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You asked in chat which diagonalizer is more powerful: I2 or I(2), and this is certainly I(2) because is equal to . Ikosarakt1 (talk ^ contribs) 08:59, April 11, 2014 (UTC)

Ah, thanks. I was still a little bit confused back then. :) King2218 (talk) 15:05, April 11, 2014 (UTC)


But I'm not sure about what I(0) and I(1) means traditionally. The matter is that , but defining I(0) either as and "I" may be correct. Ikosarakt1 (talk ^ contribs) 12:19, April 12, 2014 (UTC)
From what I know, \(I(0)=\Omega,I(1)=I\), and \(I(2)\) is first 1-weakly inaccessible cardinal. LittlePeng9 (talk) 13:07, April 12, 2014 (UTC)

Re: Blocked

that's how we deal with punsters around here

it's your pun-ishment you're.so.pretty! 07:59, May 12, 2014 (UTC)

Also, don't click this linkI want more clouds! ⛅ 08:19, May 12, 2014 (UTC)

Dependence

This talk page describes your user page. The dependencies between the talk page and the user page may be modeled by the directed graph described by the adjacency matrix [[1 0] [1 1]]. you're.so.pretty! 02:57, June 16, 2014 (UTC)

If A->B means that B describes A (because you said 'dependencies'), and the first and the second node represent my user page and my talk page respectively, then the graph should be [[1 1][0 0]]. (unless my talk page secretly describes itself) —Preceding unsigned comment added by King2218 (talkcontribs)
No, A->B means that A depends on B. I can verify that the adjacency matrix is correct:
  • The userpage describes itself, so the node representing the userpage contains a loop.
  • The talkpage describes itself, because it contains the sentence "This talk page describes your user page," so the node representing the talkpage contains a loop.
  • The userpage does not describe the talkpage.
  • The talkpage describes the userpage, because the userpage's self-description is acknowledged in the adjacency matrix.
In fact, the first sentence in my comment wasn't necessary for the matrix to work! you're.so.pretty! 04:58, June 16, 2014 (UTC)
Fine, you got me. :P King2218 (talk) 05:08, June 16, 2014 (UTC)

another 3 day coincidence thing

I made my first edit three days after you did WikiRigbyDude (talk) 01:04, June 21, 2014 (UTC)

And you joined the wiki 3 days after I did :P King2218 (talk) 01:30, June 21, 2014 (UTC)

Problem?

Problem described on your user page is incorrect, because we have \(a^n+b^n=c^n\) for \(a=b=3,c=4,n=\log_{4/3}2>\log_{4/3}16/9=2\). You forgot to state \(n\) must be an integer. LittlePeng9 (talk) 14:30, July 13, 2014 (UTC)

lol fixed King2218 (talk) 14:34, July 13, 2014 (UTC)

New formulation is incorrect - it should state "some integer", not "all integers" as it states now. Right now this is implied by easy to show statement that if equality holds, then \(c\geq n\). LittlePeng9 (talk) 15:04, July 25, 2014 (UTC)

You are a

NERDDDDDDDDD LittlePeng9 (talk) 16:19, August 8, 2014 (UTC)

I don't think nerds should say against each other that they are nerds. :P Wythagoras (talk) 16:24, August 8, 2014 (UTC)
BURNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN lol King2218 (talk) 16:56, August 8, 2014 (UTC)
/me burns LittlePeng9 (talk) 17:01, August 8, 2014 (UTC)
haha i hacked it i'm such a hardcore hacker you're.so.pretty! 19:38, August 8, 2014 (UTC)


easy :P Wythagoras (talk) 19:40, August 8, 2014 (UTC)
h4xX0rz King2218 (talk) 06:06, August 9, 2014 (UTC)


lol i hacked your box like a true hardcore hackmaster you're.so.pretty! 01:50, September 11, 2014 (UTC)
omg stop King2218 (talk) 12:34, September 14, 2014 (UTC)

peano arithmetic

I showed Peng part of an IRC log where I talked about Peano arithmetic stuff. Here are his corrections. it's vel time 18:48, October 8, 2014 (UTC)

Happy birthday

A belated happy birthday to you! Wythagoras (talk) 16:01, March 31, 2015 (UTC)

👍 LittlePeng9 likes this.
thanks guys! King2218 (talk) 16:18, March 31, 2015 (UTC)

Hi

Are you asian like me? Because i am Hypertetrakulus44 (talk) 02:04, 15 May 2021 (UTC)

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