Sum and product of a rational and irrational number 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.
 A: 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.
A: You're fine with the second part, but maybe this will help for the first part (Arpit's hint is really what you want do here; I'm just filling in the details):
Claim: Suppose $a$ is rational and $b$ is irrational. Then $a+b$ is irrational. 
Proof. Suppose $a+b$ is rational (proof by contradiction). Then
$$
a+b=\frac{p}{q},\qquad p,q\in\mathbb{Q}\space\text{(and $q\neq 0$)}.
$$
Thus, we have the following:
$$
a+b=\frac{p}{q}\Longleftrightarrow b=\frac{p}{q}-a\Longleftrightarrow b=\frac{p-a}{q}\Longleftrightarrow b=\frac{m}{q},
$$
where $m=p-a$ and $m$ is a rational number due to closure of addition/subtraction in $\mathbb{Q}$. Thus, $b\in\mathbb{Q}$, but this is a contradiction. $\Box$
