ITT we post quackery

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

WebTool

ITT we post quackery

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

9 months ago

All urls found in this thread:

http://math.stackexchange.com/questions/472957/the-continuum-hypothesis-the-axiom-of-choice

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.

9 months ago

DeathDog

a number that does not exist

Top kek.

9 months ago

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.

9 months ago

5mileys

which is actually logically inconsistent with itself.

Derive false from your favorite axiomatization of the reals of GTFO.

9 months ago

MPmaster

@WebTool

Real number are sequences of digits that are not rational.

Hows is this difficult to understand.

Brainlet.

9 months ago

Soft_member

Here's your reply, now get out.

9 months ago

whereismyname

@Methshot

Oh. But if you want it to be false you need unicity of decimal representation.

protip, there's no such a thing.

9 months ago

5mileys

@Methshot

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

9 months ago

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

9 months ago

Raving_Cute

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

9 months ago

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).

9 months ago

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.

9 months ago

Dreamworx

@Dreamworx

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

9 months ago

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.

9 months ago

FastChef

@CouchChiller

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

9 months ago

Ignoramus

@FastChef

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

9 months ago

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?

9 months ago

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.

9 months ago

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 ;)

9 months ago

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?

9 months ago

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.

9 months ago

takes2long

@WebTool

which is actually logically inconsistent with itself

go on. you asserted a claim. prove it.

9 months ago

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.

9 months ago

Carnalpleasure

@Supergrass

http://math.stackexchange.com/questions/472957/the-continuum-hypothesis-the-axiom-of-choice

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

9 months ago

viagrandad

@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.

9 months ago

Bidwell

@viagrandad

Sorry user, I don't think you know what you're talking about.

carry on

9 months ago