Contiuity tedious? probelm

[jessicaw]

Homework Statement

For each rational number x, write x as $\frac{p}{q}$
where p, q are integers with no common factors
and q > 0, and then define f(x) = $\frac{1}{q}$
. Also define f(x) = 0 for all x$\in R\Q$. Show that f
is continuous on R\Q and discontinuous at each point of Q.

Homework Equations

The Attempt at a Solution

for discontinuous i can do it.
i have difficultire doing the continuous part. i know it is a difficult task to work with the epsilon-delta definition but i think its the only option.

 

[mighty2000]

Thank you for your post. I will try to provide a clear explanation of the continuity of f(x) on R\Q.

First, let's define the function f(x) as stated in the problem:

f(x) = \frac{1}{q} for all rational numbers x = \frac{p}{q} where p, q are integers with no common factors and q > 0
f(x) = 0 for all x \in R\Q

To show that f(x) is continuous on R\Q, we need to show that for any given \epsilon > 0, there exists a \delta > 0 such that for all x \in R\Q, if |x - c| < \delta, then |f(x) - f(c)| < \epsilon.

Let's choose an arbitrary c \in R\Q. Since c is a rational number, we can write c as \frac{p}{q} where p, q are integers with no common factors and q > 0. Now, let's choose \epsilon > 0 and let \delta = \frac{\epsilon}{q}. We will show that this choice of \delta satisfies the definition of continuity.

For any x \in R\Q, if |x - c| < \delta, then |x - c| < \frac{\epsilon}{q}. This means that |p - cq| < \epsilon. Since q is a positive integer, we can always find a positive integer k such that kq > p. Therefore, we have |p - cq| = cq - p < kq - p < \epsilon. This implies that |f(x) - f(c)| = \frac{1}{q} < \frac{1}{kq} < \frac{1}{p} < \epsilon.

This shows that for any given \epsilon > 0, we can always find a \delta > 0 such that for all x \in R\Q, if |x - c| < \delta, then |f(x) - f(c)| < \epsilon. Therefore, f(x) is continuous on R\Q.

Now, let's consider the points in Q. We know that for any rational number c \in Q, there exists a sequence of irrational numbers (x_n) that converges to c. This means that for any \epsilon >