Exercise 6.3.9 ($\cos 1$ is irrational)

Question

Prove that $\cos 1$ is irrational, where ``$\,1$'' is in radian measure.

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Hint. Use the infinite series of $\cos x$ and adapt the ideas of the two preceding problems.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} \begin{proof} The infinite series of $\cos x$ for $x = 1$ gives $$\cos 1 = \sum_{j=0}^\infty \frac{(-1)^j}{(2j)!} = 1 - \frac{1}{2!} + \frac{1}{4!}- \cdots$$ We can group the terms by pairs. This gives $$\cos 1 = \sum_{k=0}^\infty \left( \frac{1}{(4k)! }- \frac{1}{(4k+2)! } ight) = \left(1 - \frac{1}{2!} ight) + \left( \frac{1}{4!} - \frac{1}{6!} ight) + \cdots$$ Put for every $n \in \N$ $$S_n = \sum_{k=0}^n \left( \frac{1}{(4k)! }- \frac{1}{(4k+2)! } ight),\qquad R_n = \sum_{k=n+1}^\infty \left( \frac{1}{(4k)! }- \frac{1}{(4k+2)! } ight),$$ so that $\cos 1 = S_n + R_n$. For all $k\in \N$, $\frac{1}{(4k)! }- \frac{1}{(4k+1)! } >0$, thus $R_n>0$, so $$ S_n < \cos 1 < S_n + R_n,$$ and $$ (4n+2)! S_n < (4n+2)! \cos 1 < (4n+2)! S_n + (4n+2)! R_n.$$ Note that $$N_n = (4n+2)! S_n = \sum_{k=0}^n \left( \frac{(4n+2)! }{(4k)!} - \frac{(4n+2)! }{(4k + 2)!} ight)$$ is an integer, such that for all $n \in \N$, $$N_n < (4n+2)! \cos 1 < N_n + (4n+2)! R_n.$$ Moreover, for all $n \in \N$, \begin{align*} (4n+2)! R_n &= (4n+2)! \sum_{k=n+1}^\infty \left( \frac{1}{(4k)! }- \frac{1}{(4k+2)! } ight)\ &\leq (4n+2)! \sum_{k=n+1}^\infty \frac{1}{(4k)! }\ & = (4n+2)! \frac{1}{(4n+4)!} \left( 1 + \frac{1}{(4n+5)(4n+6)(4n+7)(4n+8)}+ \cdots ight)\ &\leq \frac{1}{(4n+3)(4n+4)} \left( 1 +\frac{1}{ 5^4} +\frac{1}{ 5^8 } + \cdots ight)\ &\leq \frac{2}{(4n+3)(4n+4)}\ &<1. \end{align*} Therefore \begin{align} N_n < (4n+2)! \cos 1 < N_n + 1. \end{align} If $\cos 1$ was rational, then $\cos 1 = p/q$, where $q \in \N^*$. If we apply (1) with $n =q$, we obtain $$N_q < (4q+2)!\, \frac{p}{q} < N_q + 1.$$ But since $q \mid (4q+2)!$ the integer $M = (4q+2)! \frac{p}{q}$ is between $N_q$ and $N_q +1$: $$N_q < M < N_q +1,$$ where $M$ and $N_q$ are integers. This is impossible, therefore $\cos 1$ is irrational. \end{proof}

Exercise 6.3.9 ($\cos 1$ is irrational)

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Exercise 6.3.9 ($\cos 1$ is irrational)

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