Overpowered proof that π is transcendental

There is no polynomial with rational coefficients that evaluates to 0 at π. That is, π is a transcendental number, not an algebraic number. This post will prove this fact as a corollary of a more advanced theorem. There are proof that are more elementary and direct, but the proof given here is elegant.

A complex number z is said to be algebraic if it is the root of a polynomial with rational coefficients. The set of all algebraic numbers forms a field F.

The Lindemann-Weierstrass theorem says that if

α1, α2, …, αn

is a set of distinct algebraic numbers, then their exponentials

exp(α1), exp(α2), …, exp(αn)

are linearly independent. That is, no linear combination of these numbers with rational coefficients is equal to 0 unless all the coefficients are 0.

Assume π is algebraic. Then πi would be algebraic, because i is algebraic and the product of algebraic numbers is algebraic.

Certainly 0 is algebraic, and so the Lindemann-Weierstrass theorem would say that exp(πi) and exp(0) are linearly independent. But these two numbers are not independent because

exp(πi) + exp(0) = -1 + 1 = 0.

So we have a proof by contradiction that π is not algebraic, i.e. π is transcendental.

I found this proof in Excursions in Number Theory, Algebra, and Analysis by Kenneth Ireland and Al Cuoco.