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I am currently wondering if it is possible to formally derive the cardinality of a set defined as follows:

$$\left\{ k \in \mathbb{N}, k<n: \frac{n}{2}\left(1-\sqrt{1- \frac{(k-0.5),}{n}}\right)\le k < \frac{n}{2}\left(1-\sqrt{1- \frac{(k+0.5) }{n}}\right),n \in \mathbb{N}\right \} $$

Any help or suggestion would be greatly appreciated.

Thank you very much.

Gilles

Gilles Mordant
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    Welcome to MSE. I don't understand how your set is defined. What is that $\forall n\in N$ doing outside the set? And what is $N$? Did you mean $\mathbb N$? – José Carlos Santos Nov 05 '17 at 09:53
  • since it's a subset of the natural numbers its at most countable. Are you asking how to tell tht it's not finite or empty. – fleablood Nov 05 '17 at 15:43
  • Does that $n \in N$ mean the equation for $k$ will be true for all $n$ (In which case $k=0$ is the only possibility for $k$) or for some $n$ (for which case if $n$ is a multiple of 4, then $k$ is a multiple of $3$)? – fleablood Nov 05 '17 at 16:18

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if $ \frac{n}{2}\left(1-\sqrt{1- \frac{(k-0.5),}{n}}\right)\le k < \frac{n}{2}\left(1-\sqrt{1- \frac{(k+0.5) }{n}}\right)$ for all $n \in \mathbb N$ then

$\sqrt{1- \frac{(k+0.5) }{n}} > 1- \frac{2k}n \ge \sqrt{1- \frac{(k-0.5)}{n}}$

$1- \frac{(k+0.5) }{n} > 1 -\frac {4k}n + \frac {4k^2}{n^2} \ge 1- \frac{(k-0.5)}{n}$

$k - \frac 12 \le 4k -\frac {4k^2}n < k-\frac 12$

As $k \in \mathbb N$ that is simply

$k = 4k - \frac {4k^2}n$

Or $4k^2 - 3kn = k(4k-3n) = 0$. So $k=0$ or $k = \frac {3n}4$ for all $n\in \mathbb N$. No single value can equal $\frac {3n}4$ for all values of $n$.

This leads to the eternal question is $0$ natural. If it is then $n= 0$ is acceptable and we can not solve for $n=0$. So if $0$ is a natural number there is no solution for $k$ when $n=0$. If $0$ is not a natural number then $k=0$ although a solution for all natural $n$ is not an acceptable solution.

Hence your set is $\emptyset$.

fleablood
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