Exercise 1.4.1

Question

Recall that $\mathbf{I}$ stands for the set of irrational numbers.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that if $a, b \in \mathbf{Q}$, then $a b$ and $a+b$ are elements of $\mathbf{Q}$ as well.
  \item Show that if $a \in \mathbf{Q}$ and $t \in \mathbf{I}$, then $a+t \in \mathbf{I}$ and $a t \in \mathbf{I}$ as long as $a 
eq 0$.
  \item Part (a) can be summarized by saying that $\mathbf{Q}$ is closed under addition and multiplication. Is $\mathbf{I}$ closed under addition and multiplication? Given two irrational numbers $s$ and $t$, what can we say about $s+t$ and $s t$?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Trivial.
    \item Suppose $a + t \in \mathbf{Q}$, then by (a) $(a + t) - a = t \in \mathbf{Q}$ contradicting $t \in \mathbf{I}$.
    \item $\mathbf{I}$ is not closed under addition or multiplication. consider $(1 - \sqrt 2) \in \mathbf{I}$ by (b), and $\sqrt{2} \in \mathbf{I}$. the sum $(1-\sqrt2)+\sqrt2 = 1 \in \mathbf{Q} 
otin \mathbf{I}$. Also $\sqrt 2 \cdot \sqrt 2 = 2 \in \mathbf Q 
otin \mathbf I$.
   \end{enumerate}

Exercise 1.4.1

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Exercise 1.4.1

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