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For every $n\in\mathbb Z^+ = \{1, 2,\dots\}$ set of positive integers, let $r_n$ be the minimum value of $|c−d\sqrt3|$ for all nonnegative integers $c$ and $d$ with $c+d=n$.

Find, with proof, the smallest positive real number $g$ with $rn\leq g$ for all $n\in\mathbb Z^+$.

What I understood is that we need to find the maximum between the minima $r_n$, that is, the greatest $r_n$. So I reduced $|c−d\sqrt3|$ to $|n−d(1+\sqrt3)|$.

Now note that $1 \leq n-d$. If we suppose $n−d(1+\sqrt3)>0$, then $$|n−d(1+\sqrt3)| = n−d(1+\sqrt3),$$ so that $$2n − d(2+\sqrt3) > n-d > 1 \geq n−d(1+\sqrt3) \geq 1 + n + d \geq\dots$$

I stopped here because I saw that it will lead us to nothing basically, so I would appreciate any help.

Ottavio
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Gabriela Da Silva
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1 Answers1

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Writing $|c- d \sqrt 3| = |n - d(1+\sqrt 3)|$ is a nice observation. We can go a bit further:

$$|n-d (1+\sqrt 3)| = (1+\sqrt 3) \left|\frac n{1+\sqrt 3} - d\right|$$

Since $\dfrac n{1+\sqrt 3}$ is just some real number and $d \in \mathbb Z^+$, the value inside the absolute value is minimized when $d$ is the closest integer to $\dfrac n{1+\sqrt 3}$.

This shows that, for the optimal $d$, $\left|\dfrac n{1+\sqrt 3} - d\right| \le \dfrac12$ (Side note: this inequality can be strict as the fraction is always irrational.) Hence: $$r_n \le \dfrac {1+\sqrt 3}2$$

To show that this is the smallest real number that can bound this absolute value, we need to show that $\dfrac n{1+\sqrt 3}$ can be arbitrarily close to some $d \pm \dfrac12$. This will require a density argument.

player3236
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