Fractions with $a < \sigma(b)$ are dense

Question

Prove that any open interval of real numbers in $[1, \infty)$ contains a rational number $\frac{a}{b}$, $(a,b)=1$, with $b\leq a < \sigma(b)$. (Here $\sigma$ denotes sum of all positive divisors.)

No idea how to start this. Any help appreciated!

Answer 1

Let us consider an arbitrary open interval in $[1,\infty)$, say $(\alpha,\beta)$. We want: $$\alpha < \frac{a}{b}<\beta$$ $$b < a < \sigma(b)$$ Thus, it suffices that we find $a,b \in \mathbb{N}$ such that $\alpha \cdot b < a < \beta \cdot b$, $\gcd(a,b)=1$, and $\sigma(b) > \beta \cdot b$, since we know: $$\alpha \cdot b <a < \beta \cdot b < \sigma(b) \implies b < a < \sigma(b)$$ Can you complete the problem from here?