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 $|x-y|<\varepsilon$.


(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.


1 Answer 1


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.

  • $\begingroup$ 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). $\endgroup$ Aug 8, 2019 at 19:48
  • $\begingroup$ Beautifully written; I greatly appreciate it. For anyone who, like me, was a little "dense" on seeing c as an integer, let me save you the effort. Pick t first, then pick b large enough so that c=tb is an integer. $\endgroup$
    – John McVey
    Aug 9, 2019 at 18:20

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