Question

Let (N) denote the set of all natural numbers and (kN = kx x N) for any (k N). Then (8N 12N =)

• A. (4N)
• B. (10N)
• C. (24N)
• D. (40N)

Hint:

For set theory questions involving integer coordinate multiples: * Intersection (()) corresponds to finding common multiples, so: (aN bN = LCM(a, b)N). * Do not mistake intersection for finding the highest common factor (HCF). The HCF values are related to the linear combinations of sets, not direct intersections!

Answer 1

The Correct Option is C

Concept: The set (kN) represents all positive integral multiples of the number (k). When we take the intersection (()) of two such sets, say (aN bN), we are looking for elements that belong to both sets simultaneously. An element belongs to both sets if and only if it is a common multiple of both (a) and (b). The smallest positive element in this intersection set will be the Least Common Multiple (LCM) of (a) and (b). Therefore, the property follows: \[ a\mathbb{N} \cap b\mathbb{N} = \text{LCM}(a, b)\mathbb{N} \]

Step 1: Understand the composition of both individual sets explicitly. Given definition: (kN = kx x N).

• For (k = 8): \[ 8\mathbb{N} = {8 \times 1, 8 \times 2, 8 \times 3, 8 \times 4, 8 \times 5, \ldots} = {8, 16, 24, 32, 40, 48, 56, 64, 72, \ldots}
\item For (k = 12): \[ 12\mathbb{N} = {12 \times 1, 12 \times 2, 12 \times 3, 12 \times 4, \ldots} = {12, 24, 36, 48, 60, 72, 84, \ldots} \end{itemize} \end{itemize} \textbf{Step 2:} Find the elements that appear in both sets to determine the intersection set (8\mathbb{N} \cap 12\mathbb{N}): Looking at the lists of elements generated above, the common elements are: \[ 8\mathbb{N} \cap 12\mathbb{N} = {24, 48, 72, 96, \ldots} \]

Step 3: Express this common intersection set using the single multiplier set notation. We observe that every number in the set (24, 48, 72, 96, ) is a multiple of 24. We can rewrite the set as: \[ 8\mathbb{N} \cap 12\mathbb{N} = {24 \times 1, 24 \times 2, 24 \times 3, 24 \times 4, \ldots} = 24\mathbb{N} \] Mathematically, this is because the Least Common Multiple (LCM) of 8 and 12 is calculated as:

• Prime factorization of (8 = 2^3)

• Prime factorization of (12 = 2^2 3^1)

• (LCM(8, 12) = 2^3 3^1 = 8 3 = 24)
Hence, (8N 12N = 24N), which corresponds to Option (C).