Wolfram tells us $x = \arccos ( \frac{3}{5})$ is an irrational number. How can we prove it? (Without using a computer obviously)

  • $\begingroup$ A computer cannot calculate whether such a number is rational. Wolfram Alpha has some knowledge about irrational numbers and can tell whether particular types of numbers are rational. $\endgroup$ – Peter Oct 13 '17 at 13:34
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    $\begingroup$ It is not rational. See Niven's Theorem, which links to this proof. $\endgroup$ – Jaap Scherphuis Oct 13 '17 at 13:35
  • $\begingroup$ $x$ is even transcendental (See my answer) $\endgroup$ – Peter Oct 13 '17 at 13:46
  • $\begingroup$ @Peter Well, that depends on whether the result of $\arccos$ is in degrees or radians. In radians you are correct, but for degrees there are a few angles given by Niven's theorem where you have a rational angle with a rational cosine and 3/5 is not one of them. $\endgroup$ – Jaap Scherphuis Oct 13 '17 at 13:48
  • $\begingroup$ @JaapScherphuis True, but without additional context this expression should be interpreted to be in radians. By the way, do we know whether the expression (in degree) is transcendental, or can we only apply Niven's theorem to show the irrationality ? $\endgroup$ – Peter Oct 13 '17 at 13:49

Suppose $x=\arccos(\frac{3}{5})$

Then, we have $\cos(x)=\frac{3}{5}$

Obviously, we have $x\ne 0$

With the Lindemann-Weierstrass-theorem, we can show that $\cos(x)$ is transcendental for every algebraic non-zero $x$.

Therefore, $x$ cannot be algebraic, since $\frac{3}{5}$ is not transcendental.

Hence $x$ is transcendental, in particular irrational.


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