Exercise 2.2

We first check the closure law for \((\mathbb{Q}, +)\). Suppose we have two arbitrary elements of \(\mathbb{Q}\); we can write them as \(\frac{a}{b}\) and \(\frac{c}{d}\), where \(a, b, c, d \in \mathbb{Z}\).

Then:

\[\begin{split}\frac{a}{b} + \frac{c}{d} &= \frac{ad}{bd} + \frac{bc}{bd} \\ &= \frac{ad + bc}{bd}.\end{split}\]

We have an integer on the top and an integer on the bottom, so the result of adding these two values is in \(\mathbb{Q}\), and therefore the closure law is satisfied.

We could check associativity similarly to how we checked closure, but we already know that addition is associative for all real numbers; since the rational numbers are a subset of the real numbers, we can simply conclude that the associativity law holds for \((\mathbb{Q}, +)\).

The identity element in \((\mathbb{Q}, +)\) is 0, just like in \((\mathbb{Z}, +)\) and in \((\mathbb{R}, +)\).