Is pi algebraic?

No, pi (π) is not algebraic; it is transcendental.

Here's why:

A number is algebraic if it is a root of a non-zero polynomial equation with rational (or integer) coefficients. For example, the square root of 2 is algebraic because it's a root of the polynomial equation x² - 2 = 0.

A number is transcendental if it is not algebraic. In other words, it cannot be a root of any polynomial equation with rational coefficients.

The proof that pi is transcendental is complex and relies on advanced mathematical concepts. One approach involves the Lindemann–Weierstrass theorem. A simplified explanation (as mentioned in the references) involves a proof by contradiction:

1. Assume pi is algebraic.
2. If pi is algebraic, then pi multiplied by the imaginary unit i (πi) would also be algebraic.
3. The Lindemann–Weierstrass theorem states that if a is a non-zero algebraic number, then e^a is transcendental.
4. Therefore, e^πi would be transcendental.
5. However, Euler's identity states that e^πi = -1.
6. -1 is a rational number, and all rational numbers are algebraic.
7. This creates a contradiction: e^πi cannot be both transcendental and algebraic.
8. Therefore, the initial assumption that pi is algebraic must be false.

Because pi is not algebraic, it is transcendental. This was rigorously proven by Ferdinand von Lindemann in 1882.