Exercise 1.1

Question

Exercise 1: if $r$ is rational ($r 
eq 0$) and $x$ is irrational, prove that $r + x$ and $rx$ are irrational.

ghostofgarborg · Accepted Answer

Note that $\mathbb{Q}$ is closed under the arithmetic operations of addition, subtraction, multiplication and taking multiplicative inverses.
If $r+x$ were rational, so would $(r+x) - r = x$ be, a contradiction. If $rx$ were rational, so would $\frac 1 r rx= x$ be, a contradiction.

Amit Awasthi · Answer

\begin{proof}[Proof that $r+x \in \mathbb{Q}^c$]
      Assume, to get a contradiction, that $r+x \in \mathbb{Q}$. Then, $r = a/b$, $r + x = c/d$ for $a,b,c,d \in \mathbb{Z}$, by the definition of rationals. So, $a/b + x = c/d$, and $x = (bc - ad)/bd \in \mathbb{Q}$, which is a contradiction since $x \in \mathbb{Q}^c$.
    \end{proof}
    \begin{proof}[Proof that $rx \in \mathbb{Q}^c$]
      Assume, to get a contradiction, that $rx \in \mathbb{Q}$. Then, $r = a/b$, $rx + c/d$ for $a,b,c,d \in \mathbb{Z}$, by the definition of rationals. So, $ax/b = c/d$, and $x = bc/ad \in \mathbb{Q}$, which is a contradiction since $x \in \mathbb{Q}^c$.
\end{proof}

Exercise 1.1

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Exercise 1.1

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