Is $f: \mathbb Q \to \mathbb R $ given by $f(q) = \sqrt{2}q$ continuous?

This function is a simple linear function with infinitely many points removed from its domain. The question is whether it's continuous in its domain. I am not entirely sure, but my answer is that this function is in fact continuous. Rational numbers are dense in real numbers and there are infinitely many of them between any two numbers.
Let's take an arbitrary number $x_0 \in \mathbb Q$ at which we want to check continuity. Then, for all $\epsilon$ greater than $0$ there exists $\delta >0$ such that for all rational $x$, $|x-x_o|<\delta \Rightarrow |x\sqrt2-x_o\sqrt 2| < \epsilon$ and it clearly implies that $|x-x_o| < \frac{\epsilon}{\sqrt2}$ Which is not a problem, since $\epsilon / \sqrt{2}$ is a limit of a certain sequence of rational numbers. Therefore, the function is continuous.

Is my solution correct? If not, where did it go wrong?


There is a continuous function $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = \sqrt{2} x$. You have restricted its domain to the subspace $\mathbb{Q}$, so the result is continuous.

Your solution is a little confusing; make sure you understand that $\epsilon$ is "given", whereas $\delta$ is supposed to be "found". I'd argue as follows: Fix $x_0$ and $\epsilon > 0$. Then if $|x-x_0| < \delta = \epsilon/\sqrt{2}$, you find $|\sqrt{2} x - \sqrt{2} x_0| = \sqrt{2} |x-x_0| < \epsilon$. By what I've mentioned above, we needn't worry if $x_0$ is rational or not.


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