## 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}

1. thomas5267

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

so, what is the question all about. what is the union and interception ?

3. anonymous

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

actually, i mean positive integers so no negatives or zeros

5. anonymous

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

ok thanks