November 15th, 2016, 10:59 AM  #1 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0  Decide whether α must be a rational number... someone, sth?
Let $\alpha$ be a real number such that tan($\alpha$ · $\pi$) = √2. Decide whether $\alpha$ must be a rational number.
Last edited by skipjack; November 15th, 2016 at 11:37 AM. 
November 15th, 2016, 11:02 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,429 Thanks: 1314 
dude, you've posted this 3 times now. If someone had an answer for you you would have seen it. Imo this is a nontrivial problem. I've spend an hour looking at it and trying to figure out how $\alpha$ might be rational. I haven't made any headway. 
November 15th, 2016, 11:04 AM  #3 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0 
or irrational

November 15th, 2016, 11:40 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,629 Thanks: 2077 
Where did you get this problem from?

November 15th, 2016, 12:31 PM  #5 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0 
received, that was probably from an olympiad

November 15th, 2016, 01:34 PM  #6 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
Assuming for the moment that $\alpha$ is rational, there should be a repeating pattern in the decimal expansion. I looked at the first 10,000 digits of $\frac{\tan^{1}\sqrt{2}}{\pi}$ and found no repeating pattern so far. I can look at one million after I get back from work and check that as well, but I'm leaning towards $\alpha$ being irrational.

November 15th, 2016, 02:33 PM  #7 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0 
how did you get $\frac{\tan^{1}\sqrt{2}}{\pi}$ ?

November 15th, 2016, 03:03 PM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Seriously? This is an extremely difficult problem (if it is doable at all) but you seem to be saying that you do not even understand what is meant. The problem was to prove that any a such that $\displaystyle tan(a\pi)= \sqrt{2}$. To solve that equation, take the inverse tangent: $\displaystyle a\pi= tan^{1}(\sqrt{2})$. $\displaystyle a= \frac{tan^{1}(\sqrt{2})}{\pi}$. 
November 15th, 2016, 03:15 PM  #9 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0 
sorry, eclipse of mind

November 16th, 2016, 09:15 AM  #10  
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0  Quote:
Last edited by TobiWan; November 16th, 2016 at 09:54 AM.  

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