Solutions

(a) Well that's false, regardless of the given rational $a$ and irrational $b$. We shall prove that by contradiction. Since $a$ is rational, we know that $-a$ also is rational. Now, let us assume that the sum $a + b$ also is rational. Thus, since the rational numbers are closed under addition and subtraction,

$$\begin{aligned} (a + b) - a \in \mathbb{Q} \implies b \in \mathbb{Q} \end{aligned}$$

Which clearly contradicts our premise that $b$ is irrational.

(b) This amounts to determining whether the irrational numbers are closed under addition and employs a similar strategy to the previous proof. We proceed with a proof by contradiction by first assuming that $a,b \in \mathbb{Q}^{*} \implies a + b \in \mathbb{Q}$.

With that assumption, we can then also assume that $a + 2b \in \mathbb{Q}$ since $2b \in \mathbb{Q}^{*}$. As a result, if we subtract the former sum from the latter, we obtain the following contradiction:

$$\begin{aligned} (a + 2b) - (a + b) = b \in \mathbb{Q}^{*} \end{aligned}$$

Again, that is because the rational numbers are closed under addition/subtraction.

(c) Let $a \in \mathbb{Q}\wedge b \in \mathbb{Q}^{*}$ and suppose $ab \in \mathbb{Q}^{*}$. Thus, if $a \neq 0$,

$$\begin{aligned} (ab) \times a^{-1} \in \mathbb{Q}\\ b \in \mathbb{Q} \end{aligned}$$

We therefore have a contradiction in the set to which $b$ belongs. Now, if $a = 0$, then $ab = 0$. In this case, $\mathbb{Q}\ni ab \notin \mathbb{Q}^{*}$.

(d) $\exists \; a: a^{2} \in \mathbb{Q}^{*}\wedge a^{4} \in \mathbb{Q}$