# ITT we post quackery

WebTool

ITT we post quackery

I'll start us off with the field of "Real" numbers, which is actually logically inconsistent with itself.

All urls found in this thread:
http://math.stackexchange.com/questions/472957/the-continuum-hypothesis-the-axiom-of-choice
RumChicken

the field of "Real" numbers is actually logically inconsistent with itself.
Yup, that's some gold-standard quackery right there.

VisualMaster

@RumChicken
This

DeathDog

a number that does not exist

Top kek.

JunkTop

@RumChicken
@VisualMaster
The non-finitist quacks have arrived, everybody.

StrangeWizard

@WebTool
Since no one else is posting: The continuum hypothesis, which is less of an hypothesis and more of a conjecture by uneducated morons who refuse to work with other axioms than the ones their professors taught them decades ago.

5mileys

which is actually logically inconsistent with itself.
Derive false from your favorite axiomatization of the reals of GTFO.

MPmaster

@WebTool
Real number are sequences of digits that are not rational.
Hows is this difficult to understand.

Brainlet.

Soft_member

Methshot

@5mileys
0.999(9) = 1

SniperWish

@MPmaster
using a high school definition of reals
For shame

Soft_member

@MPmaster
"Real number are sequences of digits"
calls others brainlets

LuckyDusty

@WebTool
the only "real" numbers are the primes

whereismyname

@Methshot
Oh. But if you want it to be false you need unicity of decimal representation.
protip, there's no such a thing.

cum2soon

@whereismyname
So basically Dewey created one of the dankest memes

5mileys

@Methshot
Reals axioms don't require real numbers to have unique decimal representation

Dreamworx

@MPmaster
The real numbers contain the rationals
even the people hating on this post haven't pointed that out
holy shit this is why i use physicsforums

Raving_Cute

Be real OP, this is just a contrived way of making a 0.999... = 1 thread

TurtleCat

@MPmaster
Given that you want the real numbers to be a field, with a unique identity (1) and so on, you'd really have to consider an equivalence class of sequences (where e.g. the set {[0,9,9,...] and [1,0,0,...]} models the number "1") and so this isn't much better than the Cauchy sequences.
Cauchy sequences are probably chosen over "equivalence classes of sequences of digits" because it lends itself better to analysis (calculus).

Nojokur

@StrangeWizard
Is you best argument against the CH is that "its not a hypothesis its a conjecture"?

That's nitpicking. It's doubtful you even understand people's complaints about it.

Watching Wildberger does not mean you have an educated opinion on mathematical logic or set theory.

Dreamworx

@Dreamworx
There's just so much wrong with it, its hard to know where to start.

Need_TLC

@MPmaster
top lel

CouchChiller

@Nojokur
Is you best argument against the CH is that "its not a hypothesis its a conjecture"?
My best argument against CH is that any reasonable axioms you choose to perform set theoretical work in should make it possible to prove it false.

FastChef

@CouchChiller
so a reasonable axiom system for you includes the negation of CH?

JunkTop

@CouchChiller
Real analysis doesn't like CH that much.

Ignoramus

@FastChef
It doesn't necessarily have to include CH's negation, just imply it.

Spazyfool

@JunkTop
does real analysis even care about CH?
the only axiom I've really seen being discussed in a real analysis context is AOC.
@Ignoramus
what axiom would you suggest that implies it's negation?

King_Martha

@Spazyfool
AOC implies the law of excluded middle

CouchChiller

@Spazyfool
what axiom would you suggest that implies it's negation?
That's not my point. I don't propose any axiom, I just refuse to accept axioms that don't imply CH's negation.

Emberburn

@CouchChiller
I think you mean "I refuse to accept axioms that imply CH"

You don't want to bar the possibility of accepting axioms which are independent of CH ;)

Stupidasole

@Emberburn
I think Godel proved that CH can't be proved with the current axioms. If I remember right, there are proofs in real analysis that prove the axiom of choice, but CH implies AOC. Do you see the contradiction?

Fried_Sushi

@Stupidasole
I think Godel proved that CH can't be proved with the current axioms.

Actually that it's independent of the ZFC axioms. You can't prove OR DISPROVE the CH with just the ZFC axioms (which includes choice). What you remember is not correct.

takes2long

@WebTool
which is actually logically inconsistent with itself
go on. you asserted a claim. prove it.

Supergrass

The ZFC universe is a large one. If you can't prove it in ZFC + the Grothendieck add-on, I don't see how AOC can be proved so easily in real analysis. Even though it has been a while, I think they had us "prove" AOC with the well-ordering principle or some shit like that. I'm just saying AOC is a spin-off of CH, and the CH is undecidable. That's the contradiction. AOC is also undecidable.

Carnalpleasure

Continuum hypotesis doesn't imply AoC but Generalised CH implies AoC, which can Be proved in ZF.

Snarelure

@WebTool
ITT we post quackery

Oops.

SniperGod

@Carnalpleasure
Do you know what undecidable means?

@Stupidasole
there are proofs in real analysis that prove the axiom of choice
No.
CH implies AOC
No.
Please actually learn the subjects you're talking about before shitposting on Veeky Forums.

Bidwell