User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

Theorem
Let $m \in \N_{\ge 0}$.

Then $\left \{ {m, m + 1, \ldots, m + n - 1 }\right\}$ contains a number that is a multiple of $n$.

That is, in any list of $n$ successive numbers, $n$ divides one of those numbers.

Proof
Let $S_m = \left\{ { m, m + 1, \ldots, m + n -1}\right\}$ be a set containing $n$ successive numbers.

The proof proceeds by induction on $m$, the smallest number in $S_m$.

Basis for the Induction
The case $m = 1$ is verified as follows:


 * $S_1 = \left\{ { 1, 2, \ldots,n}\right\}$

This contains a multiple of $n$, namely, $n$ itself.

This is the basis for the induction.

Induction Hypothesis
Fix $m \in \N$ with $m \ge 1$.

Assume:

$S_m = \left\{ { m, m + 1, \ldots, m + n -1}\right\}$

contains a multiple of $n$.

This is our induction hypothesis.

Induction Step
This is our induction step:

Consider the set:


 * $S_{m+1} = \left\{ { m + 1, m + 2, \ldots, m + n - 1, ,m + n}\right\}$

As $n \equiv 0 \bmod n$, $m+n$ is a multiple of $n$ iff $m$ is a multiple of $n$.

That means that we can replace $m + n$ with $m$ and instead analyze:


 * ${S_{m+1}}' = \left\{ { m + 1, m + 2, \ldots, m + n - 1 ,m}\right\}$

But ${S_{m+1}}' = S_m$, which contains a multiple of $n$ by the induction hypothesis.

The result follows by the Principle of Mathematical Induction.