# Pi is irrational

^{iπ}+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.

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

Pi is irrationalby Chris Zuber is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License

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Pi is irrational