# 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,
You appear to be eligible to vote in the current Arbitration Committee election. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to enact binding solutions for disputes between editors, primarily related to serious behavioural issues that the community has been unable to resolve. This includes the ability to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate, you are welcome to review the candidates' statements and submit your choices on the voting page. For the Election committee, MediaWiki message delivery (talk) 13:49, 24 November 2015 (UTC)

## ArbCom Elections 2016: Voting now open!

 Hello, Xario. Voting in the 2016 Arbitration Committee elections is open from Monday, 00:00, 21 November through Sunday, 23:59, 4 December to all unblocked users who have registered an account before Wednesday, 00:00, 28 October 2016 and have made at least 150 mainspace edits before Sunday, 00:00, 1 November 2016. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate in the 2016 election, please review the candidates' statements and submit your choices on the voting page. MediaWiki message delivery (talk) 22:08, 21 November 2016 (UTC)