Exercise 6.3.1 ($\mathbb{R} \setminus \mathbb{Q}$ is not closed under arithmetic operations)

Question

Prove that the irrational numbers are not closed under addition, subtraction, multiplication or division.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

\begin{proof}

\begin{itemize}
\item If $x = \sqrt{2},\ y = 1 - \sqrt{2}$, then $x$ and $y$ are irrational (see Problem 2), but $x+y = 1$ is rational.
\item If $x = \sqrt{2},\ y = \sqrt{2} - 1$, then $x$ and $y$ are irrational, but $x-y = 1$ is rational.
\item If $x = \sqrt{8}, \ y = \sqrt{2}$, then then $x$ and $y$ are irrational (by Corollary 6.15), but $xy = 4$ and $x/y = 2$ are rational.
\end{itemize}
 The set $\R \setminus \Q$ of irrational numbers is not closed under addition, subtraction, multiplication or division.
\end{proof}

Exercise 6.3.1 ($\mathbb{R} \setminus \mathbb{Q}$ is not closed under arithmetic operations)

Answers

Comments

Exercise 6.3.1 ($\mathbb{R} \setminus \mathbb{Q}$ is not closed under arithmetic operations)

Answers

Comments

Add answer