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July 17th, 2014, 07:33 PM  #1 
Newbie Joined: Jul 2014 From: San Diego Posts: 2 Thanks: 0  Help: Proving 2^x = 7 is irrational
I'm finding it difficult to start the proof. I think I'm supposed to use the limit theorem but I'm not sure. The x root is throwing me off.

July 17th, 2014, 11:51 PM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 
Assume x = m/n is rational and see what happens.

July 18th, 2014, 08:40 AM  #3 
Newbie Joined: Jul 2014 From: San Diego Posts: 2 Thanks: 0  
July 18th, 2014, 01:38 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,305 Thanks: 525  Quote:
If x = m/n (m and n relatively prime), then you are getting an nth root of a power of 2, which can't be 7. It will be irrational.  
July 18th, 2014, 01:40 PM  #5 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
Suppose $x = \dfrac{m}{n}$, for integers $m,n$ with $n > 0$. Then $2^x = 2^{\frac{m}{n}} = \sqrt[n]{2^m} = 7$. Take both sides to the $n$th power, what happens? (You might want to assume $m > 0$. Why don't we need to worry about what happens when $m \leq 0$?). 

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