Some of you may have seen the video:


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").


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.

  • 5
    $\begingroup$ A simpler "proof" along the same lines: $e^{2\pi i}=1$, so $\pi=0$. $\endgroup$ – Eric Wofsey Mar 4 '18 at 5:04
  • 1
    $\begingroup$ "When the bases are identical, the powers are equal," is not true for complex numbers. In fact, the complex exponential is periodic with period $2\pi i$ $\endgroup$ – saulspatz Mar 4 '18 at 5:05
  • $\begingroup$ So, you are saying that $e^0=e^{2ai}$ does not imply $0=2ai$ $\endgroup$ – Ryan Scott Mar 4 '18 at 5:07
  • $\begingroup$ Since natural logarithm is ill defined in complex numbers, I am very uncomfortable with the derivation to $\pi=\frac{2\ln i}{i}$ by taking log on both sides. However, I’m not sure if this affects the correctness of the proof. $\endgroup$ – Szeto Mar 4 '18 at 5:08
  • 2
    $\begingroup$ There are at least 6 steps that could reasonably be called errors. Basically, none of the the familiar rules for exponents and logarithms work outside of positive real numbers, at least not without some caveats. $\endgroup$ – Eric Wofsey Mar 4 '18 at 5:13

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.


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.


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 $2\pi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.