If we have $a$ as some rational number, and $b$ as some irrational number, then are the following two always true?
i) $a+b$ is always irrational
ii) $ab$ is always irrational
I know for ii) it is false, we can let $a = \frac{0}{1}$ for example, and then $ab \not \in \mathbb{I}$. But For $a \not = 0$ this is true however. Although my book does not specify any restriction on $a$.
I feel i) is always true.
Is this the result of rational numbers being closed under addition and multiplication? Or is there something else at play.
The proof is not too difficult by contradiction, I know. I am just wondering if there is a "reason" per se.
Best Answer
You are correct that (ii) is false if $a = 0$ and (ii) is true if $a \neq 0$. Indeed, this is because $\mathbb Q$ is closed under (nonzero) divisions. If we suppose instead that $ab \in \mathbb Q$, then since $a \in \mathbb Q \setminus \{0\}$, it follows that $b = \frac{ab}{a} \in \mathbb Q$, a contradiction.
Likewise, (i) is true because $\mathbb Q$ is closed under subtractions. If we suppose instead that $a+b \in \mathbb Q$, then since $a \in \mathbb Q$, it follows that $b = (a+b) - a \in \mathbb Q$, a contradiction.