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This question already has an answer here:

So, most of us are familliar with Euler’s equation stating that $e^{i\pi}+1=0$. But I was wondering: how can an irrational number to the power of another irrational number equal a whole integer? And if that works, then how can $e^{i\pi}+1=0$ if $i$ isn’t even real?

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marked as duplicate by GNUSupporter 8964民主女神 地下教會, Gregory J. Puleo, Delta-u, José Carlos Santos, Rob Arthan May 3 '18 at 21:22

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    $\begingroup$ math.stackexchange.com/questions/341214/proof-that-ei-pi-1/… $\endgroup$ – poyea May 3 '18 at 21:15
  • $\begingroup$ Check out my blog article on the subject: dsprelated.com/showarticle/754.php $\endgroup$ – Cedron Dawg May 3 '18 at 21:16
  • $\begingroup$ To can appreciate deeply at all this equation you need to study trigonometry, complex numbers and Taylor's series. $\endgroup$ – gimusi May 3 '18 at 21:21
  • $\begingroup$ Why shouldn't an irrational power of an irrational number be an integer? And why shouldn't $e^{i\pi} + 1$ be real even though $i$ is purely imaginary? Please clarify what you don't understand about these facts? $\endgroup$ – Rob Arthan May 3 '18 at 21:22
  • $\begingroup$ Darn. I’ve studied none of those except complex numbers, but just a bit. $\endgroup$ – Detmondyou May 3 '18 at 21:22
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How can an irrational number to the power of another irrational number equal a whole integer?

You mean like $e^{\ln 2}=2$? Mind you, that's still a positive integer, unlike $-1$.

And if that works, then how can $e^{i\pi}+1=0$ if isn’t even real?

Well, $i$ is the reason the exponential doesn't have to be positive.

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Taking a slightly different approach to the other answers, because I think your confusion is elsewhere. You asked

How can an irrational number to the power of another irrational number equal a whole integer?

But this is entirely possible, even for real numbers. For instance, both $a = \sqrt{10}$ and $b = \log_{10} 4$ are irrational, but $a^b = \sqrt{10}^{\log_{10} 4}=\sqrt{10}^{2 \log_{10} 2} = 10^{\log_{10}2} = 2$, which is an integer.

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  • $\begingroup$ Yes, but you forget, $10^x$ counters $log_{10}n$. That’s like if I used the example $\sqrt{2}^2$ $\endgroup$ – Detmondyou Jun 6 '18 at 10:59
  • $\begingroup$ That's exactly the point, and $e^{i x}$ counters $\pi$ in the same way. $\endgroup$ – B. Mehta Jun 6 '18 at 11:01
  • $\begingroup$ Oh, ok. (Also realized this equation uses radians instead of degrees...) $\endgroup$ – Detmondyou Jun 6 '18 at 11:02
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It all boils down to Euler's identity

$$e^{i\theta}=\cos \theta + i \sin \theta\implies e^{i\pi}=\cos \pi + i \sin \pi=-1$$

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Since $e^{i\pi}$ = $\cos(\pi) + i\sin(\pi)$ and $\sin(\pi) = 1$, $\cos(\pi) = -1$ you get

\begin{equation} e^{i\pi}+1 = 0 \end{equation}

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