Let $G = \{ a+b \sqrt{2} \in \mathbb{R} \ |\ a,b \in \mathbb{Q} \}$.
(1) Prove that $G$ is a group under addition.
(2) Prove that the nonzero elements of $G$ are a group under multiplication. (“Rationalize the denominators” to find multiplicative inverses.)
Solution:
(1) Suppose $a+b\sqrt{2}$, $c+d\sqrt{2} \in G$. Then $$(a+b\sqrt{2}) + (c+d\sqrt{2}) = (a+c) + (b+d)\sqrt{2} \in G,$$ since $\mathbb{Q}$ is closed under addition. So $G$ is closed under addition. Addition on $G$ is associative since addition on $\mathbb{R}$ is associative. Notice that $0 = 0 + 0\sqrt{2}$, so $0 \in G$, and that $0 + z = z + 0 = z$ for all $z=a+b\sqrt{2} \in G$. Thus $0$ is an additive identity element of G. Let $a+b\sqrt{2} \in G$. Then $-a - b\sqrt{2} \in G$ and we have $$(a+b\sqrt{2}) + (-a -b\sqrt{2}) = (-a-b\sqrt{2}) + (a+b\sqrt{2}) = 0.$$ So every element of $G$ has an additive inverse in $G$. Hence $G$ is a group under addition.
(2) Suppose $a+b\sqrt{2}$, $c+d\sqrt{2} \in G$. Then $$(a+b\sqrt{2})(c+d\sqrt{2}) = (ac + 2bd) + (ad + bc) \sqrt{2} \in G,$$ since $\mathbb{Q}$ is closed under addition and multiplication. So $G$ is closed under multiplication. Multiplication on $G$ is associative since multiplication on $\mathbb{C}$ is associative. Notice that $1 = 1 + 0\sqrt{2}$, so $1 \in G$, and that $1 \cdot z = z \cdot 1 = z$ for all $z \in G$. Thus $1$ is a multiplicative identity element of $G$. Let $a+b\sqrt{2} \in G$, and note that in $\mathbb{C}$, we have $$\left( \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2} \right) = \frac{1}{a+b\sqrt{2}} \cdot \frac{a-b\sqrt{2}}{a-b\sqrt{2}} = \frac{1}{a+b\sqrt{2}}.$$ Then $G$ is closed under inversion (considered over $\mathbb{R}$), so that every element of $G$ has a multiplicative inverse in $G$. Hence $G \setminus \{0\}$ is a group under multiplication.
