User talk:Vel!/Archive 4

Ei! wie schmeckt der Coffee süße,

Lieblicher als tausend Küsse,

Milder als Muskatenwein.

Coffee, Coffee muss ich haben,

Und wenn jemand mich will laben,

Ach, so schenkt mir Coffee ein! Ah! how sweet coffee tastes!

Lovelier than a thousand kisses,

smoother than muscatel wine.

Coffee, I must have coffee,

and if anyone wants to give me a treat,

ah!, just give me some coffee!

(J.S. Bach)

Saibian's new article
Just for your information, the first article of section 17271289737475698637398725897298725 of Saibian's One to Infinity has been released! We have waiting it for many many years! It introduces functions and numbers way larger than a mere transmorgrifihgh or even Sam's Number (edit: misread the article). You should definitely go read it! ☁ I want more clouds! ⛅ 13:45, April 22, 23316241145592193160488279961352014 (UTC)
 * haha, sbiis you ol' rascal! you're.so.pretty! 16:28, April 22, 2014 (UTC)
 * Oh, and today he published the second article. -- ☁ I want more clouds! ⛅ 02:12, April 30, 2014 (UTC)

Do you recognize the song?

 * idgi you're.so.pretty! 02:08, April 25, 2014 (UTC)

Blocked
o noez i got blocked for 1 min nooooo King2218 (talk) 07:21, May 12, 2014 (UTC)

oh noes same here nooooooo LittlePeng9 (talk) 21:48, September 19, 2014 (UTC)


 * im a tyrant abusing my power. some real corrupt authoritarian shit. it's vel time 22:00, September 19, 2014 (UTC)


 * Come on, why can't you just admit that \(Q + \exists x (Sx = x)\) is consistent? -- ☁ I want more clouds! ⛅ 07:17, September 20, 2014 (UTC)


 * So is \(Q - \exists x (Sx = x)\) it's vel time 07:23, September 20, 2014 (UTC)


 * How does one subtract logic theories? LittlePeng9 (talk) 09:13, September 20, 2014 (UTC)
 * By adding their negation it's vel time 09:36, September 20, 2014 (UTC)


 * I know what you want to achieve, but intuitively, subtraction should not result in a larger theory. LittlePeng9 (talk) 10:08, September 20, 2014 (UTC)
 * You know what else isn't intuitive? \(Sx = x\). it's vel time 17:01, September 20, 2014 (UTC)


 * \(Q + \neg(\exists x (Sx = x))\) Fixed that for you. -- ☁ I want more clouds! ⛅ 12:30, September 20, 2014 (UTC)

Retired policies
Why did you delete those policies? Some of these policies seem to be appropriate on the present day. We should either mark it as "retired" or improve it to fit with our current community. -- ☁ I want more clouds! ⛅ 08:09, May 27, 2014 (UTC)
 * Most of them have been incorporated into the main policy page, so there's no real point in editing them. Feel free to restore them, but they'll probably be useless for the foreseeable future. you're.so.pretty! 16:29, May 27, 2014 (UTC)

WHAT IN THE NAME OF ANDRE JOYCE IS YOUR AVATAR
A google search was inconclusive, so what is it? Looks like some sort of muppet.....thing. WikiRigbyDude (talk) 02:51, June 17, 2014 (UTC)
 * It's Johnny T, a character from the YouTube puppet show Glove and Boots. He is a toad from Brooklyn. I might as well link you to their best work: VVS, Evolution of the Hipster, Johnny T's NYC Tourist Tips. you're.so.pretty! 03:43, June 17, 2014 (UTC)

Question
Any particular reason why you chose green as the wiki's theme color? WikiRigbyDude (talk) 00:22, July 14, 2014 (UTC)
 * does there need to be? you're.so.pretty! 05:25, July 14, 2014 (UTC)
 * There might be. LittlePeng9 (talk) 08:05, July 14, 2014 (UTC)
 * the best reason that i can give is that i wanted to replace the grey/black theme i had before. you're.so.pretty! 15:22, July 14, 2014 (UTC)

XappolBot
I've noticed that XappolBot has left the IRC channel right after you did. I don't know if you are aware of that, but I thought it's worth telling you. LittlePeng9 (talk) 17:26, July 27, 2014 (UTC)
 * I know. Since it's running off my laptop, it disconnects when I turn it off. This is a temporary problem -- within the next week I'll be getting my own webserver, which I can use to run it 24/7. you're.so.pretty! 17:59, July 27, 2014 (UTC)

What-If
Did you read the new What-If yet? You'll like the drawing at the very end! -- ☁ I want more clouds! ⛅ 01:32, August 14, 2014 (UTC) (because I should not use the word 'check' or 'cheque')
 * I did see that. I don't recognize the S-notation though you're.so.pretty! 03:48, August 14, 2014 (UTC)


 * It appears to be \(S_{BB}(1000)\), which means the maximum number of steps that a 1000-state Turing machine can take before halting. -- ☁ I want more clouds! ⛅ 13:20, August 14, 2014 (UTC)
 * Or it is S-recursion around BB function. Wythagoras (talk) 14:09, August 14, 2014 (UTC)
 * There is a possibility that this is \(S_{88}(1000)\), so we would have 88th order BB function! LittlePeng9 (talk) 14:16, August 14, 2014 (UTC)

pls add a new fatc 2 teh irc chat0
ok fb100z i hav a rely kewl math fakt u shud add to zapple boot. it is: grahms numbr, eql 2 3^^^^3, is teh larjst numbr evr. most ppl haevnt herd of grams number, so u shud add that fakt so ppl can heer of TEH REEL BIGEST NUMBRE EVAR!!!!!!!!!!!!!!!!!!!!!!!!!!!! WikiRigbyDude (talk) 18:41, August 31, 2014 (UTC)


 * lol LittlePeng9 (talk) 18:45, August 31, 2014 (UTC)
 * Did you mean: greums number King2218 (talk) 19:03, August 31, 2014 (UTC)

Wait, what? Seriously?
Have you renamed your account or have you somehow hacked something to change it for a few days? I just can't believe that you've renamed your account... Wythagoras (talk) 05:38, September 14, 2014 (UTC)


 * No, wait, I see this is seriously. Also, the leaderboard has crashed. But, why... Wythagoras (talk) 05:41, September 14, 2014 (UTC)
 * wyth, you know me. i'm not the kinda guy who jokes around. (also i thought the old name was really dorky) you're.so.pretty! 05:43, September 14, 2014 (UTC)

vel
vel


 * The question was finally answered. -- ☁ I want more clouds! ⛅ 14:47, October 16, 2014 (UTC)

vel
vel

I made a thing
Try this out: Verbatim test &mdash; ☁ I want more clouds! ⛅ 10:03, September 17, 2014 (UTC)
 * daaaaaamn it's vel time 10:14, September 17, 2014 (UTC)
 * 28 &mdash; ☁ I want more clouds! ⛅ 12:23, September 17, 2014 (UTC)
 * Or, how Cookiefonster would say it, 26. Wythagoras (talk) 15:31, September 17, 2014 (UTC)

question
did you make "A googol is a tiny dot"? Cookiefonster (talk) 15:17, September 17, 2014 (UTC)


 * yes, he did. Wythagoras (talk) 15:32, September 17, 2014 (UTC)
 * O LittlePeng9 (talk) 15:44, September 17, 2014 (UTC)
 * NO I DIDNT AAAAAAAA FUCK M EFUCK ME it's vel time 19:34, September 17, 2014 (UTC)
 * ADMIT IT (see http://googology.wikia.com/wiki/Talk:Hyper-leviathan_number) Wythagoras (talk) 10:09, September 20, 2014 (UTC)

Profile pic
You have an awesome profile picture! XD I just like it. Oh, this wikia is by far one of the coolest, and this cane definitely help me out in pre-algebra.
 * thanks. you should probably know that in general the content this wiki is seriously advanced as far as math goes. i doubt it'll be very helpful for precalc students, but who knows. it's vel time 02:16, September 28, 2014 (UTC)

Fresh Prince of Vel-Air
Now, this is a story all about how my life got flipped-turned upside down

And I'd like to take a minute

Just sit right there

I'll tell you how I became the prince of a town called Vel Air

In west Velladelphia born and raised

On the velground was where I spent most of my days

Vellin' out maxin' relaxin' all cool

And all shootin some vel-ball outside of the school

When a couple of vels who were up to no good

Started velling trouble in my neighborhood

I got in one little vell and my mom got scared

She said 'You're movin' with your auntie and uncle in Vel Air'

I velled and pleaded with her day after day

But she packed my vel case and sent me on my way

She gave me a vel and then she gave me my ticket.

I put my Velman on and said, 'I might as well kick it'.

Vel class, yo this is bad

Drinking vel juice out of a champagne glass.

Is this what the people of Vel-Air living like?

Hmmmmm this might be velright.

But wait I hear they're velly, bourgeois, all that

Is this the type of place that they just send this vel cat?

I don't think so

I'll see when I get there

I hope they're prepared for the prince of Vel-Air

Well, the vel landed and when I came out

There was a dude who looked like a vel standing there with my name out

I ain't trying to get arrested yet

I just got here

I sprang with the velness like lighting, disappeared

I whistled for a vel and when it came near

The license plate said vel and it had vel in the mirror

If anything I could say that this vel was rare

But I thought 'Nah, forget it' - 'Yo, homes to Vel Air'

I velled up to the house about 7 or 8

And I yelled to the cabbie 'Yo homes vell ya later'

I looked at my kingdom

I was finally there

To sit on my throne as the Prince of Vel Air

Cookiefonster (talk) 14:20, September 30, 2014 (UTC)

updog function
make an updog function page vel, you have used it in many sites and it's still a little ill-defined

-- A Large Number Googologist -- 22:23, October 2, 2014 (UTC)


 * it's a fucking joke Cookiefonster (talk) 01:40, October 3, 2014 (UTC)

oh.

i should stop taking things too seriously

-- A Large Number Googologist -- 01:41, October 3, 2014 (UTC)


 * not much dog what's up with you it's vel time 02:02, October 3, 2014 (UTC)

megist(r)on
can you move Megistron to the article Megiston and make megistron redirect to megiston? Cookiefonster (talk) 12:54, October 8, 2014 (UTC)

never mind Cookiefonster (talk) 13:01, October 8, 2014 (UTC)

Clarification about first order arithmetic stuff
Suppose we are only given the access to S function. This is very restricted language. We cannot even define < relation. One attempt to do this is the formula x<y which says "either y=S(x) or we have S(x)<y". Even if this was a valid definition of less-than relation, we cannot write it in our restricted language. Valid definition in formal language needs to consist out of only the symbols we are given, which means no direct recursion. So if we tried to unwind our definition, we would get a formula "either y=S(x) or y=S(S(x)) or y=S(S(S(x))) or..." which is an infinite formula, which is forbidden. It's actually not trivial at all from that example that < is undefinable, I just showed that this one idea doesn't work.

Very similar thing is with trying to define multiplication with only addition. Again, it's not clear from any examples that its impossible, but that's true.

However, if we have addition and multiplication, then this is already quite powerful set of tools. As an example, I'll give you an idea on how to define exponentiation, \(a^b\) for example. The idea is to code a list of integers as a single integer, and then use a statement that first element on the list is a, each is a times greater than the previous one, and that there are b terms on the list. Coding is the most important thing, and we do it, in this example, as follows: "There exist numbers \(N\), \(M\) and m such that \(N\equiv a \text{ mod }M\), and, for all \(0<i<b\), if \(N\equiv x \text{ mod }M+im\) and \(N\equiv y \text{ mod }M+(i+1)m\), then \(y=ax\). Then we get that \(N\equiv a^b \text{ mod }M+bm\)". It's in no way obvious that such \(N,M,m\) have to exist, but it follows from Chinese remainder theorem. Note that modulo operation is easily definable with our existing tools.

This example shows that addition and multiplication are quite expressive. We can use similar trick to define factorial, tetration, pentation, and actually all primitive recursive operations. Because we are allowed to use existential quantifiers, we can even define Turing machine halting problem and what not.

Now, the question is: how much of it can PA prove to be well-defined? After all, this definition has its power, but it's useless if we cannot prove it useful. Thankfully, PA can prove all of the primitive recursive functions total, so, for example, we can show that tetration is well defined. This is not contradictory with proof-theoretic ordinal being $\varepsilon_0=\omega\uparrow\uparrow\omega$, because we can only show that the definition is valid for finite numbers. LittlePeng9 (talk) 17:10, October 8, 2014 (UTC)


 * A recursive function f can be proven by PA to be well-defined if and only if it can be computed by a Turing Machine whose haliting time can be bounded above by \(f_\alpha\) for some \(\alpha < \varepsilon_0\). Deedlit11 (talk) 19:08, October 8, 2014 (UTC)

Summary: The inclusion of + and × in first-order arithmetic is not arbitrary, since jointly they give us access to encoding systems within natural numbers. With + and ×, we can already define all the higher hyper-operators and a lot more. it's vel time 19:34, October 8, 2014 (UTC)

Clarification on PTO stuff
When we are talking about ordinals in weak theories of PA sort, which don't have notion of infinite number defined, it's hard to talk about "existence" of ordinal numbers. We have to go a way around - we have to define an ordinal \(\alpha\) by a formula \(\phi(x,y)\), which, in some interpretation, means that $x\prec y$ in ordering of order type \(\alpha\). Now PA cannot prove that such formula exists - PA cannot quantify over predicates. However, we can talk about PA proving that such ordering is a well-order - this is actually a theorem schema, which means that for every formula \(\varphi(x)\) we can show that \((\varphi(0)\land(\forall x:(\forall y: \phi(y,x)\Rightarrow\varphi(y))\Rightarrow\varphi(x))\Rightarrow\forall x:\varphi(x)\). Here, if we assume 0 is mapped to 0 in this ordinal ordering, then this formula says that \(\varphi(0)\) holds and that if \(\varphi(y)\) is true for all ordinals \(y<x\), then it holds for \(x\) too. If \(\phi\) is an ordering of order type \(\epsilon_0\), then for some formula \(\varphi\) PA cannot prove this formula. This means that PA cannot show transfinite induction principle for \(\epsilon_0\). One can say that PA cannot show that \(\epsilon_0\) is an ordinal. However, it can prove so for some ordering for every smaller ordinal, which exactly means that PA has PTO equal to \(\epsilon_0\).

Also, King, \(\exists y\forall x: x<y\) is contradictory - we can show that \(\forall y:S(y)\not < y\), thus \(\forall y\exists x: x\not < y\), which is negation of the first statement. LittlePeng9 (talk) 18:43, October 8, 2014 (UTC)


 * Actually, I believe there are pathological formulas for well-orderings of order type \(\omega\) or \(\omega+1\) which PA cannot prove to be well-ordered. \(\varepsilon_0\) is the smallest ordinal for which there is no formula of that order type which PA proves to be well-ordered. Deedlit11 (talk) 19:13, October 8, 2014 (UTC)
 * Yeah, that's right. I'll modify the above accordingly. LittlePeng9 (talk) 19:21, October 8, 2014 (UTC)

Summary: In arithmetic theories, it doesn't make sense to talk about pure "ordinals" (which are a set theory thing), so instead we create well-orderings. The proof-theoretic ordinal of PA, therefore, is the upper bound of the order types of all the relations that are provably well-ordered in PA. For example, \(<\) is provably a well-ordering of order type \(\omega\), so the PTO of PA is greater than \(\omega\) (and indeed \(\varepsilon_0 > \omega\)). it's vel time 19:22, October 8, 2014 (UTC)

Definition of SOA
As I'm writing everything here, I'll post short definition of second-order arithmetic too: apart from quantifying over numbers, we allow quantification over sets of numbers, which we denote by capital letters. Along with axioms of PA we also add following axioms: That's it. LittlePeng9 (talk) 19:00, October 8, 2014 (UTC)
 * 1) Full induction axiom: \(\forall X((0\in X\land\forall n: (n\in X\Rightarrow S(n)\in X))\Rightarrow\forall n: n\in X)\) which says that, for every set \(X\), if it contains 0 and is closed under taking successors, then it contains all natural numbers.
 * 2) Axiom schema of comprehension: for every formula \(\phi(x)\) which can have any number of free number/set variables, which we can set to anything, we add an axiom \(\exists X\forall n:n\in X\Leftrightarrow \varphi(n)\), which says we can build a set based on any property of the numbers.

Buchholz hydra
Why did you remove that sketch of proof? Wythagoras (talk) 14:27, October 19, 2014 (UTC)


 * First, the ordinal number of the hydras get all messed up because of 0-labelled nodes, as I believe by some convention \(\Omega_0=1\). But even if we gave label \(\Omega_{k+1}\) to nodes labelled k, I think it was flawed, because if we take hydra (+(0(1(3(2))))), which has ordinal \(\Omega^{\Omega_2^{\Omega_4^{\Omega_3}}}\), and by reducing it we get (+(0(1(3(1(1(3(1))))))) (or something) which has larger ordinal. Also, there is a problem that \(\Omega_k^{\Omega_l}=\Omega_l\) for k<l. LittlePeng9 (talk) 14:46, October 19, 2014 (UTC)
 * Oh, okay. I think the proof can be fixed by using different \(\psi\) functions. (or related functions) Wythagoras (talk) 14:55, October 19, 2014 (UTC)
 * Let me say you this: Buchholz has somehow proved this result :P
 * And I believe he indeed used a collapsing function to show this. LittlePeng9 (talk) 15:10, October 19, 2014 (UTC)

twitter?
what happened to your twitter Cookiefonster (talk) 12:44, October 20, 2014 (UTC)
 * social media sucks ass it's vel time 06:47, October 22, 2014 (UTC)

how long is alejandro blocked
Cookiefonster (talk) 23:28, October 28, 2014 (UTC)
 * bro do you even Special:RecentChanges it's vel time 23:35, October 28, 2014 (UTC)