1. prove irrational

prove that sin(1) (one radian) is irrational.

2. Without Taylor-series see this on the 3rd page (7***).

3. Originally Posted by scubasteve123
prove that sin(1) (one radian) is irrational.
you first need this basic and simple fact about alternating series that if $0 < a_{i+1} < a_i, \ \forall i \in \mathbb{N},$ then $\sum_{i=1}^{\infty}(-1)^{i-1} a_i > 0. \ \ \ \ \ \ \ \ \ (*)$
now suppose $\sin 1 = \frac{m}{n},$ for some $m,n \in \mathbb{N}.$ let $A=(2n-1)! \left (\sin 1 - \sum_{i=1}^n \frac{(-1)^{i-1}}{(2i-1)!} \right)=\frac{m(2n-1)!}{n} - \sum_{i=1}^n \frac{(-1)^{i-1} (2n-1)!}{(2i-1)!}.$
thus $A$ is an integer. i'll show that this is false, i.e. $A$ cannot be an integer: using the MacLaurine expansion of $\sin 1$ we have:
$A= \sum_{i=n+1}^{\infty} \frac{(-1)^{i-1}(2n-1)!}{(2i-1)!}=(-1)^n \sum_{i=n+1}^{\infty} \frac{(-1)^{n+i-1}}{(2n)(2n+1) \cdots (2i-1)}.$ it's clear by $(*)$ that $|A| > 0.$ we also have:
$|A| \leq \sum_{i=n+1}^{\infty} \frac{1}{(2n)(2n+1) \cdots (2i-1)} \leq \sum_{i=n+1}^{\infty} \frac{1}{(2n)^{2i-2n}}=\frac{1}{4n^2 -1} < 1.$ therefore $0 < |A| < 1$ and so $A$ is not an integer. Q.E.D.