Exercise 2.4

Question

Exercise 4: Is the set of all irrational real numbers countable?

ghostofgarborg · Accepted Answer

No. We know the set of rational numbers is countable. Since 
    $\mathbb R = \mathbb Q \cup (\mathbb R \setminus \mathbb Q)$ and 
    $\mathbb R$ is uncountable, we must have that $(\mathbb R \setminus \mathbb Q)$ is uncountable
    by theorem 2.12.

Amit Awasthi · Answer

\begin{claim}
      The set of all irrational real numbers is not countable.
    \end{claim}
    \begin{proof}
      Assume the set of all irrational real numbers $\mathbb{Q}^c$ is countable. By the definition of irrational numbers, $\mathbb{R} = \mathbb{Q} \cup \mathbb{Q}^c$. $\mathbb{Q}$ is countable by the Corollary of Theorem 2.13. Then, by Theorem 2.12, $\mathbb{R}$ is a union of two countable sets, and is therefore countable. However, this is a contradiction since $\mathbb{R}$ is uncountable by the Corollary of Theorem 2.43.
    \end{proof}

Exercise 2.4

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Exercise 2.4

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