Proving Dirichelet's Function Converges to 0: A Mathematical Proof

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Homework Help Overview

The discussion revolves around proving the convergence of Dirichlet's function to 0 on the interval (0,1). The function is defined differently for rational and irrational numbers, leading to questions about its limit behavior as it approaches a given point.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the implications of the professor's proof regarding the limit of Dirichlet's function and questioning how certain values of p/q can exceed epsilon when n is chosen large enough that 1/n is less than or equal to epsilon.

Discussion Status

The discussion is ongoing, with participants expressing confusion about specific statements in the proof. They are seeking clarification on the relationship between the values of p/q and the chosen epsilon, indicating a productive exploration of the mathematical concepts involved.

Contextual Notes

Participants are grappling with the definitions and properties of Dirichlet's function, particularly in relation to rational numbers and their denominators. There is an emphasis on the constraints of the proof and the implications of choosing large values for n.

barksdalemc
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Homework Statement


Consider the Dirichelet's function defined on (0,1) by

f(x)= 0 is x is irrational
and
f(x)= 1/q if x=p/q

where p and q are positive integers with no common factors. Show that lim f(x) for any x in (0,1) is 0

The Attempt at a Solution



Here is the first line of the proof the professor gave us.

"Let eps>0 and n so large that (1/n)<=eps, where n is a natural number. The only numbers x for which f(x) can > eps are 1/2, 1/3, 2/3, ..., 1/n, ...(n-1)/n. "

I don't understand how the any p/q with n as q can be greater than epsilon is n is chosen so large that 1/n <=eps.
 
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barksdalemc said:
I don't understand how the any p/q with n as q can be greater than epsilon is n is chosen so large that 1/n <=eps.
had trouble reading this last sentence

do you mean you don't understand how it's possible that p/n > eps given that 1/n <= eps?
 
Last edited:
yes that is exactly where I am confused, but he is saying that f(p/n) > eps. f(p/n) given 1/n.
 
i'd list all those numbers to exclude them when I'm choosing delta, then any f(x)<1/n<=eps
 
The point is that, within any finite distance of a number a, there can be only a finite number of numerators for a fraction with denominator n: and so only a finite number of fractions with denominator less than or equal to n.
 

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