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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)