What are the greatest lower bound (GLB) and the least upper bound (\(LUB\)) of…
2019
What are the greatest lower bound (GLB) and the least upper bound (\(LUB\)) of the sets \(𝐴=\{3,9,12\} \) and \(𝐵=\{1,2,4,5,10\}\) if they exist in poset \((z^+, / )\)?
- A.
\(A(GLB – 3, LUB – 36); B(GLB – 1, LUB – 20)\) - B.
\(A(GLB – 3, LUB – 12); B(GLB – 1, LUB – 10) \) - C.
\(A(GLB – 1, LUB – 36); B(GLB – 2, LUB – 20) \) - D.
\(A(GLB – 1, LUB – 12); B(GLB – 2, LUB – 10)\)
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Correct answer: A
In the poset of positive integers with the divisibility relation, a lower bound of a set is an integer that divides every element of the set, and the greatest lower bound (GLB) is the largest such divisor (i.e., the greatest common divisor). An upper bound is an integer that every element of the set divides, and the least upper bound (LUB) is the smallest such integer (i.e., the least common multiple).
For A = {3, 9, 12}: GLB = gcd(3,9,12). Compute gcd(3,9) = 3, then gcd(3,12) = 3, so GLB = 3.
LUB = lcm(3,9,12). Compute lcm(3,9) = 9, then lcm(9,12) = 36, so LUB = 36.
For B = {1, 2, 4, 5, 10}: GLB = gcd(1,2,4,5,10). Because 1 is in the set and gcd with 1 is 1, GLB = 1.
LUB = lcm(1,2,4,5,10). The prime-power maxima give 2^2 (from 4) and 5 (from 5 or 10), so lcm = 2^2 * 5 = 20. Thus LUB = 20.
Final answers: For set A, GLB = 3 and LUB = 36. For set B, GLB = 1 and LUB = 20.
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