The Nineteenth Byte

Seggan

23:06

@Zionmyceliaadamancy tbh practical experience shows P != NP, why is it not solved?

Radvylf Programs

Nobody has practically experienced a number that disproves the collatz conjecture either :p

Seggan

true

but isnt the problem about proving all P = NP?

or am i bad at reading theorems

forest

23:24

P = NP means "all".

Radvylf Programs

@Seggan It's really easy to show you can do something quickly, but good luck proving there's absolutely no way to do that. Even though our experience would strongly indicate P != NP, we just don't have a way to prove that.

Zion mycelia adamancy

@Seggan Practical experience shows that all odd numbers are prime, barring a few exceptions

forest

It's provable that all primes >2 are odd.

Zion mycelia adamancy

That's not what I said :P

@Seggan Practical experience is, for lack of a better word, useless. For example, literally all of these

Seggan

mm

ok then

forest

23:34

@Seggan Pi provably never ends, but there's no proof that it never repeats. For all we know, after 10↑↑↑↑↑↑10 digits, pi just repeats 1212121212 forever. It's very, very likely that it never does repeat, but there's no proof to that effect.

Zion mycelia adamancy

@forest This is wrong

If a number is irrational, it's digits will never repeat

forest

@Zionmyceliaadamancy Isn't 1/3 an irrational number?

Zion mycelia adamancy

No

forest

Oh, what's the term then?

Zion mycelia adamancy

Irrational means that it cannot be expressed in the form p/q where p, q are integers that do not share a common factor

forest

23:36

(My math is suck)

Seggan

@forest iirc irrational is when it cannot be written as a fraction

forest

Oh yeah. What term am I mixing it up with then?

Zion mycelia adamancy

@forest You may be thinking of "normal", which means that each digit (in base 10) is distributed equally in the decimal expansion of pi

forest

Ah.

Zion mycelia adamancy

This is equivalent to saying that "all possible sequences of digits will appear in pi", which is commonly claimed, but isn't yet proven

forest

23:38

Yep.

It's very likely though, but no proof exists.

Neil

the first 100,000,000.000.000 digits of pi do look completely normal

Zion mycelia adamancy

Ye, basically every substantial decimal expansion of pi shows it to be normal

Seggan

@Zionmyceliaadamancy so does that mean eulers number appears in pi? :P

forest

@Seggan Now you're getting into "undefined" territory (stuff like infinity = infinity not being true). :P

Zion mycelia adamancy

@Seggan Not really

pppery

23:40

743

Q: Does $\pi$ contain all possible number combinations?

$\pi$ Pi Pi is an infinite, nonrepeating $($sic$)$ decimal - meaning that every possible number combination exists somewhere in pi. Converted into ASCII text, somewhere in that infinite string of digits is the name of every person you will ever love, the date, time and manner of your death, and...

Zion mycelia adamancy

"Normal" generally means that all finite sequences of digits appear in the expansion of a number

There are stronger conditions for all non-finite sequences

forest

So any of n digits of e will fit in pi (if the conjecture is true), but n can't be infinity.

Zion mycelia adamancy

It can't necessarily be infinity

forest

(If it's just "1111..." then you could say it fits, if e also ends in "1111...")

Zion mycelia adamancy

For example, it is entirely possible that after some N digits, the digits of pi are identical to the digits of e

But that's a different condition to being normal

forest

23:43

@Zionmyceliaadamancy Doesn't Euler's identity make it possible to disprove that?

Zion mycelia adamancy

@forest Of course, we know that e cannot end in 11111....., as it is also irrational

@forest Possibly? I'm not that well-read on this kind of number theory

forest

ah

(Neither am I, clearly!)

There are some conjectures that there are no proofs for and which no one even knows how to start finding a proof (except proof by counterexample using brute force, which is impractical), like a Fermat prime larger than 65537.

Zion mycelia adamancy

I suspect that, if it had been disproven, you could find it somewhat easily

forest

Whereas at least there's a lot of research into proofs relating to pi.

Seggan

was brute force ever used to disprove a theorem?

forest

23:50

Yeah.

In fact, one of the shortest papers came from that.

Euler's conjecture, I think. Lemme find it.

Zion mycelia adamancy

@Seggan I believe Legrange's number was discovered by brute force

thejonymyster

all numbers were discovered by brute force

they didnt ask to be born

forest

@thejonymyster Nah, no one discovered TREE(3) by brute force. No one could.

Brute force means starting from 0 and counting up (or otherwise counting in some pattern) until you hit upon a number.

thejonymyster

@UnrelatedString the issue is its too wordy for a cmc though itll be like, annoying as fuck

Zion mycelia adamancy

@forest Yes and no

thejonymyster

23:53

or something idk

forest

@Zionmyceliaadamancy That's the meaning in cryptography, at least (more or less).

thejonymyster

ill try anyway though cmc time let me think of how to phrase this

Zion mycelia adamancy

If something shows that "if A, then either X, Y, Z", then all you have to do is brute force proofs for "if A, then X", "if A, then Y" and "if A, then Z", which can limit your search field

forest

Yeah, I was just saying that not all numbers are found by brute force, not that you can't use brute force for proofs.

E.g. take a random number 3714147959877322721152051393610902386023347521548091461027631636332873169214. I'd bet my life that no one has ever come up with this number, but I didn't have to search for it randomly until I found it. I came up with it on the fly (well, my computer did).

thejonymyster

i think about that number all of the time

forest

23:56

lel

Zion mycelia adamancy

@forest Well, I discovered 3714147959877322721152051393610902386023347521548091461027631636332873169215 by brute force by adding 1 to that number :P

forest

That's not brute force then. :^)

thejonymyster

guys lets keep doing this bit but wiht numbers that fill up thye screen

Zion mycelia adamancy

Depending on how you define "brute force", the inductive principle implies that you can discover all numbers by brute force :P

forest

Not uncomputable numbers.

Zion mycelia adamancy

23:57

Yep, all numbers

thejonymyster

that one depends on how you defined "discover"

forest

Not without hypercomputation.

thejonymyster

if i iterate through english rayo style ill certainly define some numbers

a (nope not a number) b (nope not a number).... aaz, aba, abb, abc... zzz, aaaa...

Zion mycelia adamancy

Define positive integers by induction, define negatives by introducing subtration, define rationals by defining division, then introduce the completeness axiom to define all real numbers :P

Peano axioms ftw :P

forest

@Zionmyceliaadamancy What about a number (as a function) where calculating it with arbitrary precision would result in solving the halting problem?

lyxal

23:59

@forest time to die: libraryofbabel.info/bookmark.cgi?frpn.osug406

thejonymyster

does that number exist

Zion mycelia adamancy

@forest Sure, an arbitrary Turing machine may not be able to calculate it, but we can show that it exists

@thejonymyster Yes

thejonymyster

okay :-)

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