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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.

lulu
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Indianimperialist123
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    I'm not convinced that De-Moivre's can be used in this manner, but I'm sure someone will chime in for a definitive answer. This post and also this post and also this, which might be helpful. – pyridoxal_trigeminus May 13 '22 at 17:29
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    Also, you would be well-advised to format your question in MathJax. – pyridoxal_trigeminus May 13 '22 at 17:33
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    De-Moivre's theorem fails for rationals such as $1/2$. I believe your proof does show however that $\pi$ is not an integer. – felipeh May 13 '22 at 17:33
  • Just to say, even though this proof failed it's a nice observation to come up with and if nothing else, it's an interesting sense in which $\pi$'s irrationality is somewhat nontrivial: if De Moivre had a rational extension then we could prove it like the above, but it doesn't because complex numbers are numerous so arbitrary powers of complex numbers can be multivalued. – Isky Mathews May 13 '22 at 17:45
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    I have no idea why anyone would downvote this. It's an entirely sensible question. And the OP plainly recognizes that it's unlikely to hold water and is just asking for someone to spot the leak. Which, really, is not obvious. Granted it ought to be properly formatted, but it is a good question (+1). – lulu May 13 '22 at 17:55
  • Note: I reformatted your post, please check to make sure I didn't introduce any errors. – lulu May 13 '22 at 18:04

2 Answers2

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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,

$$\left( \cos \left(px\right) + i\sin \left(px\right) \right) ^{\frac{1}{q}} \neq \cos \left(\frac{p}{q}x\right) + i\sin \left(\frac{p}{q}x\right).$$

Adam Rubinson
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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.

Another User
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    No, but the result holds for rational exponents with the expression being multivalued. E.g. if you were to raise to the power a/b then the expression can have b values with the result holding for the principal value. – Indianimperialist123 May 13 '22 at 17:40