# Continuous Functions

• January 23rd 2010, 05:08 AM
Continuous Functions
Let:

$f(x)=\left\{\begin{matrix} \frac{1}{q}, & \mbox{if }x\mbox{ is rational} \\ 0, & \mbox{if }x\mbox{ is irrational}
\end{matrix}\right.$

while: $q>0, x=\frac{p}{q}$ , and p,q are relatively prime.

How can I prove that f is continuous in every irrational point, and noncontinuous in rational points?

It looks as if it is not continuous at all (because between every irrational number there's a rational number, and vice-verse) (Thinking)

*(Sorry for my bad English)

Can anyone please help me to find a way to solve this problem?

Thank you very much :)
• January 23rd 2010, 06:06 AM
drumist
This seems to explain this function in great detail:

Thomae's function - Wikipedia, the free encyclopedia