The set-builder form of the set {3, 6, 9, 12} can be written as
A{x : x = 3k, k ∈ ℤ, k ≠ 0}
B{x : x is divisible by 3}
C{x : 3 ≤ x ≤ 12, x ∈ ℤ}
D{x : x = 3k, k ∈ ℕ, 1 ≤ k ≤ 4}
Answer & Solution
Correct answer: D. {x : x = 3k, k ∈ ℕ, 1 ≤ k ≤ 4}
Option A names exactly {3, 6, 9, 12} via the parametrisation x = 3k for k = 1, 2, 3, 4. The other options include extra elements (negatives, all multiples of 3, all integers from 3 to 12) and so are not equal to the original four-element set.
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