# Pi is Transcendental

## Theorem

## Proof

Aiming for a contradiction, suppose $\pi$ is not transcendental.

Hence by definition, $\pi$ is algebraic.

Let $\pi$ be the root of a non-zero polynomial with rational coefficients, namely $\map f x$.

Then, $\map g x := \map f {i x} \map f {-i x}$ is also a non-zero polynomial with rational coefficients such that:

- $\map g {i \pi} = 0$

Hence, $i \pi$ is also algebraic.

From the Weaker Hermite-Lindemann-Weierstrass Theorem, $e^{i \pi}$ is transcendental.

However, from Euler's Identity:

- $e^{i \pi} = -1$

which is the root of $\map h z = z + 1$, and so is algebraic.

This contradicts the conclusion that $e^{i \pi}$ is transcendental.

Hence by Proof by Contradiction it must follow that $\pi$ is transcendental.

$\blacksquare$

## Historical Note

The transcendental nature of $\pi$ (pi) was investigated without success by Joseph Liouville in $1844$, at around the time he conjectured that Euler's number $e$ was likewise transcendental.

His ideas contributed towards work done by Charles Hermite, who proved in $1873$ that Euler's number $e$ is transcendental, but had not noticed that it was a short step from there, via Euler's Identity $e^{i \pi} + 1 = 0$, that $\pi$ is transcendental:

*I shall risk nothing on an attempt to prove the transcendence of the number $\pi$. If others undertake this enterprise, no one will be happier than I at their success, but believe me, my dear friend, it will not fail to cost them some effort.*- -- Charles Hermite, in a letter to a friend

That final step was made by Ferdinand von Lindemann, who finally achieved this proof in $1882$.

Many people believed that Ferdinand von Lindemann was a grossly inferior mathematician to Hermite, and that he achieved this result by pure luck, and that it should have been Hermite who gained the credit for it.

However, be that as it may, it was indeed Ferdinand von Lindemann and not Hermite who made that actual step of reasoning, and the result falls fair and square at his feet.

## Sources

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