Rational Points on Graph of Logarithm Function
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Theorem
Consider the graph of the logarithm function in the real Cartesian plane $\R^2$:
- $f := \set {\tuple {x, y} \in \R^2: y = \ln x}$
The only rational point of $f$ is $\tuple {1, 0}$.
Proof
Consider the graph of the exponential function in the real Cartesian plane $\R^2$:
- $g := \set {\tuple {x, y} \in \R^2: y = e^x}$
From Rational Points on Graph of Exponential Function, the only rational point of $g$ is $\tuple {0, 1}$.
By definition of the exponential function, $f$ and $g$ are inverses.
Thus:
- $\tuple {x, y} \in g \iff \tuple {y, x} \in f$
Thus for $\tuple {x, y} \in g$ to be rational, $\tuple {y, x} = \tuple {0, 1}$.
Hence the result.
$\blacksquare$
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational