CBSE Class 12 Mathematics Relations And Functions Assignment Set 19

XII - Mathematics Assignment No-04 - Relations and Functions

Q1. Let * be a binary operation on \(N\) given by \(a * b = \text{l.c.m of } (a, b)\) for all \(a, b \in N\). Find (i) \(2 * 4\) (ii) \(3 * 5\) (iii) \(7 * 21\)
Answer: The binary operation is defined as \(a * b = \text{l.c.m}(a, b)\).
(i) \(2 * 4 = \text{l.c.m}(2, 4) = 4\)
(ii) \(3 * 5 = \text{l.c.m}(3, 5) = 15\)
(iii) \(7 * 21 = \text{l.c.m}(7, 21) = 21\)
In simple words: The operation gives the least common multiple of the two given numbers.

Exam Tip: For coprime numbers, the LCM is simply their product, while for multiples, the larger number is the LCM.

Q2. Let * be a binary operation on \(Z\) defined as \(a * b = a + 3b^2\) for all \(a, b \in Z\). Find (i) \(1 * 3\) (ii) \(2 * 4\) (iii) \(3 * 2\)
Answer: The binary operation is given by \(a * b = a + 3b^2\).
(i) \(1 * 3 = 1 + 3(3)^2 = 1 + 3(9) = 1 + 27 = 28\)
(ii) \(2 * 4 = 2 + 3(4)^2 = 2 + 3(16) = 2 + 48 = 50\)
(iii) \(3 * 2 = 3 + 3(2)^2 = 3 + 3(4) = 3 + 12 = 15\)
In simple words: Substitute the values of \(a\) and \(b\) into the given algebraic expression \(a + 3b^2\).

Exam Tip: Always apply the exponent before multiplying by 3 following the order of operations (PEMDAS/BODMAS).

Q3. Find the number of binary operations on the set \(\{a, b\}\)
Answer: Let \(S = \{a, b\}\). The number of elements in \(S\) is \(n = 2\).
A binary operation on set \(S\) is a function from \(S \times S \to S\).
The total number of elements in \(S \times S\) is \(n \times n = 2 \times 2 = 4\).
The total number of binary operations is \(n^{n^2} = 2^{2^2} = 2^4 = 16\).
In simple words: A binary operation maps every pair of elements to an element in the set, giving \(2^4 = 16\) possible mappings.

Exam Tip: Remember the general formula: for a set with \(n\) elements, the number of binary operations is \(n^{n^2}\).

Q4. Define a binary operation * on the set \(\{0, 1, 2, 3, 4, 5\}\) as a * b =  a + b, & if  a + b < 6 a + b - 6, & if a + b > 6. Show that zero is the identity operation and each element 'a' of the set is invertible with \((6-a)\) being the inverse of \(a\).
Answer: Let \(S = \{0, 1, 2, 3, 4, 5\}\).
1. Show that 0 is the identity element:
For any element \(a \in S\), we know \(0 \le a \le 5\), which implies \(a + 0 = a < 6\).
Using the definition, \(a * 0 = a + 0 = a\) and \(0 * a = 0 + a = a\).
Thus, \(a * 0 = 0 * a = a\) for all \(a \in S\). Therefore, 0 is the identity element.
2. Show that each element \(a\) is invertible with inverse \((6-a)\):
For \(a = 0\), the inverse is \(6 - 0 = 6\) (Note: if \(a=0\), \(6-0=6\), but usually in modulo arithmetic, 0 is its own inverse since \(0+0 \equiv 0\). Specifically, for \(a=0\), \(0*0 = 0\). For \(a \neq 0\), let \(b = 6 - a\)).
For any \(a \in S\) (where \(a \neq 0\)), let the inverse be \(b = 6 - a\). Since \(a \in \{1, 2, 3, 4, 5\}\), \(b \in \{5, 4, 3, 2, 1\} \subset S\).
Consider \(a * (6 - a)\):
Here, \(a + (6 - a) = 6 \ge 6\), so we use the second case of the definition: \(a + (6 - a) - 6 = 6 - 6 = 0\), which is the identity element.
Similarly, \((6 - a) * a = (6 - a) + a - 6 = 0\).
Thus, \(a * (6 - a) = (6 - a) * a = 0\).
Therefore, every element \(a \in S\) is invertible, and the inverse of \(a\) is \((6 - a)\).
In simple words: Zero acts like 0 in addition, and adding \(6-a\) wraps around to 0, making it the inverse under this modulo-6 operation.

Exam Tip: This is a classic example of modulo arithmetic (\(\pmod 6\)) disguised as a piecewise function.

Q5. Which of the following binary operations on the set \(N\) are associative and which are Commutative?
(i) \(a * b = 1\) for all \(a, b \in N\)
(ii) \(a * b = \frac{a + b}{2}\) for all \(a, b \in N\)
Answer:
(i) \(a * b = 1\) for all \(a, b \in N\)
- Commutativity: \(b * a = 1\) and \(a * b = 1\). Since \(a * b = b * a = 1\), it is Commutative.
- Associativity: \((a * b) * c = 1 * c = 1\), and \(a * (b * c) = a * 1 = 1\). Since \((a * b) * c = a * (b * c) = 1\), it is Associative.
(ii) \(a * b = \frac{a + b}{2}\) for all \(a, b \in N\)
- Commutativity: \(b * a = \frac{b + a}{2} = \frac{a + b}{2} = a * b\). It is Commutative.
- Associativity: Let \(a = 1, b = 2, c = 3\).
\((1 * 2) * 3 = \left(\frac{1 + 2}{2}\right) * 3 = \frac{3}{2} * 3 = \frac{\frac{3}{2} + 3}{2} = \frac{9}{4}\)
\(1 * (2 * 3) = 1 * \left(\frac{2 + 3}{2}\right) = 1 * \frac{5}{2} = \frac{1 + \frac{5}{2}}{2} = \frac{7}{4}\)
Since \(\frac{9}{4} \neq \frac{7}{4}\), it is not Associative.
In simple words: A constant operation is always commutative and associative, whereas the arithmetic mean is commutative but fails associativity.

Exam Tip: To disprove associativity, always provide a simple counterexample using small natural numbers like 1, 2, and 3.

Q6. Consider the binary operation * on the set \(A = \{1, 2, 3, 4, 5\}\) defined by \(a * b = \min(a, b)\). Write the operation table of the operation *
Answer:
The operation table for \(a * b = \min(a, b)\) on \(A = \{1, 2, 3, 4, 5\}\) is constructed by taking the minimum of the row and column elements:

In simple words: Each cell in the table contains the smaller of the corresponding row and column numbers.

Exam Tip: Verify symmetry along the main diagonal to quickly check if the operation is commutative.

Q7. Let * be a binary operation on the set \(Q\) of rational numbers as follows:
(a) \(a * b = \frac{ab}{4}\)
(b) \(a * b = ab^2\)
Find which of the binary operations are Commutative and which are associative?
Answer:
(a) \(a * b = \frac{ab}{4}\)
- Commutativity: \(b * a = \frac{ba}{4} = \frac{ab}{4} = a * b\). It is Commutative.
- Associativity:
\((a * b) * c = \left(\frac{ab}{4}\right) * c = \frac{\left(\frac{ab}{4}\right)c}{4} = \frac{abc}{16}\)
\(a * (b * c) = a * \left(\frac{bc}{4}\right) = \frac{a\left(\frac{bc}{4}\right)}{4} = \frac{abc}{16}\)
Since \((a * b) * c = a * (b * c)\), it is Associative.

(b) \(a * b = ab^2\)
- Commutativity: Let \(a = 2, b = 3\).
\(2 * 3 = 2(3)^2 = 2(9) = 18\)
\(3 * 2 = 3(2)^2 = 3(4) = 12\)
Since \(18 \neq 12\), it is not Commutative.
- Associativity: Let \(a = 1, b = 2, c = 3\).
\((1 * 2) * 3 = (1(2^2)) * 3 = 4 * 3 = 4(3^2) = 4(9) = 36\)
\(1 * (2 * 3) = 1 * (2(3^2)) = 1 * 18 = 1(18^2) = 324\)
Since \(36 \neq 324\), it is not Associative.
In simple words: Symmetric algebraic expressions tend to be commutative and associative, while asymmetric expressions usually fail both tests.

Exam Tip: Always test commutativity with general variables first, and use numeric counterexamples to disprove associativity or commutativity quickly.

Q8. Let \(A = N \times N\) and let * be a binary operation on \(A\) defined by \((a, b) * (c, d) = (a + c, b + d)\). Show that * on \(A\) is Commutative and Associative. Also find the identity element on \(A\) if it exists.
Answer:
1. Commutativity:
Let \((a, b), (c, d) \in A\).
\((a, b) * (c, d) = (a + c, b + d)\)
\((c, d) * (a, b) = (c + a, d + b)\)
Since addition in natural numbers is commutative, \((a + c, b + d) = (c + a, d + b)\).
Thus, \((a, b) * (c, d) = (c, d) * (a, b)\). Therefore, * is Commutative.

2. Associativity:
Let \((a, b), (c, d), (e, f) \in A\).
\(((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f)\)
\((a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = (a + (c + e), b + (d + f))\)
By the associative property of addition in natural numbers, these are equal.
Thus, * is Associative.

3. Identity Element:
Let \((e_1, e_2) \in A\) be the identity element such that for any \((a, b) \in A\):
\((a, b) * (e_1, e_2) = (a, b)\)
\((a + e_1, b + e_2) = (a, b)\)
This implies \(a + e_1 = a \implies e_1 = 0\) and \(b + e_2 = b \implies e_2 = 0\).
However, \((0, 0) \notin N \times N\) because 0 is not a natural number.
Therefore, no identity element exists in \(N \times N\).
In simple words: The operation is just component-wise addition, which inherits commutativity and associativity from regular addition, but lacks an identity because zero is not in the set of natural numbers.

Exam Tip: Always check carefully whether the candidate for an identity element actually belongs to the specified domain set \(N\).