Why is this proof wrong? ( I proved the irrationality of $\pi$ using the most basic techniques so I suspect that it must have gone wrong somewhere) So it starts off by way of contradiction, supposing $\pi\in \mathbb Q$, then by De-Moivre's theorem for rational powers:
$$\left((\cos(2k\pi)+i\sin(2k\pi)\right)^{\pi}= \cos(2k\pi^2)+i\sin(2k\pi^2)=1^{\pi}$$
Here $k\in \mathbb Z$.
Then you must have $$\cos(2m\pi^2)+i\sin(2m\pi^2)=
\cos(2p\pi^2)+i\sin(2p\pi^2)$$ for all $m,p$ that are integers.
However equating real parts gives $$cos(2m\pi^2)=\cos(2p\pi^2)$$, but this implies that $$2m\pi^2=2t\pi-2p\pi^2\quad  \text {or} \quad 2m\pi^2=2t\pi+2p\pi^2$$
(for some integer $t$)
Which gives that $t= (m-p)\pi$ or $t= (m+p)\pi$; but choosing $m$ and $p$ appropriately such that for
$\pi= \frac ab$ (where $a,b$ are coprime) you have $b$ doesn't divide $m\pm p$, which gives $t$ is not an integer contrary to assumption.
The only remaining case is when $b=1$ which is trivially false as $3<\pi<4$ can be easily proven.
 A: De Moivre's formula is:
$$\left( \cos x + i\sin x\right)^n = \cos (nx) + i\sin (nx).$$
It holds true for any real $x$ and any integer $n.$
You are suggesting there is a formula such as:
$$\left( \cos x + i\sin x\right)^{\frac{p}{q}} = \cos \left(\frac{p}{q}x\right) + i\sin \left(\frac{p}{q}x\right)$$
for integers $p$ and $q.$ However, this equation in general does not hold.
What is true is that
$$\left( \cos x + i\sin x\right)^{\frac{p}{q}} = \left( \left( \cos x + i\sin x\right)^{p} \right)^{\frac{1}{q}} = \left( \cos \left(px\right) + i\sin \left(px\right) \right) ^{\frac{1}{q}}$$
but unfortunately, in general,
$$\left( \cos \left(px\right) + i\sin \left(px\right) \right) ^{\frac{1}{q}} \neq \cos \left(\frac{p}{q}x\right) + i\sin \left(\frac{p}{q}x\right).$$
A: De Moivre's formula only states that
$$
(\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin(n\theta)
$$
when $n$ is an integer. So, you have no reason to assume that the equality $(\cos(2k\pi)+i\sin(2k\pi))^\pi=\cos(2k\pi^2)+i\sin(2k\pi^2)$ holds.
