$f(x)=0$ for all $x \in\mathbb{Q}$. Prove $f(x)=0$ for all $x > \in\mathbb{R}$.
I have to use the Epsilon-Delta method. I know it starts Let $\epsilon > 0$ be arbitrary. Let y be arbitrary. Now I have to find a $\delta>0$ such that if $x \in\mathbb{R}$ and $|x−y|<\delta$ implies $|f(x)−f(y)|<\epsilon$.
But I can't figure out how to find $\delta$ and I can't figure out how the $\mathbb{Q}$ and $\mathbb{R}$ play into the whole thing because $|f(x)−f(y)|=|0−0|=0<\epsilon$ already. Please help.
You have to prove that $f(x) = 0$ for all $x \in \mathbb{R}$, so don't start there. Let $x \in \mathbb{R}$ be any real number. Then we want to show that $f(x) = 0$, or equivalently that $|f(x)| < \epsilon$ for every $\epsilon > 0$. Do you see a way to do this?
I think you're trying to answer the wrong question.
What you're (presumably) being asked to do is as follows. If $f$ is a continuous function and $f(x)=0$ for all $x \in \mathbb{Q}$ then $f(x)=0$ for all $x \in \mathbb{R}$. (You're not being asked to prove that anything is continuous.)
So suppose $f$ is continuous and $x \in \mathbb{R}$. Fix any $\varepsilon > 0$. You know that there is a $\delta > 0$ such that $|f(x)-f(y)| < \varepsilon$ whenever $|x-y|<\delta$.
By cunning choice of $y$ you can force $|f(x)|<\varepsilon$.
...but then what does this tell you about the value of $f(x)$?
