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        riddles >> medium >> x^x is irrational
        
(Message started by: NickH on Sep 7^th, 2002, 3:16am)

Title: x^x is irrational
Post by NickH on Sep 7^th, 2002, 3:16am

If x is a positive rational number, prove that x^x is irrational unless x is an integer.

Title: Re: NEW PROBLEM: x^x
Post by rugga on Sep 7^th, 2002, 2:09pm

Nice problem, NickH. How's this?

"x is a positive rational number" means x = a/b for some relatively prime positive integers a & b. If x^x is rational, then it can be written as c/d for some relatively prime integers c & d. So
c/d = x^x = (a/b)^a/b
= ((a/b)^a)^1/b
= ( a^a / b^a ) ^1/b

Raising both sides to the power of b:
c^b / d^b = a^a / b^a
or
c^b b^a = a^a d^b

Assume for a moment that b>1. Then choose some prime factor of b (call it p). Consider the number of times p occurs in the prime factorization of each of the above terms:

b^a: ka, where k is the number of times p occurs in b
a^a: 0, since a and b are relatively prime
d^b: jb, where j is the number of times p occurs in d (Since p occurs on the left-hand-side, it must also occur on the right-hand-side, so it must be a factor of d.)
c^b: 0, since c and d are relatively prime

So ka=jb since the number of occurrences has to match up on both sides of the equation. But since b is relatively prime to a, b must divide k.
In other words, the number of times p occurs in b (namely k) is a multiple of b itself, which means b is a multiple of p^b.
This is a contradiction since p^b is greater than b for any integers b,p>1. Therefore b can't be greater than 1 if x^x is rational.

Of course if b=1 then x is an integer for any integer a, and x^x is rational.

Title: Re: NEW PROBLEM: x^x
Post by NickH on Sep 7^th, 2002, 2:53pm

That's exactly the method I used!

The question came up as part of a proof that the solution of x^x = 2 is transcendental. The above result proves that x cannot be rational.

The second step was to use something called the Gelfond-Schneider Theorem. This states that if a is algebraic and not equal to 0 or 1, and b is algebraic and irrational, then a^b is transcendental. This proves that x cannot be algebraic and irrational, and therefore is transendental.

Nick