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

α_{1}, α_{2}, …, α_{n}

is a set of distinct algebraic numbers, then their exponentials

exp(α_{1}), exp(α_{2}), …, exp(α_{n})

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 π*i* 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(π*i*) and exp(0) are linearly independent. But these two numbers are not independent because

exp(π*i*) + 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.