Rational Points on Graph of Exponential Function

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