In Rudin's Principles of Mathematical Analysis, exercise number two is to prove that the set of all algebraic numbers is countable. A complex number z is said to be algebraic if there are integers a_0, ..., a_n, not all zero such that a_0*z^n + a_1*z^(n-1) + ... + a_(n-1)*z + a_n = 0 The hint given in the book is: For every positive integer N there are only finitely many equations with n + |a_0| + |a_1| + ... + |a_n| = N. I understand the gist of how the proof works. The above equation partitions the algebraic numbers into a countable collection of finite sets, proving that the algebraic numbers are countable. But my question is this: Couldn't we partition the algebraic numbers simply with the equation: |a_0| + |a_1| + ... + |a_n| = N Why add in the n when the each algebraic equation is fully characterized by the values a_0, a_1, ..., a_n? Or am I missing something?