You have been blocked indefinitely from editing in accordance with Wikipedia's blocking policy for repeated abuse of editing privileges. If you believe this block is unjustified you may contest this block by adding the text {{unblock|Your reason here}} below.

Creation of multiple accounts for the purpose of creating nonsense or vandalism is grounds for a block of all accounts. --PMDrive1061 (talk) 16:15, 10 February 2009 (UTC)Reply

{unblock|1="This account is a sock puppet of Somedudealex, and has been blocked indefinitely". This is not true. My friend, the so called "Somedudealex" created a Wikipedia-page about me without my knowledge, and showed me the link afterwards. As the person I am, I read through the page, and I saw a mistake written there. I therefore edited the page, placing one number "1" on the page, unknowing that I would get banned due to my act. I apologize, and hope that you will understand I meant no harm editing the page. I now know it was wrong of me. I am NOT a "sock puppet" of "Somedudealex", we do not even live in the same country, as IPs may show. Please reconsider the IP-ban. Thank you for your time.}

Sounds unusually plausible. I'll contact the blocking admin.  Sandstein  18:00, 10 February 2009 (UTC)Reply

  • I like to assume good faith and I agree that your reasoning is plausible. I've therefore unblocked the account. It may take a bit of time for the system to update itself; don't worry if you can't edit right away. You'll be able to soon. Thanks for letting us know what went down. Multiple new accounts editing a single article generally raises alarm bells, but nothing here is ever permanent. My apologies for the inconvenience and may I be the first to welcome you to the site. --PMDrive1061 (talk) 22:55, 10 February 2009 (UTC)Reply

Your request to be unblocked has been granted for the following reason(s):

See above.

Request handled by:  Sandstein  22:57, 10 February 2009 (UTC)Reply

Blocking administrator: Please check for active autoblocks on this user after accepting the unblock request.