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!
$\endgroup$21 Answer
$\begingroup$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.
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