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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1https://math.stackexchange.com/questions/341214/proof-that-ei-pi-1/341228 – poyea May 03 '18 at 21:15
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Check out my blog article on the subject: https://www.dsprelated.com/showarticle/754.php – Cedron Dawg May 03 '18 at 21:16
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To can appreciate deeply at all this equation you need to study trigonometry, complex numbers and Taylor's series. – user May 03 '18 at 21:21
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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? – Rob Arthan May 03 '18 at 21:22
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Darn. I’ve studied none of those except complex numbers, but just a bit. – May 03 '18 at 21:22
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@Detmondyou Then you will appreciate better that formula in the future! – user May 03 '18 at 21:23
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You should also get a kick out of: $e^{i} = i^{2/\pi}$ A real number raised to an imaginary value which equals an imaginary value raised to a real power. – Cedron Dawg May 03 '18 at 21:31
4 Answers
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It all boils down to Euler's identity
$$e^{i\theta}=\cos \theta + i \sin \theta\implies e^{i\pi}=\cos \pi + i \sin \pi=-1$$
user
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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.
J.G.
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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}
wayne
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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.
B. Mehta
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Yes, but you forget, $10^x$ counters $log_{10}n$. That’s like if I used the example $\sqrt{2}^2$ – Jun 06 '18 at 10:59
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That's exactly the point, and $e^{i x}$ counters $\pi$ in the same way. – B. Mehta Jun 06 '18 at 11:01
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