Prove to me that 1+1=2

This is an example of the addition axiom. There is no mathematical proof.
>The additive axiom in mathematics: if equal numbers are added to equal numbers, the results are equal. a+b=c+d
An axiom is an assumption. To prove it is wrong you would also need an to make an assumption. Which you would then need to prove. Then you would need to prove the assumption you needed to prove you assumption which you would then need to prove. Then you would need to prove the assumption you needed to prove you assumption which you would then need to prove. Then you would need to prove the assumption you needed to prove you assumption which you would then need to prove. Then you would need to prove the assumption you needed to prove you assumption which you would then need to prove. Then you would need to prove the assumption you needed to prove you assumption which you would then need to prove. Then you would need to prove the assumption you needed to prove you assumption which you would then need to prove. Then you would need to prove the assumption you needed to prove you assumption which you would then need to prove...

lel xD ebin engineer maymay bro *highfive*

you're fucking retarded
being the successor has nothing to do with addition; the two are defined entirely separately and, in general, ordinal addition is a pain in the ass
but you'd know that if you weren't just a pseud who browses wikipedia entries to larp

>the addition axiom
kill yourself

nice.

strongly consider suicide.

no vector spaces please

S(x) := x u {x}
x + 0 := x
x + S(y) := S(x) + y
x + sup U := sup (x + U)
HURRRRRRRRRRRR

>i call out an egregious mistake
>he corrects it
>somehow this makes ME look bad
sasuga, user

>being the successor has nothing to do with addition
>I WAS JUST PRETENDING TO BE RETARDED

>the two are defined separately
but you aren'y pretending to be illiterate

>being the successor has nothing to do with addition
>the two are defined separately
see

Depends on if the people reproduce before you count them

Also are there any negative people in the room?

>this autism

pure math homosexuals

see

Define 1, define 2, define what '+' means and define what '=' means. Then using the properties of '+' and '=', you need to show how they can lead to 1 + 1 = 2. You might say it's common sense what they mean, but a lot of things that we think are obviously true turn out to be false.

count to 2.

I have one chair {1c}, you have one chair {1c}

If I ask you to give me {1c}

It will be 0+{1c}={1c}

Then, ∑{c} = {1c} +{1c} = {2c}

==> 1+1=2

For that you need to prove that 1*chair+1*chair=2*(chair)

Fuck off

I could do that very easily but first you'd have to define mathematically what 1 and 2 and + are. See

...

>mad because his shitty physical intuitions don't help him prove things
*laughs maniacally*

clean, simple, and short
nice.

can we have the same in modern notation?

ZFC is inconsistent.

>axiom
Subhuman detected.

Prove that [math]1 \neq 0[/math] without appealing to Peano arithmetic or any other similar axiomatization.

You have one banana. You have another banana. Put them side by side and what do you see? Two bananas.
There's your proof.

Still holds, since inconsistent systems prove anything.

>Still holds
I didn't claim it doesn't hold. Read my post carefully, engineer.
> inconsistent systems prove anything.
Only assuming they are at least as powerful as intuitionistic logic.

>There's your proof.
>that's a """proof"""
you're a special kind of retard, aren't ya

How's that not a proof? You have one banana and then add on another. Now, unless some cunt's gone and stolen you bananas, you have two. SOLID PROOF

>I didn't claim it doesn't hold
Then your post is a non-sequitur.
>Only assuming they are at least as powerful as intuitionistic logic.
Which ZFC is by definition.

let's see your proof, tough guy

That's not a proof, it's an example to help illustrate the concept that 1+1=2. The equation still needs to be proven though.

>Then your post is a non-sequitur.
My post merely says that you shouldn't care about what ZFC proves since it's inconsistent.
>ZFC
>systems
Since when is ZFC a "systems"?

Your proof is that you can add one banana onto another and get two bananas.

this isn't even good bait

2x=2
x=1
QED

Oh, that mastodontic book that was shreded by Gödel in a simple elegant paper.

Easy, if we use a ring... Oh fuck!

The proof is trivial. Assume [math]Con(ZFC)[/math] holds, then the following is a long exact sequence in [math]\mathbf{Lang}[/math]:

[math]\cdots \longrightarrow 0 \longrightarrow \mathbb{Q} \otimes_\mathbb{Z} \bigoplus_{\alpha \in \mathbb{R}}^{\infty + \infty^\infty} \mathbb{Z}/p\mathbb{Z} \longrightarrow Con(ZFC) \overset{\xi^\infty} \longrightarrow Con(ZFC + AD) \longrightarrow \Lambda_\omega(H_p(ZFC)) \longrightarrow 0 \longrightarrow \cdots[/math]

Clearly [math]\Lambda_\omega(H_p(ZFC)) \cong 0[/math], since [math]\Lambda_\omega[/math] is a left-adjoint and [math]H_p(ZFC)[/math] is evidently initial in [math]\mathbf{Ab}[/math].
We also have that [math]\mathbb{Q} \otimes_\mathbb{Z} \bigoplus_{\alpha \in \mathbb{R}}^{\infty + \infty^\infty} \mathbb{Z}/p\mathbb{Z} \cong 0[/math] (trivially).

Thus the sequence below is exact as well.
[math]\cdots \longrightarrow 0 \longrightarrow 0 \longrightarrow Con(ZFC) \overset{\xi^\infty} \longrightarrow Con(ZFC + AD) \longrightarrow 0 \longrightarrow 0 \longrightarrow \cdots[/math]

Exactness trivially implies that the morphism [math]\xi^\infty[/math] is in fact an isomorphism, which trivially means [math]Con(ZFC) \cong Con(ZFC + AD) [/math], but AD is inconsistent with
AC, so [math]Con(ZFC + AD) \cong 0[/math].

Thus by transitivity and symmetry of isomorphism we trivially conclude [math]Con(ZFC) \cong 0[/math]. So we have contradiction in the form of [math]Con(ZFC) \land \neg Con(ZFC)[/math] which concludes the proof of the inconsistency of ZFC.

you're a colossal fucking retard

What field of math is this?

applied memetics by some faggot who's learning some category theory or homolical algebra

This is 100% accurate.

Define: Precision.

made me kek, thanks and nice timing. see? your posts can actually be funny instead of plain dumb

That's a rude thing to say, user. Do you not know anything?
The proof uses basic logic and category theory as well as trivial results about abelian groups.

I never believed them to be either/or.

Careful though, don't mention me 'too' directly or else someone will think we're trying to explain some sort of cosmic joke.

[math]z^*=z \exp \left(-2 i \left(\arg (z)-\pi \left\lfloor \frac{\arg (z)+\pi }{2 \pi }\right\rfloor \right)\right)[/math]

[math]\iota \unicode{f4a2}\vartheta[/math]

We literally cannot count beyond 1.

Binary is where it's at.

[math]\iota \unicode{f4a2}\vartheta \left| \pi \text{$\_$r}\right| \int \Delta f \, df[/math]

...

There. That's that thread completed.

it's schizophrenic personality disorder
quite common amongst drooling retards like that poster

For you OP: [math]\vartheta \left| \pi \text{$\_$r}\right| \int \Delta f \, df=\frac{1}{2} \pi \Delta f^2 \vartheta \left| \text{$\_$r}\right| +\text{constant}[/math]

Question : Answer

Sorry, forgot to add.

Math = Question : Answer + I (Self)

Physics = Question : Answer + Observers (Pervert)

Science = Question : Answer + C (Sight)

1 + 1 = 2

I = 1
Them = You - I
Us = I + You
We = Us + You

You + Me = Thus

Now it's your turn.

Lrn2principia fgt pls