Equivalent sets – Compare by cardinality only: For which option are sets A and B equivalent (i.e., have the same number of elements)?
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AA = {a, b, c,…, z}, B = {1, 2, 3,…, 24}
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BA = {1/3, 1/2, 3/5}, B = {x : x = n/(n+2), n ∈ N}
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CA = {2, 4, 6}, B = {(2, 4), (4, 6), (2, 6)}
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DA = {x : x = (n^3 − 1)/(n^3 + 1), n ∈ W, n ≤ 3}, B = {0, 7/9, 13/14}
Answer
Correct Answer: A = {2, 4, 6}, B = {(2, 4), (4, 6), (2, 6)}
Explanation
Introduction / Context:Two sets are equivalent if they have equal cardinalities, regardless of the nature of their elements. We only compare counts, not values or types. This question includes finite sets, an infinite set, and a formula-generated set to test careful counting.
Given Data / Assumptions:
- Option (a): 26 letters vs 24 numbers
- Option (b): a finite set of size 3 vs an infinite set over N
- Option (c): two finite sets apparently of size 3 each
- Option (d): A generated by n ∈ W, n ≤ 3; B has 3 listed rationals
Concept / Approach:Count elements in each candidate pair. Equivalent sets must have equal counts. In (c), A has 3 numbers and B has 3 ordered pairs (each pair counts as a single element), so both have cardinality 3. In (d), n = 0, 1, 2, 3 yields values −1, 0, 7/9, 13/14 → |A| = 4, but B has 3 elements, so not equivalent.
Step-by-Step Solution:(a) 26 vs 24 ⇒ not equal(b) finite 3 vs infinite ⇒ not equal(c) 3 vs 3 ⇒ equal ⇒ equivalent(d) 4 vs 3 ⇒ not equal
Verification / Alternative check:List elements explicitly where ambiguous, especially in (d) using W = {0,1,2,3,…}.
Why Other Options Are Wrong:(a) unequal counts; (b) infinite vs finite; (d) 4 vs 3 elements.
Common Pitfalls:Treating ordered pairs in (c) as more than one element each; remember they are single elements of a set of pairs.
Final Answer:A = {2, 4, 6}, B = {(2, 4), (4, 6), (2, 6)}