It's international spook day, so post some spooky math

[math]a-b = (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})[/math]
and
[math](\sqrt[2^n]{a}-\sqrt[2^n]{b})=(\sqrt[2^{n+1}]{a}+\sqrt[2^{n+1}]{b})(\sqrt[2^{n+1}]{a}-\sqrt[2^{n+1}]{b})[/math]

therefore by induction:
[math]a-b = \prod\limits_{i=1}^{\infty}(\sqrt[2^{i}]{a}+\sqrt[2^{i}]{b})\times\lim\limits_{i\rightarrow \infty}(\sqrt[2^{i}]{a}-\sqrt[2^{i}]{b})[/math]

observe that
[math]\lim\limits_{i\rightarrow \infty}(\sqrt[2^{i}]{a}-\sqrt[2^{i}]{b})=0[/math]

hence
[math]a-b=\prod\limits_{i=1}^{\infty}(\sqrt[2^{i}]{a}+\sqrt[2^{i}]{b})\times0 = 0[/math]

meaning that
[math]a=b \quad \forall a,b\in\mathbb{R}[/math]

How about some spooky physics?

>therefore by induction: [omitted]
did you type something wrong here? I don't follow this at all

you can't conclude a=b just because [math]a^{\frac{1}{\infty}} = b^{\frac{1}{\infty}}[/math], which is essentially what you've done

...

its spooky because its flawed
induction works for all n which are fixed.
but not necessarily for the limit when n to infty.
In this particular case it does not work for the limit.
why not is an exercise to you in real analisys

Can i approach infinity? I feel like infinity is just for real numbers, and only a complex number could approach a complex infinity

Oh wait nvm just skimmed the proof I get it now

en.wikipedia.org/wiki/No-ghost_theorem

*ruins your Halloween*

ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19800010907.pdf

Is this spooky math?
[eqn]\forall n\,\in\,\mathbf N,\,\forall A\,\in\, \mathscr M_n\left(\mathbf C\right),\, \left(A^*\,=\,A \,\Rightarrow\, \exists B \,\subseteq\,\mathscr M_{n,\,1}\left(C\right)^n,\, \left(\forall u\,\in\,\mathscr M_{n,\,1}\left(\mathbf C\right),\, \exists! \lambda \,\in\,\mathbf C^B,\, u\,=\,\sum_{v\,\in\,B} \left(\lambda\left(v\right)\,v\right) \,\wedge\, \forall \left(u,\,v\right)\,\in\,B^2,\, \left(u\,\neq\,v\,\Rightarrow\,u\,\bot\,v\right)\,\wedge\, \forall u\,\in\,B,\,\exists\lambda\,\in\,\mathbf R,\,A\,u\,=\,\lambda\,u\right)\right)[/eqn]

>any hermitian matrix has an orthogonal eigenbasis
wow so spooked

>he didn't get it

>with real eigenvalues
that's the easiest part

...

>that's the easiest part
sure it is, after you need it pointed out

[math]\left \langle u, A u \right \rangle
=\left \langle A^\dagger u, u \right \rangle
=\left \langle A u, u \right \rangle
=\left \langle u, A u \right \rangle^\dagger[/math]
Must hurt to be such a brainlet as to be spooked by this.

I sure am spooked by your retardation.

Euclid BTFO

Actually, no.

Spook math

You have to use coinduction and not induction or else you get an infinite loop in your theorem prover

Observe:

define Nat = Z | S Nat

isFinite :: Nat -> Bool
isFinite Z = True
isFinite (S n) = isFinite n

This doesn't work for:

infinity :: Nat
infinity = S infinity

Wow, that's a lot of precalculus. huh couldn't even have done differential geometry u lil shit?