefore, (a * b) * c = a * (b * c). Thus, * is associative.
In simple words: The operation a*b = ab/4 on rational numbers is commutative because multiplying numbers in any order and dividing by 4 gives the same result. It is also associative because grouping the numbers differently when multiplying and dividing by 4 does not change the final answer.

🎯 Exam Tip: Operations involving simple multiplication and division often exhibit both commutativity and associativity, as seen here. Algebraic simplification should confirm these properties.

 

Question. (vi) Operation on Q: a * b = \(ab^2\)

Answer:
Let a, b ∈ Q.
Commutativity: a * b = \(ab^2\). Also, b * a = \(ba^2\).
Since \(ab^2 \neq ba^2\) (in general), for example 1*2 = \(1 \cdot 2^2\) = 4, but 2*1 = \(2 \cdot 1^2\) = 2.
Therefore, a * b ≠ b * a. Thus, * is not commutative.
Associativity: Let a, b, c ∈ Q.
(a * b) * c = \((ab^2)\) * c = \((ab^2)c^2\).
a * (b * c) = a * \((bc^2)\) = \(a(bc^2)^2\) = \(a(b^2c^4)\) = \(ab^2c^4\).
Since \((ab^2)c^2 \neq ab^2c^4\) (in general, unless \(c^2 = c^4\) i.e., c=0 or c=1), for example (1*2)*3 = \((1 \cdot 2^2)\)*3 = 4*3 = \(4 \cdot 3^2\) = 4 * 9 = 36. And 1*(2*3) = 1*\((2 \cdot 3^2)\) = 1*18 = \(1 \cdot 18^2\) = 324.
Therefore, (a * b) * c ≠ a * (b * c). Thus, * is not associative.
In simple words: The operation a*b = \(ab^2\) on rational numbers is not commutative because changing the order gives different results (like \(1 \cdot 2^2\) versus \(2 \cdot 1^2\)). It is also not associative because grouping the numbers differently (like (a*b)*c versus a*(b*c)) gives different answers.

🎯 Exam Tip: Operations with exponents on only one variable (like \(ab^2\)) typically fail both commutativity and associativity. Use clear examples with small integer values to demonstrate this.

 

Question 10. उपर आपेल पैकी कइ द्विक्रियाओ माटे तटस्थ घटक प्राप्य छे ?
(i) a * b = a - b
Answer:
Operation on Q: a * b = a - b.
Let e be the identity element for *.
Then, a * e = a and e * a = a for all a ∈ Q.
a * e = a \( \implies \) a - e = a \( \implies \) e = 0.
e * a = a \( \implies \) e - a = a \( \implies \) e = 2a.
For 'e' to be an identity element, it must be unique. Here, e = 0 from the first condition, but e = 2a from the second condition, which depends on 'a'. Since these two values are not generally equal (0 ≠ 2a for a ≠ 0), a unique identity element does not exist.
Thus, an identity element does not exist for this operation.
In simple words: For the subtraction operation, an identity element 'e' would mean that doing 'a - e' gives 'a' and 'e - a' also gives 'a'. If a - e = a, then e must be 0. But if e - a = a, then e must be 2a. Since 'e' must be a single fixed number, and here it depends on 'a', there is no identity element.

🎯 Exam Tip: An identity element 'e' must satisfy both a * e = a and e * a = a, and 'e' must be a unique constant. If 'e' depends on 'a' or yields conflicting values, then no identity element exists.

 

Question. (ii) a * b = \(a^2 + b^2\)

Answer:
Operation on Q: a * b = \(a^2 + b^2\).
Let e be the identity element for *.
Then, a * e = a and e * a = a for all a ∈ Q.
a * e = a \( \implies \) \(a^2 + e^2\) = a \( \implies \) \(e^2 = a - a^2\).
For 'e' to be an identity element, it must be independent of 'a'. Here, \(e^2\) depends on 'a'. This means a unique identity element does not exist.
Thus, an identity element does not exist for this operation.
In simple words: For the operation a*b = \(a^2 + b^2\), an identity element 'e' would mean that \(a^2 + e^2\) should equal 'a'. This would mean \(e^2\) must equal \(a - a^2\). Since this value changes for different 'a', there isn't one single 'e' that works for all 'a'. So, no identity element exists.

🎯 Exam Tip: If the equation for the identity element (\(e = \dots\)) contains the variable 'a', it implies that no unique identity element exists for the operation.

 

Question. (iii) a * b = a + ab

Answer:
Operation on Q: a * b = a + ab.
Let e be the identity element for *.
Then, a * e = a and e * a = a for all a ∈ Q.
a * e = a \( \implies \) a + ae = a \( \implies \) ae = 0.
This implies either a = 0 or e = 0. If a = 0, then 0 + 0e = 0, which is 0 = 0 (true). But if a ≠ 0, then e = 0.
e * a = a \( \implies \) e + ea = a \( \implies \) e(1 + a) = a \( \implies \) e = \( \frac{a}{1+a} \).
Since 'e' must be a unique constant, and \( \frac{a}{1+a} \) depends on 'a', a unique identity element does not exist.
Thus, an identity element does not exist for this operation.
In simple words: For a*b = a + ab, an identity 'e' would mean a + ae = a, so ae = 0. This suggests e=0 if a is not zero. But also, e + ea = a, which means e(1+a) = a, so e = a/(1+a). Since these two 'e' values are not always the same and one depends on 'a', there is no single identity element.

🎯 Exam Tip: When evaluating for an identity element, check both `a * e = a` and `e * a = a`. If the `e` derived from one equation contradicts or depends on `a` from the other, then no unique identity element exists.

 

Question. (iv) a * b = \((a - b)^2\)

Answer:
Operation on Q: a * b = \((a - b)^2\).
Let e be the identity element for *.
Then, a * e = a for all a ∈ Q.
a * e = a \( \implies \) \((a - e)^2\) = a.
This means \(a - e = \pm \sqrt{a}\). So, \(e = a \mp \sqrt{a}\).
Since 'e' depends on 'a' and can have two values, it is not a unique constant.
Thus, an identity element does not exist for this operation.
In simple words: For a*b = \((a-b)^2\), an identity element 'e' would mean \((a-e)^2\) equals 'a'. This implies that e must be \(a \pm \sqrt{a}\). Since 'e' changes with 'a' and has two possible values, there is no single fixed identity element.

🎯 Exam Tip: If solving for 'e' results in an expression involving 'a' or multiple possible values, it means a unique identity element does not exist.

 

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

Answer:
Operation on Q: a * b = \( \frac{ab}{4} \).
Let e be the identity element for *.
Then, a * e = a for all a ∈ Q.
a * e = a \( \implies \) \( \frac{ae}{4} \) = a.
Assuming a ≠ 0, we can divide by a:
\( \frac{e}{4} \) = 1 \( \implies \) e = 4.
Let's check if e = 4 satisfies both conditions:
a * 4 = \( \frac{a \cdot 4}{4} \) = a.
4 * a = \( \frac{4 \cdot a}{4} \) = a.
Since a * e = a and e * a = a are both satisfied with a unique value e = 4, and 4 ∈ Q,
Thus, the identity element for * is 4.
In simple words: For a*b = ab/4, an identity element 'e' means (a times e) divided by 4 should equal 'a'. This leads to e being 4. If we check, 'a' operated with 4 (a*4) gives a, and 4 operated with 'a' (4*a) also gives a. So, 4 is the identity element.

🎯 Exam Tip: Always verify that the derived identity element works for both `a * e = a` and `e * a = a` to ensure it is indeed the identity element.

 

Question. (vi) a * b = \(ab^2\)

Answer:
Operation on Q: a * b = \(ab^2\).
Let e be the identity element for *.
Then, a * e = a and e * a = a for all a ∈ Q.
a * e = a \( \implies \) \(ae^2\) = a.
Assuming a ≠ 0, we can divide by a:
\(e^2\) = 1 \( \implies \) e = \(\pm 1\).
Now, let's check e * a = a with e = 1 and e = -1.
If e = 1, then e * a = \(1 \cdot a^2 = a^2\). For this to be 'a', \(a^2 = a\), which means a = 0 or a = 1. This does not hold for all a ∈ Q.
If e = -1, then e * a = \(-1 \cdot a^2 = -a^2\). For this to be 'a', \(-a^2 = a\), which means \(a^2 + a = 0\), so \(a(a+1) = 0\), i.e., a = 0 or a = -1. This also does not hold for all a ∈ Q.
Since a unique value of 'e' that satisfies both conditions for all 'a' does not exist,
Thus, an identity element does not exist for this operation.
In simple words: For a*b = \(ab^2\), an identity element 'e' would mean \(a \cdot e^2\) equals 'a', which gives e = \(\pm 1\). But if we check 'e * a', we get \(e \cdot a^2\). If e is 1, we get \(a^2\), which is not always 'a'. If e is -1, we get \(-a^2\), which is also not always 'a'. Since no single 'e' works for all 'a', there is no identity element.

🎯 Exam Tip: If you find multiple potential identity elements or if one side of the identity property `e * a = a` (or `a * e = a`) leads to a conditional result (e.g., `e=1` only if `a=1`), then a universal identity element does not exist.

 

Question 11. धारो के A = N × N अने A पर क्रिया *, (a, b) * (c, d) = (a + c, b + d) द्वारा व्याख्यायित छे. साजित करो के * समक्रमी छे अने जूथना नियमनुं पालन करे छे. जो * माटे A मां कोइ एकम घटक होय, तो ते शोघो.
Answer:
The operation * is defined on A = N × N as (a, b) * (c, d) = (a + c, b + d).
Let (a, b) ∈ A and (c, d) ∈ A.
Commutativity:
(a, b) * (c, d) = (a + c, b + d).
(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d).
Since (a, b) * (c, d) = (c, d) * (a, b),
Thus, the operation * is commutative.
Associativity:
Let (a, b) ∈ A, (c, d) ∈ A, and (e, f) ∈ A.
[(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f) = (a + c + e, b + d + f) ......(1)
(a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f) = (a + (c + e), b + (d + f)) = (a + c + e, b + d + f) ......(2)
From (1) and (2), we see that [(a, b) * (c, d)] * (e, f) = (a, b) * [(c, d) * (e, f)].
Thus, the operation * satisfies the associative law.
Identity Element:
Let (p, q) be the identity element for * in A.
Then, (a, b) * (p, q) = (a, b) for all (a, b) ∈ A.
(a + p, b + q) = (a, b).
This implies: a + p = a \( \implies \) p = 0.
And b + q = b \( \implies \) q = 0.
So, the identity element would be (0, 0).
However, (0, 0) is not an element of N × N, because N = {1, 2, 3, ...}, and 0 is not a natural number.
Thus, an identity element does not exist for this operation in A = N × N.
In simple words: The operation on pairs of natural numbers is defined by adding their first parts and adding their second parts. This operation is commutative because changing the order of the pairs does not change the sums. It is also associative because how you group three pairs for addition does not change the final sum of their parts. To find an identity element, we need a pair (p,q) that when added to any (a,b) gives (a,b). This means p and q must both be 0. However, 0 is not a natural number, so the pair (0,0) is not in the set N x N. Therefore, there is no identity element for this operation in N x N.

🎯 Exam Tip: When defining operations on Cartesian products (like N × N), check the properties for each component separately. For identity elements, ensure the derived element actually belongs to the specified set.

 

Question 12. નીચે આપેલાં વિધાનો સત્ય છે કે અસત્ય તે જણાવો. તમારા જવાબની યથાર્થતા ચકાસો :
(i) ગણ N પરની કોઈપણ દ્વિક્રિયા * માટે, \( a * a = a \), \( \forall a \in N \).

Answer: આ વિધાન અસત્ય છે. કારણ કે દ્વિક્રિયા '\(*\)' બે ઘટકો, જેમ કે 'a' અને 'b' વચ્ચે વ્યાખ્યાયિત થાય છે. આ એક વિધેય છે જે \( N \times N \) થી \( N \) પર જાય છે. \( a * a = a \) ની શરત એ દરેક દ્વિક્રિયા માટે સાચી હોતી નથી, જ્યાં કોઈ ઘટક પોતાની સાથે ક્રિયા કરીને તે જ ઘટક આપે.
In simple words: This statement is not true. A binary operation usually works on two different numbers, not just one number used twice to get itself. For example, \( a + a = 2a \), not \( a \).

🎯 Exam Tip: For true/false statements, provide a clear reason or a counterexample to justify your answer.

 

Question 12. નીચે આપેલાં વિધાનો સત્ય છે કે અસત્ય તે જણાવો. તમારા જવાબની યથાર્થતા ચકાસો :
(ii) જો * N પર સમક્રમી દ્વિક્રિયા હોય તો \( a * (b * c) = (c * b) * a \).

Answer: આ વિધાન સત્ય છે. દ્વિક્રિયા '\(*\)' N પર સમક્રમી હોવાથી, આપણે જાણીએ છીએ કે \( b * c = c * b \). તેથી, જો આપણે આને આપેલા સમીકરણમાં મૂકીએ, તો આપણને \( a * (b * c) = a * (c * b) \) મળે છે. સમક્રમી ગુણધર્મનો ફરીથી ઉપયોગ કરીને, \( a * (c * b) \) ને \( (c * b) * a \) તરીકે લખી શકાય છે. આથી, વિધાન સાચું ઠરે છે.
In simple words: This statement is true. If an operation is commutative, it means you can change the order of numbers without changing the result (like \( b * c = c * b \)). Using this rule, we can show that both sides of the given equation are equal.

🎯 Exam Tip: Understanding the definitions of commutative and associative properties is key to solving such problems.

 

Question 13. નીચે આપેલ વિકલ્પોમાંથી સાચો વિકલ્પ પસંદ કરો, જેથી વિધાન સાચું બને:
ગણ N પર વ્યાખ્યાયિત દ્વિક્રિયા \( * \) માટે, \( a * b = a^3 + b^3 \).
(A) \( * \) જૂથના નિયમને અનુસરે છે અને સમક્રમી બંને છે ?
(B) \( * \) સમક્રમી છે પરંતુ જૂથના નિયમને અનુસરતી નથી ?
(C) \( * \) જૂથના નિયમને અનુસરે છે પરંતુ સમક્રમી નથી ?
(D) \( * \) સમક્રમી નથી અને જૂથના નિયમને અનુસરતી નથી ?

Answer: સૌ પ્રથમ, આપણે ચકાસીએ કે ક્રિયા '\(*\)' સમક્રમી છે કે નહીં. કોઈ ક્રિયા સમક્રમી હોય છે જો \( a * b = b * a \) હોય. \( a * b = a^3 + b^3 \) માટે, આપણી પાસે \( b * a = b^3 + a^3 \) છે. \( a^3 + b^3 \) હંમેશા \( b^3 + a^3 \) બરાબર હોવાથી, આ ક્રિયા સમક્રમી છે.
આગળ, આપણે સહભાગીતા (associativity) માટે ચકાસીએ. કોઈ ક્રિયા સહભાગી હોય છે જો \( a * (b * c) = (a * b) * c \) હોય.
\( a * (b * c) = a * (b^3 + c^3) = a^3 + (b^3 + c^3)^3 \).
\( (a * b) * c = (a^3 + b^3) * c = (a^3 + b^3)^3 + c^3 \).
ચૂંકી \( a^3 + (b^3 + c^3)^3 \) સામાન્ય રીતે \( (a^3 + b^3)^3 + c^3 \) બરાબર હોતો નથી, તેથી આ ક્રિયા સહભાગી નથી.
આથી, આ ક્રિયા સમક્રમી છે પરંતુ સહભાગી નથી. આ વિકલ્પ (B) ને અનુરૂપ છે.
Answer: (B) \( * \) સમક્રમી છે પરંતુ જૂથના નિયમને અનુસરતી નથી ?
In simple words: For the operation \( a * b = a^3 + b^3 \), we see that \( a * b \) is the same as \( b * a \) because addition is commutative, so it's a commutative operation. But, when we check for associativity, \( a * (b * c) \) is not the same as \( (a * b) * c \). So, it is not an associative operation. This means option (B) is the correct answer.

🎯 Exam Tip: Always test both commutativity and associativity when asked to determine properties of a binary operation. Show all steps clearly for full marks.