Exercise 2.3

Question

Exercise 3: Prove that there exist real numbers which are not algebraic.

ghostofgarborg · Accepted Answer

This must be the case, as the set of real numbers otherwise would be countable.

Amit Awasthi · Answer

\begin{proof}
      By Exercise 2.2, we know the set of all algebraic complex numbers is countable. Since the set of algebraic real numbers is a subset of this countable set, we know that it is also countable by Theorem 2.8. This means that the complement set $A^c$ of nonalgebraic real numbers is not countable, since its union with the set of algebraic real numbers has to equal all of the reals, which are uncountable by Theorem 2.43 ($\mathbb{R} = A \cup A^c$). Thus, nonalgebraic numbers ($A^c$) exist in $\mathbb{R}$.
    \end{proof}

Exercise 2.3

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Exercise 2.3

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