Neat little problem

[nslay]

Prove for x^x=y^y where 0 < x,y < 1, x != y
That there exists infinite rational solutions.
For example:
(1/2)^(1/2)=(1/4)^(1/4)
(1/4)^(1/4)=(1/4)^(1/2*1/2)=(1/2)^(1/2)

I'll post the proof in a few days.

[Yoni]

[R.a.B.B.i.T]

I don't want to use logs, otherwise I'd attempt this.

[Yoni]

You can easily prove it using analysis. The point is to not use calculus, but solve it in an insightful way. You don't need to use logs.

[R.a.B.B.i.T]

At the level of math I'm at right now (pre-cal/trig), I'd need to use logs.

[nslay]

Here is the proof.  It's a little sloppy, it was my first proof at the time.
http://www.itsmagical.com/xtothex.doc

[dxoigmn]

Here is the proof.  It's a little sloppy, it was my first proof at the time.
http://www.itsmagical.com/xtothex.doc

Should learn latex, is fun and much better for typesetting these kind of things (i.e. mathematical proofs). Although I must say you do a nice job in word.

[Yoni]

An attempt to simplify Nate's proof:

Pick any positive integer "n".
Let x = n / (n + 1). x is obviously a rational number.
x^n and x^(n+1) are obviously rational, too (rational to the power of an integer = rational).

a = x^n
b = x^(n+1)

b/a = x
ln(a) / ln(b) = n ln(x) / (n + 1) ln(x) = n / (n + 1) = x

==> b/a = ln(a) / ln(b)
==> b ln(b) = a ln(a)
==> b^b = a^a

Better illustrated:

n = 1:
a = (1/2) ^ 1, b = (1/2)^2 = (1/4)

(1/2)^(1/2) = (1/4)^(1/4)

n = 2:
a = (2/3) ^ 2 = (4/9), b = (2/3)^3 = (8/27)

(4/9)^(4/9) = (8/27)^(8/27)

etc.
Can you generalize this result?