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While studying the GRE Math Subject test, I came across a property in a proof that for the function $$f(x)= \begin{cases} 0, & x \notin \mathbb{Q} \cap(0,1)\\ \frac{1}{n}, & x=\frac{m}{n} \end{cases} $$
with $GCD(m,n)=1, n>0$, that $f$ is continuous on every irrational point in $I=(0,1)$, and discontinuous on the rationals there.

The property in question is the following:

Given a positive integer, $q$, and an irrational number $a$ in any open sub-interval of $I$ centred at $a$, there are fewer than $1+2+ \cdots + (q-1) = \frac{q(q-1)}{2}$ rationals of the form $\frac{m}{n}<1$ with $n<q$.

Can someone help me make sense as to why this might be true?

TheMathochist
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2 Answers2

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It all starts with Dirichlet's approximation theorem (a short proof here). Everything else is just counting, with the $\frac{m}{n}<1 \Rightarrow 0<m < n$ constraint. E.g. $$n=q \Rightarrow m \in \{1,...,q-1\}$$ $$n=q-1 \Rightarrow m \in \{1,...,q-2\}$$ $$...$$ $$n=2 \Rightarrow m \in \{1\}$$ Altogether, counting all the possible values for $m$, maximum $\frac{q(q-1)}{2}$.

rtybase
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This seems like a crude estimate of the number of fractions with denominators less than or equal to $q$ in the entire interval $(0,1)$. There is 1 fraction with denominator equal to 2, 2 fractions with denominator equal to 3, 3 fractions with denominator equal to 4, up to $q-1$ fractions with denominator equal to $q$.

sharding4
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  • @Shuri2060 It should definitely read something like "fractions of the form $\frac{m}{n}<1$ with $n<q$." – sharding4 Jul 05 '17 at 20:07
  • Yes. and it's fixed. And after reading it a couple more times and messing with some values... Yes sharding4, denominators less than q. derp. For every integer $q$, there are $q-1$ fractions less than $1$ with denominator less than $q$. Obviously the number of fractions in for a given denominator will be one less than value of the denominator so the total number of such fractions, for a given $q$ will be less than $\frac{q(q-1)}{2}$. But fewer than that will be co-prime. Thanks. Apologies for asking a silly question and for the typo. – TheMathochist Jul 05 '17 at 20:56