Transcendental Slope

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The slope of a line may be transcendental.


The slope form of any number $x$ may be produced by:

\(\ds {\mathrm m}\) \(=\) \(\ds \frac {x} {1}\) (Slope Form of $x$)
\(\ds {\mathrm m}\) \(=\) \(\ds {x}\)

If $x$ is transcendental, then the slope of a line $\mathrm m$ is transcendental.


$\mathrm \pi$ is proven to be transcendental by the Lindemann-Weiersrass Theorem

\(\ds {\mathrm m}\) \(=\) \(\ds \frac {\mathrm \pi} {1}\) (Slope Form of $\pi$)
\(\ds {\mathrm m}\) \(=\) \(\ds {\mathrm \pi}\)

Slope $m$ is transcendental.