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How do you prove that the sine of 1 is not algebraic?

The sine of 1 is given by:

$$ \sin(1) = \frac{1}{1!} - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots = \sum_k \frac{(-1)^k}{(2k+1)!} $$

Proving that $\sin(1)$ is not an integer is very easy.

$\sin(1)$ must be in the range $[\frac{1}{1!}-\frac{1}{3!}, \frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}]$.

However, proving that it's not rational is slightly tricky, but I think the following argument works.

Lemma: every rational number can be written as $\frac{p}{q!}$ where $q$ is odd and there is a unique minimal $q$.

Every rational number can be written in the form $\frac{p}{q!}$ where $q$ is odd. As proof, suppose a rational number is of the form $\frac{a}{b}$, then it is equivalent to $\frac{a \frac{(2b+1)!}{b}}{(2b+1)!}$. We can then take divide both the number and the denominator by $(2b)(2b+1)$ and then $(2b-2)(2b-1)$ and then $(2b-4)(2b-3)$ and so on until the next step would make the numerator a non-integer.

End of proof of Lemma.

Suppose $\sin(1)$ is rational. Then it can be represented as $\frac{p}{q!}$ where $q$ is odd and minimal among such $q$.

$$ \frac{p}{q!} = \frac{1}{1!} - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots \pm \frac{1}{q!} \mp \cdots \\ p = \frac{q!}{1!} - \frac{q!}{3!} + \frac{q!}{5!} - \frac{q!}{7!} \cdots \pm \frac{q!}{q!} \mp R $$

In the above $R$ is a remainder.

All of the terms that precede $R$ are integers, thus their sum is an integer, call it $X$.

$$ p = X + R $$

Suppose the first term of $R$ is positive. $\frac{q!}{(q+2)!} - \frac{q!}{(q+4)!} \cdots$ is between $\frac{1}{(q+1)(q+2)}$ and $\frac{1}{(q+1)(q+2)} - \frac{1}{(q+1)(q+2)(q+3)(q+4)} = \frac{q^2-7q+11}{(q+1)(q+2)(q+3)(q+4)}$ and thus $R$ isn't an integer.

Similarly, if the first term of $R$ is negative, then $-\frac{q!}{(q+2)!} + \frac{q!}{(q+4)!}$ is between $\frac{-1}{(q+1)(q+2)}$ and $\frac{-q^2+7q-11}{(q+1)(q+2)(q+3)(q+4)}$ and thus $R$ is also not an integer when its first term is negative.

$X$ is an integer, and thus $p = X + R$ is not an integer. Therefore $\frac{p}{q!}$ is not an integral fraction, contradicting our hypothesis.

Trying to prove that this number is not algebraic is much harder.

Is there a straightforward way of proving it without invoking an absurdly powerful theorem?

Greg Nisbet
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    Maybe useful this post – Mauro ALLEGRANZA Jan 31 '23 at 07:29
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    Transcendence proofs are difficult. Lindemann-Weierstraß is your friend for tackling $\sin1$. The only “easy” transcendence proofs I know are not really easy and are for Liouville numbers and for $e$ which, due to special properties of $e$, has a fairly special proof. – FShrike Jan 31 '23 at 07:52
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    My recollection is that even the transcendence proof for $e$ is not so easy. The series for $\sin1$ looks enough like that for $e$ that maybe the $e$ proof can be modified to work for $\sin1$. – Gerry Myerson Jan 31 '23 at 08:19
  • In fact there is no easy way to prove $e$ or $\pi$ to be transcendental. The Lindemann Weierstrass theorem is very useful , not only because it immediately implies those numbers to be transcendental. The irrationality proofs for $e$ and $\pi$ are much easier , for $e$ the proof is almost trivial. – Peter Jan 31 '23 at 09:28
  • Also, the Lindemann Weierstrass theorem implies many numbers connected with sin,cos or tan to be transcendental , as pointed out. But again, there is no easy alternative. – Peter Jan 31 '23 at 09:30
  • I suggest you try to get hold of Irrational Numbers by Ivan M. Niven which gives many irrationality and transcendence proofs in a systematic fashion such that the proofs are easy to digest. – Paramanand Singh Feb 02 '23 at 05:04
  • @GerryMyerson: the transcendence of $e$ (the original by Hermite and many of its variations given later) is not based on the series for $e$ but rather on properties of the exponential function. – Paramanand Singh Feb 02 '23 at 05:08
  • @Para, the properties of the sine function look enough like those for the exponential function that maybe the $e$ proof can be tweaked to work for $\sin 1$. – Gerry Myerson Feb 02 '23 at 06:07

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