Answer to Question #285266 in Abstract Algebra for Pavan

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[\\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.


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