Transcendence

As we have already noted, finding rational solutions of polynomial equations is an important problem in number theory. This suggests a further distinction among numbers. We shall show that the number $\alpha=\sqrt{2}$ is irrational; however, it satisfies the polynomial equation $\alpha^2-2=0$, which has integral coefficients. We call a number $\alp$ ``algebraic'' if it satisfies a polynomial equation

$$c_0 \alp^n + c_1 \alp^{n-1} + \cdots + c_{n-1} \alp + c_n=0$$

in which all the coefficients c_i are integers. We call the number $\alp$ ``transcendental'' if it is not algebraic. It has been established that $\pi$, e, $2^{\sqrt{2}}$, and $e^\pi$ are all transcendental. Proofs of transcendence are generally far more difficult than those of just irrationality.