I have sets $A = \{1/n \mid n \in \mathbb{N}\}$ and $B = \left\{\frac{k}{2^n} \mid k,n \in \mathbb{N} \right\}$. Am I correct that $A \cap B$ is exactly $\frac{1}{2^n}$? Certainly I need to have numerator $1$, but I'm worried that there could be some kind of fraction simplifcation that makes, say, $\frac{1}{5}$, an element of this intersection.
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Where does the $2^n$ come from? Can you see that $\frac 13$ is in $A\cap B$ for example? – lulu Jul 19 '21 at 22:34
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I apologize. I miswrote the set $B$. The denominator should be $2^n$. – Brad G. Jul 19 '21 at 22:36
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Ah. In that case you are certainly correct. – lulu Jul 19 '21 at 22:36
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If $\dfrac k{2^n}=\dfrac 15$, then $k=\dfrac {2^n}5$, so $5$ divides $2^n$, which is a contradiction – J. W. Tanner Jul 19 '21 at 22:39
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Suppose $x \in A \cap B$. Then there is an $m \in \Bbb{N}$ such that $x = 1/m \in A$ and there are $k,n \in \Bbb{N}$ such that $x = k/2^n \in B$. Therefore, $$ 1 \cdot 2^n = m \cdot k \text{.} $$ This forces $m$ and $k$ to be powers of $2$. This resulting constraint on $m$ shows that $A \cap B \subseteq \{1/m : m = 2^n \text{ and } n \in \Bbb{N}\}$ which set simplifies to $\{1/2^n : n \in \Bbb{N}\}$.
Now for $n \in \Bbb{N}$, it is immediate that $1/2^n \in A$ and $1/2^n \in B$, so $1/2^n \in A \cap B$ and we conclude $\{1/2^n : n \in \Bbb{N}\} \subseteq A \cap B$.
Therefore, $A \cap B = \{1/2^n : n \in \Bbb{N}\}$.
Eric Towers
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Since $A$ is a subset of $B$ (just taking $k=1$), you have that $A\cap B=A$.
Cheering Rhino
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In that case, you would be correct. All the divisors of $2^n$ are powers of $2$ so, if it simplifies to a fraction of the form $\frac{1}{n'}$ then the numerator will be a power of $2$ and, after simplification, you obtain $\frac1{2^n}$ again. – Cheering Rhino Jul 19 '21 at 22:39