A Variation of "rational is dense in $\Bbb R$"

Question

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 $$|x-{a\over b}|<\epsilon.$$

This paper gives a proof: http://www.jstor.org/stable/2324814 — Anthony Carapetis, Aug 28 '13 at 01:29
If you can't access that then there's a short proof on this MO question: http://mathoverflow.net/questions/117191/using-quotient-of-prime-numbers-to-approximation-reals. It references a relatively recent number theory paper but there's no need to use so strong a result - all you need is the prime number theorem. — Anthony Carapetis, Aug 28 '13 at 01:31

Don Larynx · Answer 1 · 2013-08-28T01:19:10.100

-1

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

edited Aug 28 '13 at 01:19

answered Aug 28 '13 at 01:17

Don Larynx

4,703

Forgive me for being dense, but why is this sufficient? – Alex Wertheim Aug 28 '13 at 01:30
Since we're looking for $a,b$ primes I don't see how this even makes sense... – Anthony Carapetis Aug 28 '13 at 01:35
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. – Don Larynx Aug 28 '13 at 01:39
I think you are confused. It says ${\it less}$ than some $\epsilon > 0$. – Alex Wertheim Aug 28 '13 at 01:40
I typed "greater" mistakenly – Don Larynx Aug 28 '13 at 01:42

A Variation of "rational is dense in $\Bbb R$"

1 Answers1