Question: If An = {(n+1)k : k E N} determine the sets U{An : n E N} and N{An : n E N} Hypothesis: U{A, : n e N} = N – {1} and N{An : n e N} = Ø.
Question: If An = {(n+1)k : k E N} determine the sets U{An : n E N} and N{An : n E N} Hypothesis: U{A, : n e N} = N – {1} and N{An : n e N} = Ø.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:I'm a beginner. If I made a mistake explain everything as clearly as possible without skipping a step. (My professor insists to
not skip a step.)
Question: If An = {(n+ 1)k : k e N} determine the sets U{An : n e N} and N{An :n e N}
%3|
Hypothesis: U{A, : n e N} = N – {1} and N{An : n e N} = Ø.
Proof: Assume a E U{An : n e N}. Since An CN- {1}, for all n e N, (since if æ = (n+ 1)k, for n, k e N, then
x E N - {1}), we have An CN- {1} for some n e N. Thus U{An :n e N} C N – {1}. Moreover, if x € N – {1}, then a
can be the product of one or more natural numbers greater than one. Hence, there exists some natural numbers in N whose
product is z E N. Therefore, if æ = (n + 1)k, for some n, k E N, then a E U{An :ne N}. Thus,
N- {1} CU{An : ne N}.
Since U{An : n e N} CN - {1} and N –- {1} CU{An : n € N}, we have N – {1} =U{An : n e N}.
Suppose N{An : n e N} + 0. If æ E N{An :n E N}, then all An intersect eachother based on the lowest common multiple
of {(n + 1) : n e N}. This is determined by natural numbers with the most prime factors. Since there are infinite prime
factors, their product is oo which should be the intersection but oo is not a number. Hence it is not an element of N. Therefore,
N{An : n e N}is empty. This is a contradiction to the first statement that N{An :n e N}is not empty, thus
N{An : n e N} = Ø.
Question: Is my proof correct? What is a shorter proof that a begginer like me can write?
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