Rational Points on Graph of Exponential Function
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Theorem
Consider the graph $f$ of the exponential function in the real Cartesian plane $\R^2$:
- $f := \set {\tuple {x, y} \in \R^2: y = e^x}$
The only rational point of $f$ is $\tuple {0, 1}$.
Proof
From Exponential of Rational Number is Irrational:
- $r \in \Q_{\ne 0} \implies e^r \in \R - \Q$
Thus, apart from the point $\tuple {0, 1}$, when $x$ is rational, $e^x$ is not.
Hence the result.
$\blacksquare$
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational