# Continuity of a rational ruler like function

Consider $$f:\Bbb{R} \rightarrow \Bbb{R}$$ by$$f(x)=\begin{cases} \frac{p+\sqrt{2}}{q+\sqrt{2}}-\frac{p}{q} &\text{if}\;x=\frac{p}{q}\in\Bbb{Q}\;\text{with}\;\text{gcd}(p,q)=1\\ \\0 & \text{otherwise} \; \end{cases}$$

Prove: $$f$$ is continuous at $$(\Bbb{R} \setminus \Bbb{Q}) \cup \{1\}$$

Here's my try:

Let $$x_n \in (\Bbb{R} \setminus \Bbb{Q})$$ so that $$x_n \rightarrow x=p/q \in \Bbb{Q}$$ , so $$f(x_n)=0$$.

Therefore we make $$f$$ continuous at this $$x$$,we have $$f(x)=0$$. But $$f(x)=0$$ only when $$p=q$$ and so $$p/q=1$$. So $$f$$ is continuous at $$1$$

Let $$b$$ be an arbitrary irrational number. Now check the continuity at $$b$$:

Whatever we make $$\vert x-b \vert < \delta$$, $$\vert f(x)-f(b) \vert = \vert f(x) \vert < \epsilon$$

Since $$x \in (b-\delta,b+\delta)$$ is irrational, then $$f(x)=0<\epsilon$$

and $$x \in (b-\delta,b+\delta)$$ is rational except $$1$$, so $$x=p/q$$ and note that $$f(x)=\frac{p+\sqrt{2}}{q+\sqrt{2}}-\frac{p}{q}$$ is irrational,since $$p \neq q$$, we conclude $$f(x)=0< \epsilon$$

Summary: For every $$x \in N_\delta(b)$$, $$\vert f(x)-f(b) \vert < \epsilon$$

Am I right? Any Thoughts?

• I agree with if $x \in (b-\delta,b+\delta)$ is irrational, then $f(x)=0<\epsilon$ ... but if $x \in (b-\delta,b+\delta)$ is rational, so $x=p/q$ and note that $f(x)=\frac{p+\sqrt{2}}{q+\sqrt{2}}-\frac{p}{q}$ is irrational, we conclude $f(x)=0< \epsilon$ how? You will probably have to look at $\left|\frac{p+\sqrt{2}}{q+\sqrt{2}}-b+b-\frac{p}{q}\right|\leq \left|\frac{p+\sqrt{2}}{q+\sqrt{2}}-b\right|+\left|b-\frac{p}{q}\right|$ to conclude that? – rtybase Sep 5 '18 at 10:18
• @rtybase: Oh! sorry! only $\frac{p+\sqrt{2}}{q+\sqrt{2}}-\frac{p}{q}$ is irrational not zero! What I do? – user444830 Sep 5 '18 at 10:23
• Look at the inequality above, in my comment, and see if you can reduce it to $<\varepsilon$ – rtybase Sep 5 '18 at 10:26
• @rtybase: the second inequality $\vert b-\frac{p}{q} \vert < \delta=\frac{\epsilon}{2}$, since $x=p/q$ is in the neighbourhood of $b$. Am I right? – user444830 Sep 5 '18 at 10:30
• @rtybase: No! I mean to prove $f$ is continuous at all irrational points and discontinuous at all rationals except $1$. – user444830 Sep 6 '18 at 9:12

First we try to simplify the function $$f(x)=\begin{cases} \frac{p+\sqrt{2}}{q+\sqrt{2}}-\frac{p}{q} &\text{if}\;x=\frac{p}{q}\in\Bbb{Q}\;\text{with}\;\text{gcd}(p,q)=1\\ \\0 & \text{otherwise} \; \end{cases}=\begin{cases} \dfrac{\sqrt{2}(q-p)}{q(q+\sqrt{2})}=\dfrac{\sqrt 2(1-x)}{q+\sqrt 2} &\text{if}\;x\in\Bbb{Q}\\ \\0 & \text{otherwise} \; \end{cases}$$for some $x_0\in \Bbb Q^c$ it is quite obvious that for $x\in \Bbb Q^c$ and $x\to x_0$ we have $f(x)=0$. If $x\in\Bbb Q$ notice that $x\to x_0$ and $q$ grows unbounded (why?), so the numerator of $\dfrac{\sqrt 2(1-x)}{q+\sqrt 2}$ remains bounded around $\sqrt 2(1-x_0)$ and the denominator goes to $\infty$. This means that $\dfrac{\sqrt 2(1-x)}{q+\sqrt 2}\to 0$ as $x\to x_0$ and therefore the function is continuous over $\Bbb R-\Bbb Q$ so in $x_0=1$ because $x=1$ is a root of $\dfrac{\sqrt 2(1-x)}{q+\sqrt 2}$ so that even if $q\not\to\infty$ we have $1-x\to 0$ and $f(x)\to 0$ which completes our proof.
First, let's rewrite the formula for $f(x)$ when $x=p/q\in \mathbb Q$: $$f(x)=\sqrt{2}\frac{1-x}{q^2(q+\sqrt{2})}.$$ Continuity at the irrationals. Fix any irrational $y\in \mathbb R\setminus \mathbb Q$ and fix $\varepsilon>0$. To establish continuity at $y$, we must find $\delta>0$ such that $|f(x)|<\varepsilon$ for all $x\in (y-\delta,y+\delta)$. Here is how to find such a $\delta$. Fix an integer $N>2|1-y|/\varepsilon$ and let $\delta_1$ be the distance from $y$ to the closest rational number with denominator at most $N$. Let $\delta_2>0$ be sufficiently small such that $$\max(|1-y-\delta_2|,|1-y+\delta_2|)<\sqrt{2}|1-y|.$$ (Such a $\delta_2$ exists since $y\not=1$.) Finally, set $\delta=\min(\delta_1,\delta_2)$. Then for all $x\in (y-\delta,y+\delta)$ either $x$ is irrational (in which case $f(x)=0$) or, by the above formula, $$|f(x)|=\sqrt{2}\frac{|1-x|}{q^2(q+\sqrt{2})}\leq \sqrt{2}\frac{|1-x|}{N^3}\leq \sqrt{2}\frac{|1-x|}{N}<\epsilon\frac{|1-x|}{\sqrt{2}|1-y|}<\varepsilon.$$ This establishes continuity at the irrationals.
Continuity at $x=1$. Note that $f(1)=0$ and $|f(x)|\leq \sqrt{2}|1-x|$. Continuity follows by the squeeze theorem.
Discontinuity at all remaining points of $\mathbb R$. It remains to show that $f(x)$ is discontinuous at every $x\in\mathbb Q\setminus \{1\}$. Indeed, note that at any such $x$ we have that $f(x)\not=0$. On the other hand, let $x_1,x_2,\ldots$ be a sequence of irrational numbers converging to $x$. Then $f(x_n)=0$ for all $n$, and in particular $\lim_{n\to\infty}f(x_n)=0$ which does not equal $f(x)$. Hence $f$ is discontinuous at $x$.
• sir...your proof is ok but $\epsilon-\delta$ argument for this proof is hard to understand. Is there any other equivalent way to show this like the inequality in the rtybase first comment – user444830 Sep 8 '18 at 8:11