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The equation $\sin{x} = x^2$ has a unique positive real root. I wonder if there is any standard technique how to show that this number is irrational (rational), preferably a technique which works also in other similar scenarios.

I tried an inverse symbolic calculator on the numerical approximation of the root, but it didn´t find anything whatsoever.

JaCkO91
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Adam
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4 Answers4

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In fact $x$ is irrational. See Corollary 2.7 in Niven's book. If the root $x$ in question was rational, then $\sin(x)$ is irrational and so $\sin(x)-x^2$ could not be zero (since this would be an irrational minus a rational).

Casteels
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It is known that $\sin(1)$ is transcendental (see Wikipedia), and it follows that $\sin(r)$ must be irrational and even transcendental for any nonzero rational number $r$ (since there is a nontrivial algebraic relation between $\sin(r)$ and $\sin(1)$). Hence $\sin(r) = r^2$ cannot hold for any nonzero $r$.

user43208
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If you plot the graph of sin x and $x^2$ you will see that it intersect at two point x=0 and the equation can also be written as $\frac{e^{-ix}-e^{ix}}{2}=x^2 $ by Euler form of complex number . So other root will be irrational .

jimjim
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Surya
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    That doesn't necessarily hold. $x=0$ is also a solution to this (since it's just a rearrangement of the original equation) and it is not irrational – Dan Oct 08 '13 at 12:45
  • I am using $x is not n\pi$ to rearrange the equation – Surya Oct 08 '13 at 12:48
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    My last comment (deleted) was patently false, sorry. But it's not clear that every value of $\sin(x)$ s.t. $x\neq n\pi$ is irrational if $x \in \mathbb Q$. If so you should either prove it or show where it is proved. – Dan Oct 08 '13 at 12:54
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It can be shown that the sines and cosines of the acute angles in a pythagorean triangle are irrational. So know if $x$ would be rational ...

Michael Hoppe
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