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+b2∣a,b∈Q} . Note that Q[2]⊂R , and the operations of + and . on Q[2] are the usual + and . of real numbers. Not only is Q[2] closed under + and . but Q[2] is a field (a subfield of R ).
- Q[2] is closed under addition. (a+b2)+(c+d2)=(a+c)+(b+d)2 .
- Q[2] is closed under multiplication. (a+b2)(c+d2)=ac+ad2+bc2+bd22=(ac+2bd)+(ad+bc)2 .
- Addition is associative and commutative on Q[2], since it is associative and commutative in R.
- Identity for addition: 0=0+02∈Q[2].
- Inverses for addition: The inverse of a+b2 is −(a+b2)=−a+−b2∈Q[2] .
- Multiplication is associative and commutative on Q[2] , since it is associative and commutative in R.
- Distributivity holds in Q[2] , since it holds in R.
- Identity for multiplication: 1=1+02∈Q[2].
- Inverses for multiplication: Given a,b∈Q with a+b2=0 ,( either a=0orb=0 ), we need to find c,d∈Q such that (a+b2)(c+d2)=1 .
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