Proof for the irrationality of $eπ$ ('complex'ly?).*

Question

Some of you may have seen the video:

https://youtu.be/DLWpj34UNRk

It is a proof of the irrationality of $eπ$ by Ron (14years) , presented by 'blackpenredpen' Youtuber.

The proof goes something like this:

We know that $e^{iπ}=-1$ ("Euler's famous identity").

$e^{iπ}=i^2$

So, $ π=\frac{2\ln{i}}{i}$

We must prove that $eπ$ is irrational.

Let us assume, on the contrary, that $eπ$ is rational.

$eπ=\frac{a}{b}$ where $a$ and $b$ are rational numbers. $$eπ=e\frac{2\ln{i}}{i}=\frac{a}{b}$$ $$2eb\ln{i}=ai$$ $$\ln{i^{2eb}}=ai$$ $$\ln{(-1)^{eb}}=ai$$ $$(-1)^{eb}=e^{ai}$$ Squaring on both sides, $$1^{eb}=e^{2ai}$$ $$1=e^{2ai}$$ $$e^0=e^{2ai}$$ When the bases are identical, the powers are equal. $$0=2ai$$ $$a=0$$ Hence, $$eπ=\frac{0}{b}$$ $$eπ=0$$ This is not possible and therefore gives us a contradiction.

This contradiction has arisen because of our incorrect assumption the $eπ$ is rational.

We conclude that $eπ$ is irrational.

(*) Is this proof legitimate? Or is there something dubious about it? If so, please point it out.

Answer 1

This proof reminds me of a very famous mathematical paradox concerning complex logarithm: $$e^0=e^{2\pi i}$$

$$\ln(e^0)=\ln(e^{2\pi i})$$ $$0=2\pi i$$

In fact, we shall use only one branch of the complex logarithm, and stay with it always. There is no single branch of logarithm defined for both input of argument $0$ and $2\pi$.

Answer 2

Look at the proof starting here:

$e \pi =\frac{a}{b}$ where $a$ and $b$ are relatively prime integers.

The rest of the proof never uses that $a,b$ are integers. If you replace $a,b$ by real numbers, the rest of the "proof" still holds.

Actually, if you write at this point in the proof $e \pi =\frac{a}{b}$ where $a=e \pi $ and $b=1$, the rest of the "proof" still derives a contradiction.

Answer 3

The first mistake I see is the claim that $2eb\ln{i}\stackrel{?}{=}\ln{i^{2eb}}$. If we're taking the principal branch of the logarithm and power functions, there's no reason for that to be true! Consider the simplest nontrivial example: $b=1$. The left-hand side is just:

$$2e\ln i=e\pi i$$

But the right-hand side is:

$$\ln{i^{2e}}=\ln e^{2e\ln i}=\ln{e^{e\pi i}}=e\pi i+2n\pi i=(e+2n)\pi i$$

...where $n$ is the integer that makes the whole expression as close as possible to $0$. Since $2<e<3$, this will be $n=-1$, so: $$\ln{i^{2e}}=(e-2)\pi i\neq e\pi i.$$

As Eric Wofsey points out, there are other errors after that, which only compound the incorrectness of the argument.