**poulo@mtuxo.UUCP (r.poulo)** (05/29/85)

I saw (in Rudin's Principles of Mathematical Analysis, I think) an interesting function showing what strange things can happen when discussing continuity and what a powerful restriction it is to say that a function is continuous. Unlike all examples I have seen of functions that are everywhere continuous but nowhere differentiable, this function is very simple. It has the property that it is continuous at all irrational points and discontinuous at all rational points. If x is irrational let f(x) = 0. If x is rational then suppose x = p/q in lowest terms. Let f(x) = 1/q. In any interval f(x) jumps an infinite number of times. Sample values for some rationals are: f(0) = f(1) = f(2) = ... = 1. f(1/2) = f(3/2) = f(5/2) = ... = 1/2. f(1/5) = f(2/5) = f(3/5) = f(4/5) = 1/5; f(5/5) = 1. Problems for you mathematicians out there: 1. Prove that f(x) is continuous at irrational points and discontinuous at rational points. (This one is real simple -- if your attempts take more than 30 seconds to explain you are doing it wrong.) 2. See if you can devise a simple function that is discontinuous at irrational points and continuous at the rational points. I have not succeeded, but that proves nothing. 3. I have wondered if f(x) is possibly differentiable at the irrational points. (It is immediate that it is not differentiable at the rational points since it is not continuous there.) This question immediately leads to number theory and particularly to the theory of rational approximations, about which I know little. Perhaps someone can complete the discussion below. Consider the expression f(x+dx)-f(x) -------------- dx Since we are considering irrational values of x we have that f(x) = 0, so we can consider the expression f(x+dx)/dx. If we let dx-->0 through rational values the sum x+dx will always be irrational, f(x+dx) will be 0, and the limit (taken through rational points) will be 0. If f(x) is differentiable we must have a limit of 0 when dx-->0 through appropriately chosen irrational points that make the sum x+dx assume rational values. In particular let p / q be a rational sequence that converges on x and suppose n n that x+dx assumes these values. Then f(x+dx) = 1 / q and dx = (p / q ) - x. n n n The ratio f(x+dx)/dx takes on the values 1 ----------- p - xq n n Does anyone know enough about rational approximations to tell what happens to the value p - xq as n --> infinity? As the ratio of two expressions that both --> 0 n n it is not obvious what should happen. If it is differentiable at these points it would be a fascinating example / counterexample to give to freshman calculus classes. However I am 2/3 convinced that this is too much to ask and that the function is not differentiable anywhere. As a final comment note that the integral of f(x) over any interval is 0. (If you don't know measure theory this won't make sense, but no flames please.)