Rational Points on Graph of Logarithm Function

From ProofWiki
Jump to navigation Jump to search

Theorem

Consider the graph of the logarithm function in the real Cartesian plane $\R^2$:

$f := \left\{ {\left({x, y}\right) \in \R^2: y = \ln x}\right\}$


The only rational point of $f$ is $\left({1, 0}\right)$.


Proof

Consider the graph of the exponential function in the real Cartesian plane $\R^2$:

$g := \left\{ {\left({x, y}\right) \in \R^2: y = e^x}\right\}$

From Rational Points on Graph of Exponential Function, the only rational point of $g$ is $\left({0, 1}\right)$.

By definition of the exponential function, $f$ and $g$ are inverses.

Thus:

$\left({x, y}\right) \in g \iff \left({y, x}\right) \in f$

Thus for $\left({x, y}\right) \in g$ to be rational, $\left({y, x}\right) = \left({0, 1}\right)$.

Hence the result.

$\blacksquare$


Sources