1. ## continous proof problem

Hi Everyone,

I'm new to the forums, but this place sounds awesome. I have a group assignment due tomorrow and we've gotten all of the problems except this last one. Let me know if anyone has any input on it -- it would be greatly appreciated

f(x) = 0 if x is rational
x if x is irrational

With epsilon = .1, find a delta that makes plausible that f is continuous at 0.

Is this even possible? Because as you get closer and closer to x (0) there will always be at least one rational and at least one irrational number in the interval? Thanks so much for your help!!!

Jason

2. If $\displaystyle \varepsilon = 0.1$ then choose $\displaystyle \delta = 0.1$ thus we have $\displaystyle \left| {x - 0} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 0} \right| < \varepsilon$.
This is because $\displaystyle f(x) = \left\{ {\begin{array}{lr} 0 & {x \in Q} \\ x & {x \in \Re \backslash Q} \\ \end{array}} \right.$ and in either case $\displaystyle \left| {f(x)} \right| < \varepsilon$

3. Thanks so much!!!! I understand it now too