Function f Continuity at Irrational and 0: Sketch

[iNCREDiBLE]

Let f be that function defined by setting:

f(x) = x if x is irrational
= p sin(1/q) if x = p/q in lowest terms.

At what point is f continuous?

Continuous for irrational x, and for x = 0. Sketch:
p*sin(1/q) - p / q
= p(sin(1/q) -1/q)
But sin x - x = o(x^2) when x -> 0
So, for large q,
|p(sin(1/q) - 1/q)| < p (1/q)^2 = (p/q) / q

Is this correct?

 

[code.master]

First off, I am no master of number theory or modern algebra even. I have only observed said courses, and I have not researched my own opinion enough to give it as fact. That said, this is what I think.

Id watch my wording to prevent myself from saying continuous for irrational x, and say rather, continuous at every point x for x~irrational. the function itself is, i believe, nowhere-continuous. you might find some more searching about dirichlets function, its pretty similar.

oh, and your sketch is probably right. the function does go to 0 at x=0, and

I think the correct answer to this is that the function is continuous at x=0, since the limit of the two individual functions as x->0 from both sides is all the same value, 0. for all others, you might say that the point is continuous within the irrational space, but that in terms of the function, which lies in R, it is discontinuous.

id say it presents a problem since the sum of two irrational numbers has not been explicitly proven to be an irrational number, so there are numbers between every other number that have yet to be defined, and since many of those potentially rational numbers involves $\pi$ (like constant multiples of eulers constant) it might bring the possibility of creating a locally continuous point in the original space. anyhow, ill shut up since I am probably totally wrong.