GSEB Class 12 Maths Solutions Chapter 1 Relations and Functions Exercise 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 \( a * b = ab^2 \).
(iv) On \( Z^{+} \), define \( * \) by \( a * b = |a - b| \).
(v) On \( Z^{+} \), define \( * \) by \( a * b = a \).
Answer:
(i) If \( a > b \), then \( a * b = a - b \) will be greater than 0, meaning it is a member of \( Z^{+} \). However, if \( a < b \), then \( a * b = a - b \) will be less than 0, which is not a member of \( Z^{+} \).
\( \implies \) Therefore, the given operation \( * \) is not considered a binary operation.
(ii) For any numbers \( a \) and \( b \) that belong to \( Z^{+} \), their product \( ab \) will also belong to \( Z^{+} \).
\( \implies \) Thus, the operation \( * \), defined as \( a * b = ab \), is a binary operation.
(iii) For all real numbers \( a \) and \( b \), the expression \( ab^2 \) will also be a real number.
\( \implies \) So, the operation \( * \), defined by \( a * b = ab^2 \), is a binary operation.
(iv) For any numbers \( a \) and \( b \) that belong to \( Z^{+} \), the absolute difference \( |a - b| \) will also belong to \( Z^{+} \).
\( \implies \) Thus, the operation \( * \) defined as \( a * b = |a - b| \) is a binary operation.
(v) When defined on \( Z^{+} \), if \( a * b = a \), and \( a, b \) are members of \( Z^{+} \), it means \( a \) is also a member of \( Z^{+} \).
\( \implies \) Hence, this is a binary operation.
In simple words: An operation is binary if its result always stays within the original set. We check each given rule to see if the answers always fit back into the specified set (like all positive integers or all real numbers). If not, it's not binary.

Exam Tip: Remember that for an operation to be binary on a set, the result of the operation on any two elements from that set must also be an element of the same set.

 

Question 2. For each binary operation * defined below, determine whether * is commutative or associative:
(i) On \( Z \), define \( * \) by \( a * b = a - b \).
(ii) On \( Q \), define \( * \) by \( a * b = ab + 1 \).
(iii) On \( Q \), define \( * \) by \( a * b = \frac{a}{2} \).
(iv) On \( Z^{+} \), define \( * \) by \( a * b = 2^a \).
(v) On \( Z^{+} \), define \( * \) by \( a * b = a^b \).
(vi) On \( R - \{-1\} \), define \( * \) by \( a * b = \frac{a}{b+1} \).
Answer:
(i) On \( Z \), the operation is defined as \( a * b = a - b \).
(a) Commutativity:
\( b * a = b - a \).
Since \( a - b \neq b - a \) (unless \( a=b \)), it implies that \( a * b \neq b * a \).
\( \implies \) Therefore, the operation is not commutative.
(b) Associativity:
\( a * (b * c) = a * (b - c) = a - (b - c) = a - b + c \).
\( (a * b) * c = (a - b) * c = (a - b) - c = a - b - c \).
Since \( a - b + c \neq a - b - c \) (unless \( c=0 \)), it implies \( a * (b * c) \neq (a * b) * c \).
\( \implies \) Consequently, the binary operation defined is not associative.

(ii) On \( Q \), the operation \( * \) is defined as \( a * b = ab + 1 \).
(a) Commutativity:
\( b * a = ba + 1 \).
Since \( ab + 1 \) is the same as \( ba + 1 \) because multiplication of numbers is commutative,
\( \implies \) So, \( a * b = b * a \).
Therefore, the defined binary operation is commutative.
(b) Associativity:
\( a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1 \).
\( (a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1 \).
Since \( abc + a + 1 \neq abc + c + 1 \) (unless \( a=c \)), it means \( a * (b * c) \neq (a * b) * c \).
\( \implies \) Hence, the binary operation defined is not associative.

(iii) On \( Q \), the operation \( * \) is defined as \( a * b = \frac{a}{2} \).
(a) Commutativity:
\( b * a = \frac{b}{2} \).
Since \( \frac{a}{2} \neq \frac{b}{2} \) in general (unless \( a=b \)), it means \( a * b \neq b * a \).
\( \implies \) Thus, the binary operation defined is not commutative.
(b) Associativity:
\( a * (b * c) = a * \left( \frac{b}{2} \right) = \frac{a}{2} \).
\( (a * b) * c = \left( \frac{a}{2} \right) * c = \frac{\left( \frac{a}{2} \right)}{2} = \frac{a}{4} \).
Since \( \frac{a}{2} \neq \frac{a}{4} \) (unless \( a=0 \)), it implies \( a * (b * c) \neq (a * b) * c \).
\( \implies \) Therefore, the defined binary operation is not associative.

(iv) On \( Z^{+} \), the operation \( * \) is defined as \( a * b = 2^a \).
(a) Commutativity:
\( b * a = 2^b \).
Since \( 2^a \neq 2^b \) in general (unless \( a=b \)), it implies \( a * b \neq b * a \).
\( \implies \) Therefore, the binary operation defined is not commutative.
(b) Associativity:
\( a * (b * c) = a * (2^b) = 2^a \).
\( (a * b) * c = (2^a) * c = 2^{2^a} \).
Since \( 2^a \neq 2^{2^a} \) in general, it means \( a * (b * c) \neq (a * b) * c \).
\( \implies \) Thus, the binary operation defined is not associative.

(v) On \( Z^{+} \), the operation \( * \) is defined as \( a * b = a^b \).
(a) Commutativity:
\( b * a = b^a \).
In general, \( a^b \neq b^a \) (e.g., \( 2^3 = 8 \) but \( 3^2 = 9 \)).
\( \implies \) Hence, \( a * b \neq b * a \).
Therefore, the operation \( * \) is not commutative.
(b) Associativity:
\( (a * b) * c = (a^b) * c = (a^b)^c = a^{bc} \).
\( a * (b * c) = a * (b^c) = a^{(b^c)} \).
Since \( a^{bc} \neq a^{(b^c)} \) in general, it follows that \( (a * b) * c \neq a * (b * c) \).
\( \implies \) Thus, the operation \( * \) is not associative.

(vi) On \( R - \{-1\} \), the operation \( * \) is defined as \( a * b = \frac{a}{b+1} \).
(a) Commutativity:
\( b * a = \frac{b}{a+1} \).
In general, \( \frac{a}{b+1} \neq \frac{b}{a+1} \) (e.g., if \( a=1, b=2 \), then \( \frac{1}{2+1} = \frac{1}{3} \) but \( \frac{2}{1+1} = \frac{2}{2} = 1 \)).
\( \implies \) Therefore, \( a * b \neq b * a \).
Hence, the operation defined is not commutative.
(b) Associativity:
\( a * (b * c) = a * \left( \frac{b}{c+1} \right) = \frac{a}{\frac{b}{c+1} + 1} = \frac{a}{\frac{b+c+1}{c+1}} = \frac{a(c+1)}{b+c+1} \).
\( (a * b) * c = \left( \frac{a}{b+1} \right) * c = \frac{\frac{a}{b+1}}{c+1} = \frac{a}{(b+1)(c+1)} \).
Since \( \frac{a(c+1)}{b+c+1} \neq \frac{a}{(b+1)(c+1)} \) in general, it means \( a * (b * c) \neq (a * b) * c \).
\( \implies \) Consequently, the binary operation defined is not associative.
In simple words: For commutativity, check if \( a * b \) gives the same result as \( b * a \). For associativity, check if \( a * (b * c) \) gives the same result as \( (a * b) * c \). If they are different in general, the property doesn't hold.

Exam Tip: To prove an operation is NOT commutative or associative, a single counterexample with specific numbers is sufficient. To prove it IS, you need to show it holds for all general elements.

 

Question 3. Consider the infimum binary operation \( \land \) on the set \( \{1, 2, 3, 4, 5\} \) defined by \( a \land b = \min \) of \( a \) and \( b \). Write the multiplication table of the operation \( \land \).
Answer: The multiplication table for operation \( \land \) on the set \( \{1, 2, 3, 4, 5\} \), where \( a \land b = \min(a, b) \), is provided below. This table shows the smaller value between any two elements when the operation is applied.

In simple words: This table shows the smaller number when you compare any two numbers from the set \( \{1, 2, 3, 4, 5\} \). You find the row for the first number and the column for the second, and where they meet is the minimum.

Exam Tip: When creating a multiplication table for a binary operation, ensure all possible pairs of elements are covered and the operation rule is correctly applied for each cell.

 

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 \( (2 * 3) * (4 * 5) \).
Answer:
(i) Using the given table, we can compute the expressions.
First, for \( (2 * 3) * 4 \):
\( 2 * 3 = 1 \) (from the table).
Then, \( (2 * 3) * 4 = 1 * 4 = 1 \) (from the table).
Next, for \( 2 * (3 * 4) \):
\( 3 * 4 = 1 \) (from the table).
Then, \( 2 * (3 * 4) = 2 * 1 = 1 \) (from the table).
So, both computations result in 1.
(ii) To check if the operation \( * \) is commutative, we need to see if \( a * b = b * a \) for all elements \( a, b \in \{1, 2, 3, 4, 5\} \).
From the given table:
We notice that for any \( a, b \), the value of \( a * b \) is the same as \( b * a \).
For instance, \( 2 * 4 = 2 \) and \( 4 * 2 = 2 \).
Also, \( 3 * 5 = 1 \) and \( 5 * 3 = 1 \).
This pattern holds throughout the table.
\( \implies \) Therefore, the given binary operation \( * \) is commutative.
(iii) We need to compute \( (2 * 3) * (4 * 5) \).
From the table, \( 2 * 3 = 1 \).
Also from the table, \( 4 * 5 = 1 \).
So, \( (2 * 3) * (4 * 5) = 1 * 1 \).
Finally, \( 1 * 1 = 1 \) (from the table).
Thus, the result is 1.
In simple words: For (i) and (iii), use the table to find the result of each step. For (ii), check if the table is symmetrical across its main diagonal to confirm commutativity.

Exam Tip: When reading a multiplication table, the first number refers to the row, and the second number refers to the column. The intersection gives the result of the operation.

 

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 question 4 above? Justify your answer.
Answer: The given set is \( \{1, 2, 3, 4, 5\} \), and the operation \( *' \) is defined as \( a *' b = HCF(a, b) \).
Let us construct the multiplication table for this operation:

Comparing this table with the one provided in Question 4, we observe that all the corresponding entries are identical.
\( \implies \) Therefore, the operation \( *' \) is indeed the same as the operation \( * \) defined in Question 4.
In simple words: We make a new table using the Highest Common Factor (HCF) for each pair of numbers in the set. Then we check if this new table looks exactly like the table from Question 4. If it does, the operations are the same.

Exam Tip: To compare two operations, the most reliable way is to construct their respective operation tables (if the set is small) and check for identical entries.

 

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) Are elements of \( N \) invertible for the operation \( * \)?
Answer:
(i) The operation \( * \) is defined as the Least Common Multiple (L.C.M.) of \( a \) and \( b \).
For \( 5 * 7 \):
\( 5 * 7 = L.C.M.(5, 7) = 35 \).
For \( 20 * 16 \):
\( 20 * 16 = L.C.M.(20, 16) = 80 \).
(ii) To check for commutativity, we need to verify if \( a * b = b * a \).
By definition, \( a * b = L.C.M.(a, b) \).
Also, \( b * a = L.C.M.(b, a) \).
Since \( L.C.M.(a, b) \) is always equal to \( L.C.M.(b, a) \), it follows that \( a * b = b * a \).
\( \implies \) Therefore, the binary operation \( * \) is commutative.
(iii) To check for associativity, we need to compare \( a * (b * c) \) and \( (a * b) * c \).
\( a * (b * c) = a * L.C.M.(b, c) = L.C.M.(a, L.C.M.(b, c)) = L.C.M.(a, b, c) \).
\( (a * b) * c = L.C.M.(a, b) * c = L.C.M.(L.C.M.(a, b), c) = L.C.M.(a, b, c) \).
Since both expressions are equal to \( L.C.M.(a, b, c) \), it implies \( a * (b * c) = (a * b) * c \).
\( \implies \) Therefore, the given binary operation \( * \) is associative.
(iv) An identity element \( e \) for the operation \( * \) in \( N \) must satisfy \( a * e = e * a = a \) for all \( a \in N \).
We check for \( e = 1 \):
\( 1 * a = L.C.M.(1, a) = a \).
\( a * 1 = L.C.M.(a, 1) = a \).
Since \( 1 * a = a * 1 = a \), the identity element for \( * \) in \( N \) is \( 1 \).
(v) For an element \( a \in N \) to be invertible, there must exist an element \( b \in N \) such that \( a * b = e \), where \( e \) is the identity element (which is 1).
So, we need \( L.C.M.(a, b) = 1 \).
The only way the L.C.M. of two natural numbers can be 1 is if both numbers are 1.
Thus, only \( a = 1 \) has an inverse (which is 1 itself).
\( \implies \) Therefore, not all elements of \( N \) are invertible for the operation \( * \); only the element 1 is invertible.
In simple words: For L.C.M. operations: (i) find the L.C.M. of the numbers; (ii) L.C.M. is always commutative; (iii) L.C.M. is always associative; (iv) 1 is the identity because \( L.C.M.(a, 1) = a \); (v) only 1 is invertible because \( L.C.M.(a, b) = 1 \) only if \( a=1 \) and \( b=1 \).

Exam Tip: Remember that commutativity and associativity are often inherent properties of operations like L.C.M. and H.C.F. The identity element is crucial for checking invertibility.

 

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.
Answer: The given set is \( \{1, 2, 3, 4, 5\} \). The binary operation \( * \) is defined as \( a * b = L.C.M.(a, b) \).
For an operation to be a binary operation on a set, the result of the operation for any two elements in the set must also be in that same set.
Let's consider two elements from the set, for example, 4 and 5.
\( 4 * 5 = L.C.M.(4, 5) = 20 \).
However, \( 20 \) does not belong to the given set \( \{1, 2, 3, 4, 5\} \).
\( \implies \) Therefore, this operation is not a binary operation on the specified set.
In simple words: No, this operation is not binary on the given set. If you take 4 and 5 from the set and find their L.C.M., you get 20, which is not in the set. For an operation to be binary, all results must stay inside the original set.

Exam Tip: A key condition for a binary operation on a set is closure: the result of the operation on any two elements of the set must also be an element of that set.

 

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 there exist identity for this operation on \( N \)?
Answer:
(a) Commutativity:
To check for commutativity, we need to verify if \( a * b = b * a \).
By definition, \( a * b = H.C.F.(a, b) \).
We know that \( H.C.F.(a, b) \) is always equal to \( H.C.F.(b, a) \).
\( \implies \) So, \( a * b = b * a \).
Therefore, the binary operation \( * \) is commutative.
(b) Associativity:
To check for associativity, we need to compare \( a * (b * c) \) and \( (a * b) * c \).
\( a * (b * c) = a * H.C.F.(b, c) = H.C.F.(a, H.C.F.(b, c)) = H.C.F.(a, b, c) \).
\( (a * b) * c = H.C.F.(a, b) * c = H.C.F.(H.C.F.(a, b), c) = H.C.F.(a, b, c) \).
Since both expressions are equal to \( H.C.F.(a, b, c) \), it means \( a * (b * c) = (a * b) * c \).
\( \implies \) Therefore, the binary operation \( * \) as defined is associative.
(c) Identity Element:
An identity element \( e \) for the operation \( * \) in \( N \) must satisfy \( a * e = e * a = a \) for all \( a \in N \).
This means \( H.C.F.(a, e) = a \).
If \( e \) were the identity, then \( H.C.F.(a, e) = a \). This implies that \( e \) must be a multiple of \( a \) for all \( a \in N \). The only number that is a multiple of all natural numbers is not a natural number itself.
For example, if \( e=1 \), \( H.C.F.(a, 1) = 1 \), which is only equal to \( a \) if \( a=1 \). It does not hold for all \( a \in N \).
\( \implies \) Therefore, there does not exist an identity element for this operation on \( N \).
In simple words: For H.C.F. operations: it is commutative because \( HCF(a, b) \) is the same as \( HCF(b, a) \). It is associative because \( HCF(a, HCF(b, c)) \) is the same as \( HCF(HCF(a, b), c) \). There is no identity element because no single number \( e \) exists such that \( HCF(a, e) \) always equals \( a \) for all natural numbers \( a \).

Exam Tip: Operations like H.C.F. and L.C.M. often share similar properties. Always formally check the definitions for commutativity, associativity, and identity element for each new operation.

 

Question 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) \( a * b = \frac{ab}{4} \)
(vi) \( a * b = ab^2 \)
Find which of the above binary operations are commutative and which are associative.
Answer:
(i) The operation is defined as \( a * b = a - b \).
(a) Commutativity:
\( b * a = b - a \).
Since \( a - b \neq b - a \) in general (e.g., \( 5-3=2 \) but \( 3-5=-2 \)),
\( \implies \) Therefore, the operation \( * \) is not commutative.
(b) Associativity:
\( a * (b * c) = a * (b - c) = a - (b - c) = a - b + c \).
\( (a * b) * c = (a - b) * c = (a - b) - c = a - b - c \).
Since \( a - b + c \neq a - b - c \) in general (e.g., if \( c \neq 0 \)),
\( \implies \) Therefore, the operation \( * \) is not associative.

(ii) The operation is defined as \( a * b = a^2 + b^2 \).
(a) Commutativity:
\( b * a = b^2 + a^2 \).
Since \( a^2 + b^2 \) is always equal to \( b^2 + a^2 \) (due to commutativity of addition),
\( \implies \) Therefore, \( a * b = b * a \).
This binary operation is commutative.
(b) Associativity:
\( a * (b * c) = a * (b^2 + c^2) = a^2 + (b^2 + c^2)^2 \).
\( (a * b) * c = (a^2 + b^2) * c = (a^2 + b^2)^2 + c^2 \).
Since \( a^2 + (b^2 + c^2)^2 \neq (a^2 + b^2)^2 + c^2 \) in general,
\( \implies \) Therefore, the given operation \( * \) is not associative.

(iii) The operation is defined as \( a * b = a + ab \).
(a) Commutativity:
\( b * a = b + ba \).
In general, \( a + ab \neq b + ba \) (e.g., \( 1*2 = 1+1 \cdot 2 = 3 \), but \( 2*1 = 2+2 \cdot 1 = 4 \)),
\( \implies \) Therefore, \( a * b \neq b * a \).
This operation \( * \) is not commutative.
(b) Associativity:
\( a * (b * c) = a * (b + bc) = a + a(b + bc) = a + ab + abc \).
\( (a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c = a + ab + ac + abc \).
Since \( a + ab + abc \neq a + ab + ac + abc \) in general (e.g., if \( c \neq 0 \) and \( a \neq 0 \)),
\( \implies \) Therefore, the given binary operation \( * \) is not associative.

(iv) The binary operation \( * \) is defined as \( a * b = (a - b)^2 \).
(a) Commutativity:
\( b * a = (b - a)^2 \).
Since \( (b - a)^2 = (-(a - b))^2 = (a - b)^2 \),
\( \implies \) Therefore, \( a * b = b * a \).
This binary operation is commutative.
(b) Associativity:
\( a * (b * c) = a * (b - c)^2 = (a - (b - c)^2)^2 \).
\( (a * b) * c = (a - b)^2 * c = ((a - b)^2 - c)^2 \).
Since \( (a - (b - c)^2)^2 \neq ((a - b)^2 - c)^2 \) in general,
\( \implies \) Therefore, the operation \( * \) is not associative.

(v) The binary operation \( * \) is defined as \( a * b = \frac{ab}{4} \).
(a) Commutativity:
\( b * a = \frac{ba}{4} \).
Since \( ab = ba \), it follows that \( \frac{ab}{4} = \frac{ba}{4} \).
\( \implies \) Therefore, \( a * b = b * a \).
This operation \( * \) is commutative.
(b) Associativity:
\( a * (b * c) = a * \left( \frac{bc}{4} \right) = \frac{a \left( \frac{bc}{4} \right)}{4} = \frac{abc}{16} \).
\( (a * b) * c = \left( \frac{ab}{4} \right) * c = \frac{\left( \frac{ab}{4} \right) c}{4} = \frac{abc}{16} \).
Since both expressions are equal, \( (a * b) * c = a * (b * c) \).
\( \implies \) Thus, the given operation \( * \) is associative.

(vi) The binary operation \( * \) is defined as \( a * b = ab^2 \).
(a) Commutativity:
\( b * a = ba^2 \).
In general, \( ab^2 \neq ba^2 \) (e.g., if \( a=1, b=2 \), then \( 1 \cdot 2^2 = 4 \), but \( 2 \cdot 1^2 = 2 \)).
\( \implies \) Therefore, \( a * b \neq b * a \).
This operation \( * \) is not commutative.
(b) Associativity:
\( a * (b * c) = a * (bc^2) = a(bc^2)^2 = a(b^2 c^4) = ab^2 c^4 \).
\( (a * b) * c = (ab^2) * c = (ab^2)c^2 = ab^2 c^2 \).
Since \( ab^2 c^4 \neq ab^2 c^2 \) in general (e.g., if \( c \neq 0, 1 \)),
\( \implies \) Therefore, \( a * (b * c) \neq (a * b) * c \).
The given binary operation \( * \) is not associative.
In simple words: We check each operation for two main properties: commutativity (if \( a*b = b*a \)) and associativity (if \( a*(b*c) = (a*b)*c \)). We perform the calculations for both sides of these equations. If the results are always the same for any rational numbers \( a, b, c \), then the property holds. Otherwise, it doesn't. Sometimes, simple examples (counterexamples) can quickly prove that a property does not hold.

Exam Tip: Remember that commutativity and associativity are independent properties. An operation can have one, both, or neither. Always test them separately using the definitions.

 

Question 10. Show that none of the operations given above has identity.
Answer: The binary operation \( * \) is on the set \( Q \) of rational numbers.
(i) Defined as \( a * b = a - b \).
For an identity element \( e \) to exist, it must satisfy \( a * e = a \) and \( e * a = a \) for all \( a \in Q \).
If \( a * e = a \), then \( a - e = a \implies e = 0 \).
If \( e * a = a \), then \( e - a = a \implies e = 2a \).
Since \( 0 \neq 2a \) (unless \( a=0 \)), there is no single element \( e \) that works for all \( a \).
\( \implies \) Therefore, there is no identity element for this operation.

(ii) The operation is defined as \( a * b = a^2 + b^2 \).
For an identity element \( e \), we need \( a * e = a \) and \( e * a = a \).
If \( a * e = a \), then \( a^2 + e^2 = a \implies e^2 = a - a^2 \). This value of \( e \) depends on \( a \), so a general identity element does not exist.
If \( e * a = a \), then \( e^2 + a^2 = a \implies e^2 = a - a^2 \). Again, \( e \) depends on \( a \).
\( \implies \) Therefore, this operation \( * \) has no identity element.

(iii) The binary operation \( * \) is defined as \( a * b = a + ab \).
For an identity element \( e \), we need \( a * e = a \) and \( e * a = a \).
If \( a * e = a \), then \( a + ae = a \implies ae = 0 \). For this to be true for all \( a \in Q \), we need \( e = 0 \).
If \( e * a = a \), then \( e + ea = a \implies e(1 + a) = a \). So, \( e = \frac{a}{1+a} \). This value of \( e \) depends on \( a \), so a general identity element does not exist.
Since \( e=0 \) and \( e=\frac{a}{1+a} \) are contradictory,
\( \implies \) Therefore, there is no identity element.

(iv) The binary operation \( * \) is defined as \( a * b = (a - b)^2 \).
For an identity element \( e \), we need \( a * e = a \) and \( e * a = a \).
If \( a * e = a \), then \( (a - e)^2 = a \). This means \( a - e = \pm \sqrt{a} \implies e = a \pm \sqrt{a} \). This value of \( e \) depends on \( a \).
If \( e * a = a \), then \( (e - a)^2 = a \). This means \( e - a = \pm \sqrt{a} \implies e = a \pm \sqrt{a} \). Again, \( e \) depends on \( a \).
\( \implies \) Therefore, there is no identity element.

(v) The operation \( * \) is defined as \( a * b = \frac{ab}{4} \).
For an identity element \( e \), we need \( a * e = a \) and \( e * a = a \).
If \( a * e = a \), then \( \frac{ae}{4} = a \). If \( a \neq 0 \), then \( \frac{e}{4} = 1 \implies e = 4 \).
If \( e * a = a \), then \( \frac{ea}{4} = a \). If \( a \neq 0 \), then \( \frac{e}{4} = 1 \implies e = 4 \).
If \( a = 0 \), then \( 0 * e = 0 \), which is \( \frac{0 \cdot e}{4} = 0 \), so \( 0 = 0 \). This holds for any \( e \).
So, \( e=4 \) works for all \( a \in Q \).
\( \implies \) Therefore, the identity element for this operation is \( 4 \).

(vi) The operation \( * \) is defined as \( a * b = ab^2 \).
For an identity element \( e \), we need \( a * e = a \) and \( e * a = a \).
If \( a * e = a \), then \( ae^2 = a \). If \( a \neq 0 \), then \( e^2 = 1 \implies e = \pm 1 \).
If \( e * a = a \), then \( ea^2 = a \). If \( a \neq 0 \), then \( e = \frac{a}{a^2} = \frac{1}{a} \).
Since \( e = \frac{1}{a} \) depends on \( a \), there is no single fixed identity element. Also, \( e=\pm 1 \) is not consistent with \( e=\frac{1}{a} \).
\( \implies \) Therefore, there is no identity element for this operation.
In simple words: An identity element \( e \) is a special number that, when combined with any other number \( a \) using the operation, leaves \( a \) unchanged (e.g., \( a * e = a \)). We check each operation by setting \( a * e = a \) and \( e * a = a \) and solving for \( e \). If \( e \) has a single, constant value that works for all \( a \), it's an identity. Otherwise, there is none. In most cases here, \( e \) either depends on \( a \) or leads to a contradiction. Only for \( a*b = ab/4 \) is there an identity, which is 4.

Exam Tip: To find an identity element, solve both \( a * e = a \) and \( e * a = a \). For an identity to exist, a unique value of \( e \) must satisfy both equations for all \( a \) in the set.

 

Question 11. Let \( A = N \times N \) 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 \( * \) on \( A \), if any.
Answer: The set is \( A = N \times N \), and the binary operation \( * \) is defined as \( (a, b) * (c, d) = (a + c, b + d) \).
(a) Commutativity:
To show that \( * \) is commutative, we need to prove that \( (a, b) * (c, d) = (c, d) * (a, b) \).
Given \( (a, b) * (c, d) = (a + c, b + d) \).
Now, let's compute \( (c, d) * (a, b) \):
\( (c, d) * (a, b) = (c + a, d + b) \).
Since addition of natural numbers is commutative (\( a+c = c+a \) and \( b+d = d+b \)), we have \( (a + c, b + d) = (c + a, d + b) \).
\( \implies \) Therefore, \( (a, b) * (c, d) = (c, d) * (a, b) \).
This means the operation \( * \) is commutative.
(b) Associativity:
To show that \( * \) is associative, we need to prove that \( (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e, f) \).
First, let's compute \( (a, b) * [(c, d) * (e, f)] \):
\( (a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f) \)
\( = (a + (c + e), b + (d + f)) \)
\( = (a + c + e, b + d + f) \).
Next, let's compute \( [(a, b) * (c, d)] * (e, f) \):
\( [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) \)
\( = ((a + c) + e, (b + d) + f) \)
\( = (a + c + e, b + d + f) \).
Since both sides are equal,
\( \implies \) The given binary operation \( * \) is associative.
(c) Identity Element:
For an identity element \( (e_1, e_2) \) to exist in \( A = N \times N \), it must satisfy \( (a, b) * (e_1, e_2) = (a, b) \) for all \( (a, b) \in A \).
Applying the definition: \( (a + e_1, b + e_2) = (a, b) \).
This implies \( a + e_1 = a \) and \( b + e_2 = b \).
For \( a + e_1 = a \), we get \( e_1 = 0 \).
For \( b + e_2 = b \), we get \( e_2 = 0 \).
So, the potential identity element is \( (0, 0) \).
However, \( N \) represents the set of natural numbers, which typically does not include \( 0 \). Assuming \( N = \{1, 2, 3, \ldots\} \), the element \( (0, 0) \) does not belong to \( N \times N \).
\( \implies \) Therefore, an identity element does not exist for this operation on \( A = N \times N \).
In simple words: This operation adds the first parts together and the second parts together. It's commutative because changing the order of pairs still gives the same sum. It's associative because grouping the additions differently still gives the same total sum. However, there's no identity element because it would need to be \( (0,0) \), and natural numbers usually don't include zero.

Exam Tip: When dealing with operations on ordered pairs, apply the operation component-wise. Remember that the definition of natural numbers (N) usually excludes zero unless explicitly stated otherwise.

 

Question 12. State whether the following statements are true or false. Justify.
(i) For any arbitrary binary operation \( * \) on a set \( N \), \( a * a = a \), \( \forall a \in N \).
(ii) If \( * \) is commutative binary operation on \( N \), then \( a * (b * c) = (c * b) * a \).
Answer:
(i) The statement claims that for any arbitrary binary operation \( * \) on a set \( N \), \( a * a = a \) for all \( a \in N \).
This statement is false.
Justification: A binary operation only requires that for any \( a, b \in N \), \( a * b \in N \). It does not impose any condition like \( a * a = a \). For example, consider the operation of addition on \( N \). \( a + a = 2a \). Clearly, \( 2a \neq a \) for any \( a \neq 0 \). So, the condition \( a * a = a \) is not universally true for all binary operations.
(ii) The statement says if \( * \) is a commutative binary operation on \( N \), then \( a * (b * c) = (c * b) * a \).
This statement is true.
Justification:
Given that \( * \) is commutative, we know that \( b * c = c * b \).
So, the right-hand side of the equation, \( (c * b) * a \), can be rewritten as \( (b * c) * a \).
Also, since \( * \) is commutative, we can swap the order of elements in \( (b * c) * a \) to \( a * (b * c) \).
Therefore, \( a * (b * c) = (c * b) * a \) holds true.
In simple words: (i) False, because an operation like addition means \( a+a=2a \), which is not \( a \). (ii) True, because if an operation is commutative, you can swap the order of numbers. So \( (c * b) \) is the same as \( (b * c) \), and then you can swap \( a \) and \( (b * c) \) too.

Exam Tip: When evaluating statements about properties like commutativity or specific conditions like \( a * a = a \), always test them against basic counterexamples such as addition or multiplication to verify their truthfulness.

 

Question 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?
Answer: The binary operation \( * \) on the set \( N \) is defined as \( a * b = a^3 + b^3 \).
Let's check for commutativity:
\( b * a = b^3 + a^3 \).
Since addition is commutative, \( a^3 + b^3 = b^3 + a^3 \).
\( \implies \) Thus, \( a * b = b * a \). The operation \( * \) is commutative.
Next, let's check for associativity:
\( a * (b * c) = a * (b^3 + c^3) = a^3 + (b^3 + c^3)^3 \).
\( (a * b) * c = (a^3 + b^3) * c = (a^3 + b^3)^3 + c^3 \).
In general, \( a^3 + (b^3 + c^3)^3 \neq (a^3 + b^3)^3 + c^3 \).
For example, let \( a=1, b=2, c=3 \).
\( 1 * (2 * 3) = 1 * (2^3 + 3^3) = 1 * (8 + 27) = 1 * 35 = 1^3 + 35^3 = 1 + 42875 = 42876 \).
\( (1 * 2) * 3 = (1^3 + 2^3) * 3 = (1 + 8) * 3 = 9 * 3 = 9^3 + 3^3 = 729 + 27 = 756 \).
Since \( 42876 \neq 756 \), the operation is not associative.
\( \implies \) Therefore, the operation \( * \) is commutative but not associative.
The correct option is **(B) Is * commutative but not associative?**
In simple words: This operation involves cubing numbers and adding them. It's commutative because \( a^3 + b^3 \) is the same as \( b^3 + a^3 \). But it's not associative because if you group the operations differently, the results won't generally be the same; the powers make a big difference.

Exam Tip: For complex operations, always work out the expressions for commutativity and associativity step-by-step. If a property doesn't hold, a single counterexample with specific numbers is enough to prove it.