$f(x)=\left\{ x^2+x, x \in \Bbb Q\right\}, f(x)=\left\{ x^3 + 1, x \notin \Bbb Q \right\}$
I want to prove that $f$ is discontinuous at $x \ne 1$.
What I have so far is:
Fix $\delta > 0$. We need to find an $\epsilon > 0$ s.t. $|f(x)-f(p)| \geq \epsilon$, $\forall |x-p| < \delta, p \ne 1$.
Consider $p \in \Bbb Q$, $x \notin \Bbb Q$. Then, $|f(x)-f(p)|=|(x^3+1-p^2-p)|$. We need to find an $\epsilon(\delta, p)$ s.t. $0<|x-p|<\delta \implies |(x^3+1-p^2-p)| \geq \epsilon$.
I think I need to algebraically manipulate the left side of the equation so that I can pull out a $|x-p|$, which would give epsilon as a function of delta and $p$. However, I can't seem to find how to do this.
Let $a$ be a continuity point of $f$. Since rational and irrationls are dense in $\mathbb{R}$, let $(q_n)$ and $(r_n)$ be sequences of rationals and irrationals, respectively, converging to $a$. Then by continuity \begin{equation} \lim_\limits{n\to \infty}f(q_n) = \lim_\limits{n\to \infty}f(r_n), \end{equation} i.e., \begin{equation} a^2+a = a^3 + 1. \end{equation} But you can see that the polynomial $a^3-a^2-a+1$ has only two real roots, namely, $1$ and $-1$, and this shows that $f$ is discontinuous at $x\neq \pm1$.