Is pi in base e is transcendental or algebraic?

Question

inkyvoyd · Answer

Or do we not know?

kinggeorge · Answer

I propose that $\pi$ is transcendental. From http://en.wikipedia.org/wiki/Transcendental_number#Numbers_proven_to_be_transcendental We can see that if $\ln(a)$ is transcendental if, for some $a$, $a$ is algebraic.

kinggeorge · Answer

Nevermind. I thought I had it. :(

inkyvoyd · Answer

D:

kinggeorge · Answer

Try this on for size. Let$$a=e^{i \pi} +1 + e^b \qquad \qquad b\in\mathbb{N}, \;\;\;b \neq 1$$Now, $$\ln(a) = \ln(e^{i \pi} +1+e^b)=\ln(e^b)=b$$Since we're in $\mathbb{Q}(e)$ $b$ must be algebraic. Using the property I linked to, we know that if b is algebraic, this implies that either $b=0, 1$ or $\ln(a)$ is algebraic. However, $b\neq 0, 1$ so $\ln(a)=b$ is algebraic.

And we've gone in a circle! Thank you and good night. I shall inquire about this question to my professor tomorrow.

kinggeorge · Answer

This could however be used as a proof that all rational numbers are algebraic in this field we're working with.

inkyvoyd · Answer

Wait what? You're saying that pi is rational?

inkyvoyd · Answer

I'm confuzzled.

anonymous · Answer

number is coled transcendental if it's  not a solution of algebraic equation like a polinomial

anonymous · Answer

$$e ^{\pi} $$
is transcendental

inkyvoyd · Answer

but, I don't see what it has to do with pi in base e...

anonymous · Answer

you mean base e logarithm?

inkyvoyd · Answer

No, base e as in base 2 (binary) base 10 (decimal)

anonymous · Answer

Oo

anonymous · Answer

no idea

inkyvoyd · Answer

pi in base e=
10(base e) +pi-e in base e
pi-e in base e is the same as pi in binary?

inkyvoyd · Answer

*pi-e in binary

kinggeorge · Answer

To end your confuzzlement @inkyvoyd , I proved nothing. I showed that if a number $b$ is algebraic, then it follows that it's algebraic.