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If G={a+b√2/a,b€Q}then prove that G is a group with respect to usual a

If G={a+b√2/a,b€Q}then prove that G is a group with respect to usual addition

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A)
Let Q[ √ 2] = {a + b √ 2 | a, b ∈ Q}

Q[ √ 2] is closed under addition. (a + b √ 2) + (c + d √ 2) = (a + c) + (b + d) √ 2

Q[ √ 2] is closed under multiplication. (a + b √ 2)(c + d √ 2) = ac + ad√ 2 + bc√ 2 + bd√ 2^ 2 = (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 + 0√ 2 ∈ Q[ √ 2].

• Inverses for addition: The inverse of a + b √ 2 is −(a + b √ 2) = −a + −b √ 2 ∈ 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 + 0√ 2 ∈ Q[ √ 2].

Not only is Q[ √ 2] closed under + and ·, but Q[ √ 2] is a field (a subfield of R).
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