Question

prove that there exist real numbers which is not algebraic

Answer 1

Prove that there exist real numbers which are not algebraic.

A complex number $z$ is said to be algebraic if there are integers $a_0, \ldots, a_n$, not all zero, such that

$$a_0 z^n + a_1 z^{n-1} + \ldots + a_{n-1^z} + a_n = 0$$

Prove that the set of all algebraic numbers is countable. Hint: For every positive integer $N$ there are only finitely many equations with

$$n + |a_0| + |a_1| + \ldots + |a_n| = N.$$

Proof: For every positive integer $N$ there are only finitely many equations with

$$n + |a_0| + |a_1| + \ldots + |a_n| = N.$$

(since $1 \leq n \leq N$ and $0 \leq |a_0| \leq N$). We collect those equations as $C_N$. Hence $\cup C_N$ is countable. For each algebraic number, we can form an equation and this equation lies in $C_M$ for some $M$ and thus the set of all algebraic numbers is countable.

Proof: If not, $R^1 = \{ \text{all algebraic numbers} \}$ is countable, a contradiction.