I have a function defined as follows: Let $t=\frac{p}{q}$ be fully reduced for $t\in \mathbb Q$

$$ f(t)=\left\{ \begin{array}{lll} 1/q & \text{if} & t\in\mathbb Q\\ 0 & \text{if} & t\in\mathbb R\setminus\mathbb Q \\ 1 & \text{if} & t = 0 \end{array} \right. $$

I have proved that the given function is not continuous for $t \in \mathbb Q$ but I'm having trouble proving that for $t \in \mathbb R \setminus \mathbb Q$ the function is continuous.

Can someone point me in the right direction? I'm thinking about using a decimal approximation for $t \in \mathbb R \setminus \mathbb Q$ but I don't know where to go.

  • $\begingroup$ $x$ is $t$, right? $\endgroup$ – user67133 Nov 5 '13 at 0:00
  • $\begingroup$ Yeah, oops! Fixed that. $\endgroup$ – druckermanly Nov 5 '13 at 0:03

Show that the denominators of rationals approximating a given irrational must become larger and larger the better the approximation is.

A bit more formally, given $\epsilon>0$, an irrational number $x$, and any rational number $p/q$ such that $|p/q-x|<\epsilon$, then $q\rightarrow\infty$ as $\epsilon\rightarrow 0$.

I suggest a proof by contradiction: if the denominators are all bounded, could you actually get an arbitrarily good approximation? Also be sure to cover the possibility of "maybe some denominators get really large, but others don't." This doesn't happen, but why?

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  • $\begingroup$ I tried to do so, but I'm stuck trying to prove that the denominator of a rational approximation of an irrational number is monotone increasing. I know that I can say it will tend towards $\infty$, but the monotone part is where I'm stuck. $\endgroup$ – druckermanly Nov 5 '13 at 0:59
  • $\begingroup$ @user2899162 It doesn't really need to be monotone, as long as the smallest denominator in the $\epsilon$ ball is still diverging to $\infty$. The basic idea is that if $M$ is the smallest denominator in a given ball, then all rationals with denominator at most $M$ are a minimum non-zero distance away from the irrational point. So you move into the corresponding small ball and get denominators all at least $M+1$. Rinse and repeat. In this sense you get a monotonically increasing sequence of "smallest denominators", I guess. $\endgroup$ – zibadawa timmy Nov 6 '13 at 0:31

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