How to prove the irrationality of this number. Prove that $\cos\frac{1}{\sqrt{2019}}$ is irrational.
Would someone give me a hand on this question? Even a hint is appreciated. I don't have any clue.
 A: Following the suggestions of @Carlo, I solved this problem myself. It is indeed similar to the proof that $e$ is irrational. Thinking it might help someone else, so I just post it in the following.
Suppose it is rational number, such that
$$\cos\frac{1}{\sqrt{2019}} = p/q,$$
where $p$ and $q$ are co-primes. Expand the cosine as sum of power series, which is
$$\cos\frac{1}{\sqrt{2019}} = \sum_{i=0}^\infty \frac{(-)^i}{(2i)!}\frac{1}{2019^i}.$$
Define the sum 
$$S_n = \sum_{i=0}^nx_i,\quad \mathrm{and}\quad x_i=\frac{(-)^i}{(2i)!}\frac{1}{2019^i}$$
It is reasonable to assume for some integer large enough $N\gg1$, we have 
$$x_{2m}+x_{2m+1}>0,\ \forall\ m\geqslant N,$$
also
$$x_{2m+1}+x_{2m+2}<0,\ \forall\ m\geqslant N,$$
So it is easy to see that
$$0<p/q-S_{2N-1}<x_{2N}=\frac{1}{(4N)!2019^{2N}}.$$
Thus if we choose, for example, that $4N>q$, we will have
$$0<(4N)!2019^{2N}\times(p/q-S_{2N-1})<1.$$
But on the other hand, $(4N)!2019^{2N}\times(p/q-S_{2N-1})$ must be an integer. Thus by contradiction, it must be irrational.
A: If $x\in \Bbb R$ and for every $r>0$ there exists $<a,b>\in \Bbb Z\times \Bbb Z^+$ such that $0<|x-a/b|<r/b,$ then $x \not \in \Bbb Q.$
Proof: If $x=c/d$ with $c\in \Bbb Z$ and $d\in \Bbb Z^+$  then for any $<a,b>\in \Bbb Z\times \Bbb Z^+$ we have $$0<|x-a/b|\implies 0<|c/d-a/b|\implies 0<|cb-ad|\implies$$ $$\implies 1\le |cd-ab|\implies (1/d)/b\le (1/bd)|cb-ad|=|x-a/b|.$$ (I.e.  if $x=c/d$ then the "$r$" of my 1st sentence cannot be less than $1/d$.)
Apply this with $n\in \Bbb Z^+$ large enough that $r<(2n)!(2019)^n\
\cdot|\sum_{j=n+1}^{\infty}(-1)^j/((2j)!2019^j)\,|$
with $b=(2n)!(2019)^n$ and $a/b=\sum_{j=0}^n (-1)^j/((2j)!2019^j).$
