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November 15th, 2016, 10:59 AM   #1
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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.
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November 15th, 2016, 11:02 AM   #2
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dude, you've posted this 3 times now.

If someone had an answer for you you would have seen it.

Imo this is a non-trivial 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.
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November 15th, 2016, 11:04 AM   #3
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or irrational
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November 15th, 2016, 11:40 AM   #4
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Where did you get this problem from?
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November 15th, 2016, 12:31 PM   #5
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received, that was probably from an olympiad
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November 15th, 2016, 01:34 PM   #6
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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.
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November 15th, 2016, 02:33 PM   #7
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how did you get $\frac{\tan^{-1}\sqrt{2}}{\pi}$ ?
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November 15th, 2016, 03:03 PM   #8
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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: $\displaystyle a\pi= tan^{-1}(\sqrt{2})$. $\displaystyle a= \frac{tan^{-1}(\sqrt{2})}{\pi}$.
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November 15th, 2016, 03:15 PM   #9
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sorry, eclipse of mind
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November 16th, 2016, 09:15 AM   #10
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Quote:
Originally Posted by Compendium View Post
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.
the thing is that the problem says $\alpha$ has to be rational number, but not necessarily, it can be irrational as well

Last edited by TobiWan; November 16th, 2016 at 09:54 AM.
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