# regarding the sets ${A}_{n}$ from the traveling hump sequence

In this entry, $\lfloor \cdot \rfloor $ denotes the floor function.

Following is a proof that, for every positive integer $n$, $[{\displaystyle \frac{n-{2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}}{{2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}}},{\displaystyle \frac{n-{2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}+1}{{2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}}}]\subseteq [0,1]$.

###### Proof.

Note that this is equivalent^{} (http://planetmath.org/Equivalent) to showing that, for every positive integer $n$,

$n-{2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}\ge 0$ and $n-{2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}+1\le {2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}$. This in turn is equivalent to showing that, for every positive integer $n$, ${2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}\le n$ and $n+1\le {2}^{\lfloor {\mathrm{log}}_{2}n\rfloor +1}$.

The first inequality^{} is easy to prove: For every positive integer $n$, ${2}^{\lfloor {\mathrm{log}}_{2}n\rfloor}\le {2}^{{\mathrm{log}}_{2}n}=n$.

Now for the second inequality. Let $n$ be a positive integer. Let $k$ be the unique positive integer such that

${2}^{k-1}\le n\le {2}^{k}-1$. Then $n+1\le {2}^{k}={2}^{k-1+1}={2}^{\lfloor k-1\rfloor +1}={2}^{\lfloor {\mathrm{log}}_{2}{2}^{k-1}\rfloor +1}\le {2}^{\lfloor {\mathrm{log}}_{2}n\rfloor +1}$. ∎

Title | regarding the sets ${A}_{n}$ from the traveling hump sequence |
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Canonical name | RegardingTheSetsAnFromTheTravelingHumpSequence |

Date of creation | 2013-03-22 16:14:28 |

Last modified on | 2013-03-22 16:14:28 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 6 |

Author | Wkbj79 (1863) |

Entry type | Proof |

Classification | msc 28A20 |