A question about analysis...

In this pdf document:

http://math.berkeley.edu/~scanez/courses/math104/fall11/homework/hw7-solns.pdf

For the proof of quesiton 2, the text says "First, let r in Q. Since the irrationals are dense in R, there exists a sequence $(y_n)$ of irrational numbers converging to r. Since the sequence $f(y_n) = 0$ does not converge to $f(r) \not=0$, f is not
continuous at r."

I'm a bit confused about this. Becuase the irrationals are dense, we know that there infinetely many irrationals around r...but how does that there there is a "sequence" that "converges" to r?

Also how do we know that $f(y_n) \not=0$?

Thank you indvance.

Well, if you take say $\frac{1}{2}$. You know that for each n there are infinitely many rationals inside $(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}+\frac{1}{n})$. If you pick one of these rationals for each n, then the sequence of rationals that you pick will converge to 1/2. Actually, you can do the same for any rational number, not just 1/2 and that gives you that any rational number has a sequence of irrationals that tend towards it.


Where does it say that? If you mean $f(r) \neq 0$ then this is because r is rational and $\frac{1}{q} \neq 0$ for any $q \in \mathbb{Z}$