Question #285266

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

1
Expert's answer
2022-01-10T12:36:17-0500

Solution:

Let Q[2]={a+b2a,bQ}Q[\sqrt{2}]=\{a+b \sqrt{2} \mid a, b \in Q\} . Note that Q[2]RQ[\sqrt{2}] \subset R , and the operations of + and . on Q[2]Q[\sqrt{2}] are the usual + and . of real numbers. Not only is Q[2]Q[\sqrt{2}] closed under + and . but Q[2]Q[\sqrt{2}] is a field (a subfield of R ).


- Q[2]Q[\sqrt{2}] is closed under addition. (a+b2)+(c+d2)=(a+c)+(b+d)2(a+b \sqrt{2})+(c+d \sqrt{2})=(a+c)+(b+d) \sqrt{2} .


- Q[2]Q[\sqrt{2}] is closed under multiplication. (a+b2)(c+d2)=ac+ad2+bc2+bd22=(ac+2bd)+(ad+bc)2(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[2]Q[\sqrt{2}], since it is associative and commutative in R.


- Identity for addition: 0=0+02Q[2].0=0+0 \sqrt{2} \in Q[\sqrt{2}].


- Inverses for addition: The inverse of a+b2a+b \sqrt{2} is (a+b2)=a+b2Q[2]-(a+b \sqrt{2})=-a+-b \sqrt{2} \in Q[\sqrt{2}] .


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


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


- Identity for multiplication: 1=1+021=1+0 \sqrt{2} \in Q[2]Q[\sqrt{2}].


- Inverses for multiplication: Given a,bQa, b \in Q with a+b20a+b \sqrt{2} \neq 0 ,( either a0 or b0a \neq 0\ or\ b \neq 0 ), we need to find c,dQc, d \in Q such that (a+b2)(c+d2)=1(a+b \sqrt{2})(c+d \sqrt{2})=1 .

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


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