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I know that the theorem that the set $\Bbb Q$ of rationals is dense in $\Bbb R$ says:

For every $x\in \Bbb R$ and every $\epsilon>0$, there exist $a$, $b\in \Bbb Z$ with $b\ne0$ such that $$|x-{a\over b}|<\epsilon.$$

But what about if I change the condition "$a,b\in \Bbb Z$ with $b\ne0$" by "$a$ and $b$ are both primes". Does the theorem still hold? In other words, I wonder whether the following is true:

For every $x\in \Bbb R^+$ and every $\epsilon>0$, there exist $a$, $b\in \Bbb P$, where $\Bbb P$ is the prime numbers set, such that $$|x-{a\over b}|<\epsilon.$$

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Since $\frac{a}{b} > 0$, yes.

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  • $\begingroup$ Forgive me for being dense, but why is this sufficient? $\endgroup$ – Alex Wertheim Aug 28 '13 at 1:30
  • $\begingroup$ Since we're looking for $a,b$ primes I don't see how this even makes sense... $\endgroup$ – Anthony Carapetis Aug 28 '13 at 1:35
  • $\begingroup$ If the distance between x and a rational a/b is less than some $\epsilon > 0$, then the equation is true. Hence the primes also fit under this subset. $\endgroup$ – Don Larynx Aug 28 '13 at 1:39
  • $\begingroup$ I think you are confused. It says ${\it less}$ than some $\epsilon > 0$. $\endgroup$ – Alex Wertheim Aug 28 '13 at 1:40
  • $\begingroup$ I typed "greater" mistakenly $\endgroup$ – Don Larynx Aug 28 '13 at 1:42

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