I understand what that statement means, but for some reason I can't succinctly (because it is a question worth only 1 mark) put into words why it is the case. Could I say that: $\frac{1}{p}\rightarrow 0$ as $p\rightarrow\infty$ and so $\frac{1}{p}=0 This question is in the context of countable sets, if that is relevant. Thanks in advance.
Explain why it is the case that for each $r > 0$, there is a prime $p$ with $\frac{1}{p}
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real-analysis
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4This follows from the Archemedian Principle. Since we have a $1/n < r$ and for any $n$ there is a prime $p > n$ we know that $1/p < 1/n$ so $1/p < r$. – 2017-02-28
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As I said in the comments this property follows from the Archimedean principle.
Given some $r$ by AP we know that there is an $n$ such that $\frac{1}{n} < r$. And by the infinitude of the primes we know that given any $n$ there is a prime $p > n$ which means $\frac{1}{p} < \frac{1}{n}$. So by transitivity we know $\frac{1}{p} < r$ for any $r>0$.
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0That's a nice way of putting it. I'll mark your answer as correct when I'm allowed to. Thank you. – 2017-02-28