Worked solution: Lindemann's transcendence proof of $\pi$ settles squaring the circle (1882)

Frequently asked questions

Is Euler's identity itself something that needed to be proved by Lindemann?

No — Euler's identity was already well established (from complex power series, decades before Lindemann) and is unconditionally true. Lindemann's actual new contribution, covered in the next step, is the harder statement that $e^\alpha$ cannot be algebraic for any nonzero algebraic $\alpha$ .

Source

Ferdinand von Lindemann, 1882

Über die Zahl π

Step 3 of 6: Setting up the contradiction with Euler's identity

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In plain words

To prove $\pi$ is transcendental, Lindemann used the classic move of proof by contradiction: assume the opposite, that $\pi$ is algebraic, and derive something absurd. Euler's famous identity $e^{i\pi}+1=0$ is the perfect lever for this, because it links $\pi$ directly to a power of $e$ .

If $\pi$ were algebraic, multiplying it by the algebraic number $i=\sqrt{-1}$ would give another algebraic number, $i\pi$ . Euler's identity then says $e$ raised to this algebraic power $i\pi$ equals $-1$ — itself algebraic. But Hermite's method, suitably extended, says that can never happen.

e^{i\pi} + 1 = 0

Detailed analysis

Assume, for contradiction, that $\pi$ is algebraic. Since $i=\sqrt{-1}$ is a root of $x^2+1=0$ , it is algebraic, and the product of two algebraic numbers is algebraic (their combined field extension over $\mathbb{Q}$ is finite), so $i\pi$ would be a nonzero algebraic number under this assumption.

Euler's identity, $e^{i\pi}+1=0$ , i.e. $e^{i\pi}=-1$ , is a theorem of complex analysis, true unconditionally regardless of what $\pi$ turns out to be: it follows from the power-series definitions of $e^x$ , $\cos x$ , and $\sin x$ , giving $e^{ix}=\cos x+i\sin x$ and hence $e^{i\pi}=\cos\pi+i\sin\pi=-1$ . Combined with the assumption that $\pi$ is algebraic, this would mean $e^\alpha=-1$ for the nonzero algebraic number $\alpha=i\pi$ , and $-1$ is itself algebraic (a root of $x+1=0$ ).

So the assumption 'π is algebraic' forces a very specific consequence: some nonzero algebraic number $\alpha$ (namely $i\pi$ ) has $e^\alpha$ algebraic. The next step shows this consequence is exactly what Lindemann proved impossible, closing the contradiction.