wu :: forums
« wu :: forums - x^x is irrational »

Welcome, Guest. Please Login or Register.
Apr 19th, 2024, 5:22am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   medium
(Moderators: william wu, ThudnBlunder, towr, Eigenray, SMQ, Grimbal, Icarus)
   x^x is irrational
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: x^x is irrational  (Read 3271 times)
NickH
Senior Riddler
****





   
WWW

Gender: male
Posts: 341
x^x is irrational  
« on: Sep 7th, 2002, 3:16am »
Quote Quote Modify Modify

If x is a positive rational number, prove that x^x is irrational unless x is an integer.
« Last Edit: Oct 23rd, 2003, 7:44pm by Icarus » IP Logged

Nick's Mathematical Puzzles
rugga
Newbie
*





   


Gender: male
Posts: 21
Re: NEW PROBLEM: x^x  
« Reply #1 on: Sep 7th, 2002, 2:09pm »
Quote Quote Modify Modify

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 xx is rational, then it can be written as c/d for some relatively prime integers c & d.  So
c/d = xx = (a/b)a/b  
  = ((a/b)a)1/b
  = ( aa / ba ) 1/b
 
Raising both sides to the power of b:
cb / db = aa / ba
or
cb ba = aa db
 
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:
 
ba:  ka, where k is the number of times p occurs in b
aa:  0, since a and b are relatively prime
db:  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.)
cb:  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 pb.
This is a contradiction since pb is greater than b for any integers b,p>1.  Therefore b can't be greater than 1 if xx is rational.
 
Of course if b=1 then x is an integer for any integer a, and xx is rational.
IP Logged
NickH
Senior Riddler
****





   
WWW

Gender: male
Posts: 341
Re: NEW PROBLEM: x^x  
« Reply #2 on: Sep 7th, 2002, 2:53pm »
Quote Quote Modify Modify

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
IP Logged

Nick's Mathematical Puzzles
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board