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22 Answers
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 ex is irrational for rational x isnt a trivial result!
In fact while reading about this, i found that wikipedia has a stronger result that ex 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
2q = ep
but ex is always irrational for x - natural nos
( frm the expansion of ex can be proved )
celestine was pretty quick to say that ln2 is irrational. have we got anything to support that hypothesis?
there is even one mor einteresting example,
celestine why dont u have a try ?
celestine u have not exactly got what i meant.
ln2=q (an irrational)
this means eq=2 ( a rational).
so (irrational)irrational=2
i think it is needless to say that e is irrational.
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
lets open up another interesting branch.
If p and q are irrational numbers can pq be rational?
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 ?