Prove $[\forall n \in \mathbb{N}, \cos(n!) \neq 1]$, without using $\pi$ is irrational.
Using $\pi \in (\mathbb R-\mathbb Q)$, I can prove it...
Thanks everyone!
Asked
Active
Viewed 102 times
6
-
if $\pi \in \mathbb{Q}$,let $\frac{p}{q}=\pi$ and put $k = 2p$.then $\cos(k)=\cos(2q\pi)=1$? – Jan 04 '18 at 03:28
-
I changed $\pi\in \mathbb{(R - Q)}$ to $\pi\in(\mathbb R - \mathbb Q).$ The minus sign should not be within \mathbb{}. $\qquad$ – Michael Hardy Jan 04 '18 at 03:29
-
4Wouldn't such a proof be equivalent to proving $\pi$ is irrational? If $\pi = a/b$ then $1 = \cos((a-1)!b\pi) = \cos(a!)$ for $a \geq 3$. The cases $a = 1$ and $2$ can be dealt with easily. – Tob Ernack Jan 04 '18 at 03:29
-
@mochix Where did you get this problem ? – Sandeep Silwal Jan 04 '18 at 03:39
-
I think you could use the taylor series for cos. – Q the Platypus Jan 04 '18 at 03:45
2 Answers
4
If $\cos(n!) = 1$ for some $n\in\mathbb N,$ then $n! = 2\pi m$ for some $m\in\mathbb N,$ so $\pi = m/(n!)$ and thus $\pi$ is rational.
Conversely if $\pi$ is rational then $\pi = m/\ell$ for some $m,\ell\in\mathbb N,$ and $\ell$ is a divisor of $n!$ for some $n\in\mathbb N,$ so then $\cos(n!)=1.$
Therefore the only way to prove $\cos(n!)\ne1$ for every $n\in\mathbb N$ is by proving that $\pi$ is irrational. Various ways of doing that exist: https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational
-
yes, I want to prove "pi is irrational" with using "$cos(n!)\neq 1$". I think "$cos(n!)\neq 1$" is easier than "pi is irrational". There is no way to prove "$cos(n!)\neq 1$" by pi is irrational? – Jan 04 '18 at 03:42
-
@mochix : Finding a novel proof that $\pi$ is irrational is a somewhat ambitious thing to do. – Michael Hardy Jan 04 '18 at 15:06
1
This is not a full answer, just a comment written to give someone more skilled an idea of possible proof.
If we have $\cos (n_0!)=1$ for some $n_0 \in \mathbb N$ then we have $(\cos (n_0!))^2=1$. This implies $\sin(n_0!)=0$ and this implies $$\sum_{n=0}^{+ \infty}\frac{(-1)^n \cdot (n_0!)^{2n+1}}{(2n+1)!}=0$$ and this probably could be shown to be impossible.