[Calculus]Hard continutity problem.

[jalvarado]

Homework Statement

Define $G$ as follows:

$G(x) = \left\{ \begin{array}{c l} x, & \mbox{if } x \mbox{ is irrational} \\ \sqrt{\frac{1+p^2}{1+q^2}}, & \mbox{if } x = \frac{p}{q} \mbox{where } gcd(p,q) = 1 \end{array} \right.$

Show that $G$ is discontinuous at each negative number and also at each nonnegative rational number, but is continuous at each positive irrational number.

Homework Equations

$\lim_{x \rightarrow a} f(x) = f(a)$

The Attempt at a Solution

I found this question in an old calculus book at the end of the limits and continuity section.
I tried the approach using $\lim_{x \rightarrow a} G(x) = G(a)$ and trying the different cases for $a$ and it makes sense intuitively but I'm thinking and epsilon-delta approach is what is needed here. Any help?

 

[Dick]

Show what your reasoning is so far.

 

[jalvarado]

Sorry I didn't put it before. It takes a while to post from my phone.

I first test the case where x goes to an irrational number. Say b.
G(b) = b, so now we must look at the limit as x--->b of G(x).
My reasoning is that there is rationals around G(b) that approach this value as x gets closer and closer to b.

So for nonnegative irrationals continuity holds.

For the cases of positive or negative rationals we let wlog some number r = s/t where s and t are coprime.

Then G(r) = √(1+s^2)/(1+t^2). Now the limit as x--->r of G(x) wouldn't exist as x has to go through irrational values as it gets closer to r from either positive or negative side. So it wouldn't be continuous as
lim as x--->r of G(x) is not equal to G(r).