1
$\begingroup$

Show that if $f:\mathbb{R}\rightarrow \mathbb{R}$, such that

$f(p-q\sqrt{3})=p-q\sqrt{2},\qquad \forall\; p,q\in \mathbb{N}$,

then $f$ is not continuous on $\mathbb{R}$.

  • 1
    the statement is incomplete: not all real numbers are of the form $p-q\sqrt{3}$, how is $f$ defined on others? – 2017-02-27
  • 0
    I made a slight edit, to help clarify the problem, please let me know if this was not the intended question. – 2017-02-27
  • 1
    Hint:(1) f(0) must be what? (2) Can you find a sequence $x_1,x_2,x_3,...$ with $x_n=p_n-q_n\sqrt{3}$ and $q_n$ approaching $\infty$ as $n$ approaches $\infty$, such that $x_n$ approaches $0$ as $n$ approaches $\infty$? (3) For that sequence, what must be true of the sequence $f(x_1),f(x_2),f(x_3),...$? – 2017-02-27
  • 0
    I've tried this, but I can't reach to the final result. Can you be more clear, please? – 2017-02-27
  • 0
    Let's see you deal with my first two hints. I suggest that you edit your question to show what you've tried. – 2017-02-27

1 Answers 1

2

Hint: You can choose $p$ and $q$ such that $p$ is arbitrarily large and $p-\sqrt{3}q$ is less than some $\varepsilon>0$. Since limits commute with continuous functions, what can we say about $f(0)$?

Let $p_n$ be a sequence of natural numbers diverging to infinity which are the integer part of multiples of $\sqrt{3}$. Then for each $n$, there exists some $q_n>\frac{p_n-\frac{1}{n}}{\sqrt{3}}$. We may pick the smallest such $q_n$, so that $p_n-q_n\sqrt{3}>0$, which exists for $p_n$ close to a multiple of $\sqrt{3}$, and hence $\lim p_n - q_n\sqrt{3} =0$.

Now, $q_n < \frac{\frac{1}{n}+p_n}{\sqrt{3}}$, so that $p_n-\sqrt{2}q_n> \left(1-\sqrt{\frac{2}{3}}\right)p_n -\frac{\sqrt{2}}{\sqrt{3}n}$ which clearly diverges.

  • 0
    Could you, please, alaborate the last implication? You choose some $q_n$ greater than $l=\frac{p_n-\frac 1n}{\sqrt 3}$, which itself is smaller than $g=\frac{p_n+\frac 1n}{\sqrt 3}$ – but how $q_n>l$n=10^6,\ p_n=\lfloor n\sqrt 3\rfloor = 1732050,\ \frac{p_n-\frac 1n}{\sqrt 3}\approx 999999.533749$ implies $q_n=1000000$ which is _not_ less than $\frac{p_n+\frac 1n}{\sqrt 3}\approx 999999.533750$. Can you show $q_n<\frac{p_n+\frac 1n}{\sqrt 3}$ holds e.g. for $n$ big enough? – 2017-02-27
  • 0
    One approach is to use continued fractions to solve Pell equation $x^{2}-3y^{2}=1$ and the values of $x, y$ will act as $p_n, q_n$ such that they both tend to $\infty$ and $p_n-q_n\sqrt{3}\to 0$. And then clearly the sequence $p_n-q_n\sqrt{2}=(p_n-q_n\sqrt{3})+q_n(\sqrt{3}-\sqrt{2})$ diverges. – 2017-02-27
  • 0
    Thus we can set $$p_1=2,q_1=1,p_{n+1}=2p_n+3q_n,q_{n+1}=p_n+2q_n$$ and our job is done. – 2017-02-27