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Statement 4.114: Let x exist in [0,∞), then x exists in the union, n…
Statement 4.114: Let x exist in [0,∞), then x exists in the union, n exist in ∞, for [0,n].
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Since o≤ x < m, m exists in the set of natural numbers.
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Since 0 ≤ x ≤ m, x exists in [0,m]
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Thus x exists in the Union, n exists in the set of natural number, for [0,n].
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Statement 4.117: Let x exist in (0,1). Then x does not exist in the intersection, n exists in the set of natural numbers, [0,1/n]
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AFSOC x exists in the intersection, n exists in the set of natural numbers, [0,1/n]
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Then for all n in the set of natural number, we have e exists in [0,1/n]
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Since 1/x >1 and m > 1/x, we have m >1
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Since m exists in the set of integers and m >1, m exists in the set of natural numbers
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Thus x does not exist in [0,1/m].
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But we know from earlier that for all n that exists in the set of natural numbers, x exists in [0,n].
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This contradicts the assumption that x exists in the intersection, x exists in the set of natural numbers, [0,n].
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Thus we conclude that x does not exist in the intersection, n exists in the set of natural numbers, [0,n].
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Statement 4.119: For all m in the set of natural numbers, A_m = {x exists in the the set of integers such that x divides 2m}, then the set of natural numbers is a subset of the Union, k exists in the set of natural number, A_k.
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Since n exists in the set of natural numbers, i exists in the set of natural numbers.
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Since 2i = 6n and n exist in the set of natural number, n divides 2 i.
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Thus n exists in the union, k exists in the set of natural numbers, A_k
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Thus the set of natural numbers is a subset of the Union, k exists in the set of natural numbers, A_k
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