Given that $f(x) = -1$ if $f$ is irrational and $f(x)=1$ if $f$ is rational, show that $f$ is not continuous anywhere. 
Given that $f(x) = -1$ if $f$ is irrational and $f(x)=1$ if $f$ is rational, show that $f$ is not continuous anywhere.



*

*Let's show that between any 2 rationals there is always an irrational number


Let's consider $a,b \in Q$ such that $a<b$, there is an infinite number of rational $r$ such that $a<r<b$. Where $r=\frac{a+b}{n}$ where $n \in Z$
Also $$a<r<b$$, $$ a +\sqrt{2}<r<b+\sqrt{2}$$ , $$a<r- \sqrt{2}<b$$
It follows that we have always an irrational between any two rationals


*Between any two irrationals, we always have a rational.


Here I am not sure how to proceed.


*One can find a rational in between two irrationals and vice versa an irrational between two rationals.


Therefore as the value of $x$ approaches a value from the left or right of $r$ , $x$ will oscillates between a rational and irrational infinitely. Therefore limits will oscillate between $1$ and $-1$ infinitely. 
it shows that $$f(r^-) \neq f(r^+) \neq f(r)$$
it follows that there no continuity anywhere on $D_f$
Is this correct? Is there a more efficient (clean) way to show the discontinuity?
Any input is much appreciated

Pick a real number $x$.  We will show that $f$ is discontinuous at $x$.  
Side Remark: Discontinuity of $f$ at $x$ can be shown by exhibiting at least once sequence $y_{n}$ that converges to $x$, yet violates the requirement
$$
\lim_{n \rightarrow \infty} f(y_{n}) = f(\; \lim_{n \rightarrow \infty} y_{n} \;).
$$
(Essentially, $f$ is continuous at $x$ if and only if $f$ commutes with the limit operation of convergence to $x$.)
Now, the proof:
Case 1: $x$ is rational.
Then there exists a sequence $y_{n}$ of irrational numbers that converges to $x$.  Thus,
$$
\lim_{n \rightarrow \infty} f(y_{n}) = -1 \neq 1 = f(\; \lim_{n \rightarrow \infty} y_{n} \;).
$$
Case 2: $x$ is irrational.
Then there exists a sequence $y_{n}$ of rational numbers that converges to $x$.  By an argument analogous to that in Case 1,
$$
\lim_{n \rightarrow \infty} f(y_{n}) \neq f(\; \lim_{n \rightarrow \infty} y_{n} \;).
$$
