Expert: Sherry Wallin - 12/5/2009

Question
Given a function f(x) = 0    if x is irrational
                     = 1/q  if x is rational and is in the form
                            p/q, where p and q have no common
                            factors
(e.g. f(sqrt2)=0 and f(17/29)= 1/29)
Prove that f is continuous for all irrational values and discontinuous for all rational values

Answer
Greg~
    If you are looking at rational functions in their reduced form there are irrational numbers everywhere between any two rational numbers, the graph would be discrete and not continuous. Now if the number is irrational and the function f maps the irrational number to 0 then the line y = 0 is continuous. As far as a proof goes for the irrational part assume that f is not continuous for all irrational values so there exists an irrational value such that the graph has a discontinuity in it, but all irrational numbers are mapped to 0 so there is your contradiction and therefore all irrational values graph as a continuous function. Either that or if the graph is discontinuous then there has to be an exact representation p/q and the function maps to q, but q is an integer which contradicts the fact that the value you are looking at is irrational. For the rational part assume that the graph is continuous for at least one rational value. That means your number will be mapped to 0, but zero can't be q (since we don't divide by 0) thus contradicting that f is continuous for a rational value.

Math Prof