Give $f:\mathbb{R} \rightarrow \mathbb{R}$ defined as $f(x)=x$ if $x$ is irrational, $\displaystyle f\left(\frac{p}{q}\right)=q$ if $\displaystyle \frac{p}{q}$ is a irreductible fraction with $p>0$, $f(0)=0$. Show that $f$ is not bounded at any interval nondegenerate.
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Looked bounded to me. On $[a,b]$, $\min(a,-1)\leq f(x)\leq \max(b,1)$. – Thomas Andrews Feb 04 '15 at 14:29
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2Are you sure that $f(\frac pq) = \frac1q$ and not $f(\frac pq) = q$ or $f(\frac pq) = p$? Your function is bounded. – AlexR Feb 04 '15 at 14:32
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This function is continuous at irrational points, thus locally bounded around them – Petite Etincelle Feb 04 '15 at 14:35
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sorry, is $f(p/q)=q$ – sebastian azocar Feb 04 '15 at 14:42
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show that every interval contains a rational number of arbitrarily large denominator. – hunter Feb 04 '15 at 14:42
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I show that exists for every interval $(a,b)$ an irrational $m/q$, arguing that $\exists q_1 \in \mathbb{N}, b-a>1/q_1>0$ and given arbitrarily $A>0$, $\exists q_2 \in \mathbb{N}, q_2>A$, then, $q=max{q_1,q_2}, \exists m \in \mathbb{N}$ such that $a<m/q<b$ and $f(m/q)=q>A$, but my problem is, happens if $m$ divides $q$ – sebastian azocar Feb 04 '15 at 14:51
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Hint Pick some interval $[a,b]$ and some integer $n$.
Prove that for each $1 \leq k \leq n$ there are finitely many fractions with denominator $k$ in the interval. As there are infinitely many rational numbers, there must be one with denominator larger than $n$.
N. S.
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Let $I=(a,b)$, with $a<b$, and $q$ be a prime number, such that $$ q>\frac{1}{b-a}. $$ Such $q$ exists, as the primes are infinitely many.
Then, $$ a=\frac{aq}{q}<\frac{\lfloor aq\rfloor+1}{q}\le \frac{aq+1}{q}=a+\frac{1}{q}<a+(b-a)=b, $$ and hence $\displaystyle \frac{\lfloor aq\rfloor+1}{q}\in(a,b) $ and clearly $$ f\left(\frac{\lfloor aq\rfloor+1}{q}\right)=q. $$ Thus in every interval $f$ takes arbitrarily large values.
Yiorgos S. Smyrlis
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