3.2.1

Let $$S = \{a+b\sqrt2: a,b \in \mathbb{Q}\} $$, then we will show that together with the usual operations of addition and multiplication inherited from $$ \mathbb{R}$$, S forms a field. Note S is a subset of the real numbers.

First, recall that sums, products, additive inverses and multiplicative inverses of rational numbers are also rational numbers, since $$\mathbb{Q} $$ is a field. The addition operation on S is associative since it is associative in $$\mathbb{R} $$. Then, given $$a+b\sqrt2, c+d\sqrt2 \in S $$, we have $$ (a+b\sqrt2) + (c+d\sqrt2) = (a+c) + (b+d)\sqrt2 $$, using associativity of addition. Since $$ a+c$$ and $$ b+d$$ are both rational, this shows that S is closed under addition. The additive identity 0 is in S, since $$0 = 0 + 0\sqrt2 $$, and every element $$a+b\sqrt2$$ of S has an additive inverse $$-a-b\sqrt2 \in S $$. S is hence a group under addition, and is abelian because the addition operation is commutative in $$ \mathbb{R}$$.

Similarly, we can show that $$S/\{0\} $$ is closed under multiplication by observing that given two nonzero elements of S, $$(a+b\sqrt2)(c+d\sqrt2) = (ac + 2bd) + (ad+bc)\sqrt2 $$. Multiplication is associative in $$\mathbb{R}$$ and hence in S, the multiplicative identity 1 is in S since $$1 = 1 + 0\sqrt2 $$, and every nonzero element $$a+b\sqrt2 $$ of S has a multiplicative inverse $$(a-b\sqrt2)/(a^2 - 2b^2) $$. Note that if a and b are not both 0, then$$a^2 - 2b^2 \neq 0 $$, otherwise we could write $$\sqrt2 = \sqrt{a^2/b^2} $$, implying $$\sqrt2 $$ is rational, a contradiction, thus the inverse is well-defined for nonzero elements. $$S/\{0\} $$ is a group under multiplication, and is abelian since multiplication commutes in $$\mathbb{R} $$.

Finally, addition and multiplication are compatible with the distributive law since they are distributive in $$\mathbb{R} $$, showing S is a field.