Algebra Solutions: The rational numbers with the square root of 2 adjoined form a group

Let $G = \{ a+b \sqrt{2} \in \mathbb{R} \ |\ a,b \in \mathbb{Q} \}$ .

Prove that $G$ is a group under addition.
Prove that the nonzero elements of $G$ are a group under multiplication. (“Rationalize the denominators” to find multiplicative inverses.)

- 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 $a+b\sqrt{2} \in G$ . 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 is a group under addition.
- 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 is a group under multiplication.

Algebra Solutions

The rational numbers with the square root of 2 adjoined form a group

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