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.
:'(
I don't want to use logs, otherwise I'd attempt this.
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.
At the level of math I'm at right now (pre-cal/trig), I'd need to use logs.
Here is the proof. It's a little sloppy, it was my first proof at the time.
http://www.itsmagical.com/xtothex.doc
Quote from: nslay on April 29, 2005, 07:04 PM
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.
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?