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I am working on this exercise:

$$ forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$

To clarify, r is a real number, q is a rational number. This is what I have so far:

$$ 0 < |r - q| < varepsilon $$
$$ -varepsilon < (r-q) < varepsilon $$
$$ r - varepsilon < q < r + varepsilon $$

Then making use of a given theorem that "Between any two distinct real numbers there is a rational number and an irrational number," I conclude the proof by claiming that the inequality must hold by the theorem and given that $r - epsilon$ and $r + epsilon$ are two distinct real numbers.

Is this sufficient for the proof? Otherwise, would appreciate some help on how to proof the above inequality more rigorously.

I recall the Archimedean Property that for each positive real number r, there exists a positive integer n such that $frac{1}{n} < r$. There is a hint to use this condition, but I am unsure how to apply it into the proof.

Thanks in advance for any help!

If you're using the density of the rationals, then your proof is upside down, i.e., to fix it, go from $r - epsilon < q < r + epsilon$ to your claim.

Nevertheless, you're using a rocket launcher against a fly; in fact, the two statements are pretty much equal, so it's almost circular reasoning. Suppose WLOG $r > 0$; since you have the Archimedian property, fix $epsilon$ and pick the least $n$ such that $epsilon geq 1/n$. If $r < epsilon$ pick $q=0$ and you're done. Otherwise, pick the biggest $m$ such that $r - m/n geq 0$; since $r - (m+1)/n < 0$ (because $m$ is maximal), $r-m/n < 1/n leq epsilon$. Pick $q = m/n$ and we're done.