So apparently it's not known if [math]pi^{pi^{pi^pi}}[/math] is an irrational number

So apparently it's not known if [math]\pi^{\pi^{\pi^\pi}}[/math] is an irrational number.
ITT: We find out.

(-1) = -1
(-1)(-1) = 1
(-1)(-1)(-1) = -1

Is that suposed to go on forever or is it just pi^pi^pi^pi?

OP here, it's just 4.

Why should we care whether a number is irrational or not?

It's not. Here's a proof by contraction: Suppose it is rational, let pi^pi^pi^pi = a/b, where a and b are """real""". Then we have (a/b)^(a/b) must be rational, which is a contraction as proved by Barnett in his 2013 seminal paper on exact ∞-groupoids in noncommutative cohomolgy Frobenioids in the framework of synthetic homotopical Grothendieck étale contravariants.

You don't need to care about it. It's a question for those who do care. If you don't care, don't enter the thread.

but what about [math]\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^{\pi^\pi}}}}}}}}}}}}}}}}}}}}}}[/math]?

Well, it's just one of the easiest examples of the bigger problem;
We don't know if there exists a power tower of [math]\pi[/math] and [math]e[/math] that is an integer.

But why do you care?

>[math]\pi[/math] and [math]e[/math]
*or

SOLUTION COMING

I was wrong

...

WOW IT'S FUCKING NOTHING

>windows 10
>facebook
>/lgbt/
>google calculator
is this bait?

Only way it could be better is if it had a reddit tab open

not that guy but
things like this

if it went forever then it would be equal to +oo

There must be a bijection between the rational (positive) numbers and the subset of the irrational numbers that , when pi is raised to that irrational number equal a rational number.

therefore their cardinality is the same.

and so it is almost certain that pi^a is irrational if a is irrational.

almost certain aint good enough bruh, uncountable infinities are big

is pi^pi irrational? Then what's to stop any order of pi to be irrational?

OP, we don't even know if Pi times e is algebraic. We have a loooong way to go.

An irrational number to an irrational power isn't necessarily irrational.

√(-1)2 for example

Give an example within the reals

A = 2^(sqrt(2))
B = sqrt(2)
Both irrational.
A^B = 4

No, we don't

I wonder if theres a neat proof for pi^pi being transcendental, like e^pi being transcendental as
e^pi=(e^i*pi)^-1=(-1)^-i which is transcendental by gelfond schneider theorem

Once I saw /lgbt/, I thought the next tab was gonna be gay porn.