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anonymous

  • one year ago

For each n that belongs N, let A subscript n={(n+1)k: k belong to N}. (a) What is A subscript 1 intersection A subscript 2? (b) Determine the union of the sets {A subscript n: n belong to N} and the intersection {A subscript n: n belong to N}

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  1. thomas5267
    • one year ago
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    So: \[ \text{Assume }\mathbb{N}=\{1,2,3,4,5,6,7,\dots\},\,\text{i.e. }0\notin \mathbb{N}\\ A_n=\{(n+1)k,\,k\in \mathbb{N}\}\\ A_1=\{2k,\, k\in \mathbb{N}\}\\ A_2=\{3k,\, k\in \mathbb{N}\}\\ A_1\cap A_2=?\\ \bigcup_{r\in \mathbb{N}}A_r=A_1\cup A_2 \cup A_3 \cdots\\ \bigcap_{r\in\mathbb{N}}A_r=A_1\cap A_2 \cap A_3 \cdots \]

  2. anonymous
    • one year ago
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    so, what is the question all about. what is the union and interception ?

  3. anonymous
    • one year ago
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    so we're told that \(A_n\) is the set of integer multiples of \(n+1\), so intersection of \(A_1,A_2\) is $$\{\dots,-2,0,2,\dots\}\cap\{\dots,-3,0,3,\dots\}=\{\dots,-6,0,6,\dots\}$$ i.e. the set of integers divisible by both \(2\) and \(3\), or, equivalently, integer multiples of \(6\)

  4. anonymous
    • one year ago
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    actually, i mean positive integers so no negatives or zeros

  5. anonymous
    • one year ago
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    using this logic it should be obvious now why $$\bigcap_{n\in\mathbb{N}}A_n=\emptyset$$. similarly, it should be clear that if we take the union of all the positive multiples of positive integers bigger than \(1\), we'll get every nonnegative integer *except* one: $$\bigcup_{n\in\mathbb{N}}A_n=\mathbb{N}\setminus\{1\}$$

  6. anonymous
    • one year ago
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    ok thanks

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