Where is the error in this proof for showing that $f:\mathbb{R}\rightarrow\mathbb{R},x\mapsto\sin{(2\pi x)}$ is constant?

Question

Show that $f:\mathbb{R}\rightarrow\mathbb{R},x\mapsto\sin{(2\pi x)}$ is constant.

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$$1=1^x=(e^{2\pi i})^x=e^{2\pi ix}=\cos{(2\pi x)}+i\sin{(2\pi x)}$$ Where is the error in this "proof"? Does it not hold for all $x\in\mathbb{R}$, if yes which ones?

Answer 1

The mistake is $$ \left(e^{2 \pi i }\right)^{x}=e^{2 i \pi x} $$ holds only for $x \in \mathbb{Z}$.

Answer 2

Notice that $(-2)^2 = 4$. So $[(-2)^2]^{\frac 12} = 4^{\frac 12} = 2$.

But $[(-2)^2]^{\frac 12} = (-2)^{2\frac 12} = (-2)^1 = -2 \ne 2$.

What went wrong?

The assumption that $(a^m)^n = a^{mn}$ only holds if either $m, n \in \mathbb Z$ or if $a > 0; a \in \mathbb R; m,n \in \mathbb R$.

If we define $a^m = a*.... *a; m$ times then $(a^m)^n = a^{mn}$ follows by associativity.

If we define $b^{\frac mn} = \sqrt[n]{b^m}; b > 0; \frac mn \in \mathbb Q$ then $(b^q)^r = b^{qr}$ follows, by putting the fractions under a common denominator. This reduces to a radical base and integer powers.

If we define $b^x = \lim_{q\to x} b^q; b > 0$ then $(b^x)^y = b^{xy}$ can be deduced by the continuity of limits.

But when we define a fundimentally different definite for complex powers as $e^{iy} = \cos y + i \sin y$ then $(e^{iy})^z = (\cos y + i \sin y)^z \ne \cos yz + i \sin zy= e^{iyz}$ just plain doesn't follow.