# Is this set of fractions dense in the interval $\big[\frac 13,\frac 12\big)$?

I have an interest in the set $$A= \bigg\{\frac{ab+c}{(2a+1)b+c}\,\bigg|\, a \in {\mathbb Z}^+, b\in{\mathbb Z}^+~\text{is $$(a+1)$$-smooth}, 0\leq c\leq ab\bigg\}.$$ In particular, is $$A$$ dense in the interval $$\big[\frac 13,\frac 12\big)$$?

The question is pretty much self-explanatory (given I mean by $${\mathbb Z}^+$$ the positive integers, and $$c$$ is an integer), though I'll explain $$a$$-smooth. Given integers $$a$$ and $$b$$, $$b$$ is $$a$$-smooth when every prime dividing $$b$$ is bounded above by $$a$$. Said another way, $$b$$ has a factorization in the integers where each factor is bounded above by $$a$$.

Now for some background. I am working with a family $$\mathcal G$$ of (finite, solvable) groups for which, for each $$G\in\mathcal G$$ the set $${\rm cd}(G)$$ of character degrees is the disjoint union of sets $$X$$ and $$Y$$ where $$|X|=(a+1)b$$ and $$|Y|=ab+c$$ (the values $$a$$, $$b$$, and $$c$$ are parameters which determine a particular group $$G\in\mathcal G$$). The conjecture pertains to what percentage of the set $${\rm cd}(G)=X\cup Y$$ the subset $$Y$$ itself is, i.e., what is the ratio $$\frac{|Y|}{|X|+|Y|}=\frac{ab+c}{(2a+1)b+c}$$?

For what it's worth, the smoothness criterion is coming from the group theory. Without that requirement, two things are left unknown: (i) that $$|Y|=ab+c$$, and (ii) that $$|Y|$$ is the correct numerator.

Among things I would like to see happen, the best outcome is, given an arbitary rational number $$x\in[\frac13,\frac12)$$, there is a group $$G$$ for which $$\frac{ab+c}{(2a+1)b+c}=x$$, i.e., $$x\in A$$. Next best, for that $$x$$, is satisfying the formal definition of dense, for each $$\varepsilon>0$$ there exists $$y\in A$$ so that $$|x-y|<\varepsilon$$.

So...

(1) Is the set $$A$$ dense in $$\big[\frac13,\frac12\big)$$?

(2) If so, does $$A$$ contain all of $$\big[\frac13,\frac12\big)\cap\mathbb Q$$?

Work: I can prove (2) is true if the condition of $$b$$ being $$(a+1)$$-smooth is removed. Also, I have performed enough computer runs to be (heuristically) convinced that, for a given rational $$x$$ in the interval, the number of candidate $$a$$'s is sufficiently small so that finding a value $$a$$ which is large enough to allow an appropriate $$b$$ to be $$(a+1)$$-smooth seems unlikely; I would be very surprised were (2) true.

Take $$a=1$$, so $$b = 2^k$$, and let $$c = t 2^k$$ where $$t$$ is a dyadic rational in $$[0,1]$$. Then $$\frac{ab+c}{(2a+1)b+c} = \frac{t+1}{t+3}$$ The dyadic rationals are dense in $$[0,1]$$ and the function $$f:\; t \mapsto (t+1)/(t+3)$$ is continuous from $$[0,1]$$ onto $$[1/3, 1/2]$$, so these values are indeed dense in $$[1/3, 1/2]$$.
As for (2), note that $$\frac{ab+c}{(2a+1)b + c} \ge \frac{a}{2a+1}$$ so your fraction can't be less than $$2/5$$ unless $$a=1$$. Thus the only values in $$[1/3, 2/5]$$ are the images of dyadic rationals under $$f$$, which are not all the rationals in that interval.
• And for a similar argument for (2) that is uniform in $a$ I propose to rewrite the original fraction as (1-s)/(2-s), where s=(1-c/b)/(a+1) is some nonnegative rational number with (a+1)-smooth denominator and absolute value at most 1/(a+1). – Luca Ghidelli Aug 8 '19 at 19:48