# map

• Apr 17th 2009, 11:16 AM
manjohn12
map
Suppose $\displaystyle f(x) = \begin{cases} 0 \ \ \ \text{if} \ \ x \notin \mathbb{Q} \\ 1/q \ \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = p/q \ \text{in lowest terms} \end{cases}$

1. Fix $\displaystyle t \in \mathbb{R}$ and $\displaystyle n \in \mathbb{N}$. Show that $\displaystyle f$ maps only finitely many elements of $\displaystyle (t-1/2, t+1/2)$ to $\displaystyle 1/n$.

So the length of this interval is $\displaystyle 1$. I think its $\displaystyle n-1$. But that is intuition. How do you make it rigorous?

2. Prove that $\displaystyle f$ is continuous at every irrational number. So $\displaystyle \forall \epsilon >0, \exists \delta >0$ such that $\displaystyle |x-a| < \delta \implies |f(x)| < \epsilon$. Do we want to elements that dont map to $\displaystyle 1/n$?
• Apr 17th 2009, 06:43 PM
manjohn12
Basically we are trying to show that $\displaystyle f$ maps a finite number of rational numbers in lowest terms in the interval $\displaystyle I = (t- \frac{1}{2}, t+\frac{1}{2})$ to $\displaystyle \frac{1}{n}$. Let $\displaystyle x = p/n \in I$ and in lowest terms. Then $\displaystyle p$ and $\displaystyle n$ are coprime. Can we somehow use this to deduce that the maximum number is $\displaystyle n-1$? number of fractions $\displaystyle p/n$ in interval is $\displaystyle n$. number in lowest terms is $\displaystyle \leq n-1$.