Answer to Question #285266 in Abstract Algebra for Pavan

Question

real of the form a + b 2 , a; b ∈ Z form s a ring.

Answer 1

Solution:

Let $Q[\sqrt{2}]=\{a+b \sqrt{2} \mid a, b \in Q\}$ . Note that $Q[\sqrt{2}] \subset R$ , and the operations of + and . on $Q[\sqrt{2}]$ are the usual + and . of real numbers. Not only is $Q[\sqrt{2}]$ closed under + and . but $Q[\sqrt{2}]$ is a field (a subfield of R ).

- $Q[\sqrt{2}]$ is closed under addition. $(a+b \sqrt{2})+(c+d \sqrt{2})=(a+c)+(b+d) \sqrt{2}$ .

- $Q[\sqrt{2}]$ is closed under multiplication. $(a+b \sqrt{2})(c+d \sqrt{2})=a c+a d \sqrt{2}+b c \sqrt{2}+ b d \sqrt{2}^{2}=(a c+2 b d)+(a d+b c) \sqrt{2}$ .

- Addition is associative and commutative on $Q[\sqrt{2}]$, since it is associative and commutative in R.

- Identity for addition: $0=0+0 \sqrt{2} \in Q[\sqrt{2}].$

- Inverses for addition: The inverse of $a+b \sqrt{2}$ is $-(a+b \sqrt{2})=-a+-b \sqrt{2} \in Q[\sqrt{2}]$ .

- Multiplication is associative and commutative on $Q[\sqrt{2}]$ , since it is associative and commutative in R.

- Distributivity holds in $Q[\sqrt{2}]$ , since it holds in R.

- Identity for multiplication: $1=1+0 \sqrt{2} \in$ $Q[\sqrt{2}]$.

- Inverses for multiplication: Given $a, b \in Q$ with $a+b \sqrt{2} \neq 0$ ,( either $a \neq 0\ or\ b \neq 0$ ), we need to find $c, d \in Q$ such that $(a+b \sqrt{2})(c+d \sqrt{2})=1$ .

Hence, $Q[\sqrt{2}]$ is a ring.