For a proof of quantum universality, I need to show that $\tan^{-1}\left(2\right)$ is not a rational multiple of $\pi$. How do I show this? I feel like showing algebraic independence over the rationals is hard in general, but is it possible for $\pi$ and an awkward trigonometric value?
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1The only rational $q$ with $\arctan(q)$ a rational multiple of $\pi$ are $0,\pm 1.$ This follows from unique factorization in $\mathbb Z[i].$ – Thomas Andrews Oct 09 '21 at 22:21
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You can show that if $x$ is a rational multiple of $\pi$, then $2\cos{x}$ is a algebraic integer (it’s the sum of $e^{ix}$ and $e^{-ix}$, both roots of unity). But the cosine of $\tan^{-1}(2)$ is $\frac{1}{\sqrt{5}}$, and $\frac{2}{\sqrt{5}}$ is not an algebraic integer (because its square is a rational non-integer).
Aphelli
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1I guess this follows from the fact that $\cos(p/q \pi) + i \sin(p/q \pi)$ is algebraic from roots of unity being algebraic, and then you add that to the (also algebraic) conjugate? – 416E64726577 Oct 09 '21 at 21:51
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This works for any $\arctan(p/q),$ $p,q$ integers, aside from a few cases. Your argument gives $p^2+q^2\mid 4q^2.$ You can show $p=\pm q$ or $p=0.$ – Thomas Andrews Oct 09 '21 at 22:42
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Let $\theta = \tan^{-1}2$. If $\theta$ is a rational multiple of $\pi$, say $\theta = \frac{p}{q}\pi$ for $p,q \in \mathbb{Z}_{+}$, we have
$$e^{2iq\theta} = e^{2\pi pi} = 1$$ Since $e^{2i\theta} = \frac{1+i\tan\theta}{1-i\tan\theta} = \frac{1+2i}{1-2i}$. Above condition is equivalent to
$$\left(\frac{1+2i}{1-2i}\right)^q = 1\quad\iff\quad (1+2i)^q - (1-2i)^q = 0\tag{*1}$$ For $n \in \mathbb{N}$, let $T_n = \frac1i((1+2i)^n - (1-2i)^n)$. $T_n$ is an integer sequence satisfying:
$$T_0 = 0,\; T_1 = 4,\; T_n = 2T_{n-1} - 5T_{n-2}\quad\text{ for } n \ge 2$$ Looking at everything module $5$, we find:
$$T_n \equiv 2T_{n-1} \pmod 5, \forall n \ge 2\quad\implies\quad T_n \equiv 2^{n-1}T_1 \equiv 2^{n+1} \pmod 5, \forall n \ge 1$$ This means for all $q \in \mathbb{Z}_{+}$, $T_q$ is never divisible by $5$ and cannot be $0$. So RHS of $(*1)$ can never be satisfied and hence $\theta$ cannot be a rational multiple of $\pi$.
achille hui
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You can get that $(1+2i)^q\neq(1-2i)^q$ if you know unique factorization in $\mathbb Z[i].$ That will give you all cases of $\arctan(q)$ for $q$ rational other than $0,\pm1.$ – Thomas Andrews Oct 09 '21 at 22:28
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@ThomasAndrews I know $\mathbb{Z}[i]$ is a UFD but I want a proof that doesn't need anything from algebraic number theory. – achille hui Oct 09 '21 at 22:37
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Here's an alternate method, inspired from this answer.
Let $\theta = \tan^{-1}(2)$. For the sake of contradiction, assume that $n\theta = m\pi$ for some positive integers $n, m$. Then, we have $$1 = e^{2 \iota n \theta} = (e^{2\iota \theta})^n = \left(\frac{1 + \iota \tan\theta}{1 - \iota \tan\theta}\right)^n = \left(\frac{1 + 2 \iota}{1 - 2 \iota}\right)^n.$$ Rearranging, we get $$(1 + 2 \iota)^n = (1 - 2 \iota)^n,$$ which is a contradiction since $1 + 2 \iota$ and $1 - 2 \iota$ are non-associate primes in the UFD $\Bbb Z[\iota]$.
Aryaman Maithani
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