interesting

yes it is

for more information see wolfram mathworld.

prophet sir , i did not get your point, what do u mean by

have we got anything to support that hypothesis?

its a standard result.

that e^x is irrational for rational x isnt a trivial result!

In fact while reading about this, i found that wikipedia has a stronger result that e^x is irrational for any algebraic number x (Lindemann–Weierstrass theorem )

Ok thats enuf math trivia i guess :D

ln2 is transcedental

or

if
ln2=p/q

2^q = e^p

but e^x is always irrational for x - natural nos

( frm the expansion of e^x can be proved )

gelfund ... a russian mathematician

post it b555

nice one :)

yeah, this was demonstrated by Gelfund. Your knowledge is quite deep for your age

proof by eg :

consider

y = (√2)^√2

if y is rational then satisfied

else if y is irrational

we have y^√2 = 2 which is rational

hence solution exists

the ans is yes

ive already read about that

yes u r right sir

Yeh I have got i^{i}=e^{-\Pi /2} which makes it real and what anant sir has done is also right!!

well actually iⁱ is multi-valued. What I considered was only the principal branch. In general,
i = e^{i(2n\pi +\frac{\pi}{2})}
So, i^i = e^{i^2(2n\pi +\frac{\pi}{2})}=e^{-(2n\pi +\frac{\pi}{2})}

is

iⁱ only equal to e^-pi/2 ?

can u provide a general solution ?