# Pi is irrational

After writing my MathML demo, I wanted to make a quick note of something that I found kinda interesting. First, it uses Euler's Identity (e +1 = 0), written in terms of pi, which results in a fraction of two number which simply don't exist. Then, if you assume that pi can be expressed as a fraction if integers (a/b), it leads to what seems to be a contradiction.

Pi has already been proven to be irrational in many ways, but this one might be the simplest, if it qualifies as a proof at all. The conclusion seems impossible to me, but might not actually be a contradiction.

$\begin{array}{c}{e}^{i\mathrm{\pi }}+1=0\\ {e}^{i\mathrm{\pi }}=-1\\ \mathrm{ln}\left({e}^{i\mathrm{\pi }}\right)=\mathrm{ln}\left(-1\right)\\ i\mathrm{\pi }=\mathrm{ln}\left(-1\right)\\ \mathrm{\pi }=\frac{\mathrm{ln}\left(-1\right)}{i}\\ \mathrm{\pi }=\frac{\mathrm{ln}\left(-1\right)}{\sqrt{-1}}\\ \\ a\in \mathrm{ℤ},b\in \mathrm{ℤ}\\ \mathrm{\pi }=\frac{a}{b}\\ \frac{\mathrm{ln}\left(-1\right)}{\sqrt{-1}}=\frac{a}{b}\\ \therefore a×\sqrt{-1}=b×\mathrm{ln}\left(-1\right)\end{array}$