# User talk:Xario

Welcome!

Hello, Xario, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

I hope you enjoy editing here and being a Wikipedian! Please sign your messages on discussion pages using four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question on this page and then place {{helpme}} before the question. Again, welcome! Katr67 (talk) 18:20, 24 September 2008 (UTC)

## ${\displaystyle g_{n}<{\sqrt {p_{n+1}}}+{\sqrt {p_{n}}}.}$ Reverted.

Here is a correction I did and why.

${\displaystyle g_{n}<{\sqrt {p_{n+1}}}+{\sqrt {p_{n}}}.}$ Reverted.

If ${\displaystyle g_{n}=p_{n+1}-p_{n}}$ denotes the nth prime gap, then Andrica's conjecture can also be rewritten as ${\displaystyle g_{n}<2{\sqrt {p_{n}}}+1}$.

Proof:

${\displaystyle {\sqrt {p_{n+1}}}-{\sqrt {p_{n}}}<1}$
${\displaystyle {\sqrt {p_{n+1}}}<1+{\sqrt {p_{n}}}}$
squared is: ${\displaystyle ({\sqrt {p_{n+1}}})^{2}<(1+{\sqrt {p_{n}}})^{2}}$
${\displaystyle p_{n+1}<1^{2}+2*{\sqrt {p_{n}}}+{\sqrt {p_{n}}}^{2}}$
${\displaystyle p_{n+1}<1+2*{\sqrt {p_{n}}}+p_{n}}$
${\displaystyle g_{n}=p_{n+1}-p_{n}}$, so
${\displaystyle g_{n}<2{\sqrt {p_{n}}}+1}$
Done. If you have a better way to write it to make it clear, let me know. John W. Nicholson (talk) 23:42, 5 September 2012 (UTC)

## von Hagens video

I replied on the article talkpage but here is the Daily Planet 2007 plastination video on youtube. Staircase to secret lab is at 5:17. Δρ.Κ. λόγοςπράξις 22:34, 17 August 2015 (UTC)

## ArbCom elections are now open!

Hi,
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