prove that sin(1) (one radian) is irrational.
Hint start with taylor series expansion
you first need this basic and simple fact about alternating series that if $\displaystyle 0 < a_{i+1} < a_i, \ \forall i \in \mathbb{N},$ then $\displaystyle \sum_{i=1}^{\infty}(-1)^{i-1} a_i > 0. \ \ \ \ \ \ \ \ \ (*)$
now suppose $\displaystyle \sin 1 = \frac{m}{n},$ for some $\displaystyle m,n \in \mathbb{N}.$ let $\displaystyle A=(2n-1)! \left (\sin 1 - \sum_{i=1}^n \frac{(-1)^{i-1}}{(2i-1)!} \right)=\frac{m(2n-1)!}{n} - \sum_{i=1}^n \frac{(-1)^{i-1} (2n-1)!}{(2i-1)!}.$
thus $\displaystyle A$ is an integer. i'll show that this is false, i.e. $\displaystyle A$ cannot be an integer: using the MacLaurine expansion of $\displaystyle \sin 1$ we have:
$\displaystyle A= \sum_{i=n+1}^{\infty} \frac{(-1)^{i-1}(2n-1)!}{(2i-1)!}=(-1)^n \sum_{i=n+1}^{\infty} \frac{(-1)^{n+i-1}}{(2n)(2n+1) \cdots (2i-1)}.$ it's clear by $\displaystyle (*)$ that $\displaystyle |A| > 0.$ we also have:
$\displaystyle |A| \leq \sum_{i=n+1}^{\infty} \frac{1}{(2n)(2n+1) \cdots (2i-1)} \leq \sum_{i=n+1}^{\infty} \frac{1}{(2n)^{2i-2n}}=\frac{1}{4n^2 -1} < 1.$ therefore $\displaystyle 0 < |A| < 1$ and so $\displaystyle A$ is not an integer. Q.E.D.