Prove that is ({a + b√2|a, b ∈ Q}, +) abelian group?
Answers
Step-by-step explanation:
18.12. Let Q[
√
2] = {a + b
√
2 | 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 + 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 com-
mutative 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].
• Inverses for multiplication: This is the interesting one. Given a, b ∈ Q with a+b
√
2 6=
0, (either a 6= 0 or b 6= 0), we need to find c, d ∈ Q such that (a+b
√
2)(c+d
√
2) = 1.
Solution 1: (a+b
√
2)(c+d
√
2) = (ac+2bd)+ (ad+bc)
√
2, so we need ac+2bd = 1
and ad + bc = 0. Let’s do some linear algebra.
a 2b
b a c
d
=
1
0
The matrix on the left has determinant a
2 − 2b
2
, so it’s invertible unless a
2 = 2b
2
.
Fortunately, if a
2 = 2b
2
, then a = ±
√
2b, but a, b ∈ Q, so the only way this could
happen is if a = b = 0.
Multiplying by the inverse matrix
1
a
2 − 2b
2
a −2b
−b a
,
we have
c
d
=
1
a
2 − 2b
2
a −2b
−b a 1
0
=
1
a
2 − 2b
2
a
−b
,
so the inverse of a + b
√
2 is
a
a
2 − 2b
2
−
b
a
2 − 2b
2
√
2 ∈ Q[
√
2].
Solution 2: We can invert a+b
√
2 in R:
1
a + b
√
2
. Rationalizing the denominator,
1
a + b
√
2
=
a − b
√
2
(a + b
√
2)(a − b
√
2)
=
a − b
√
2
a
2 − 2b
2
=
a
a
2 − 2b
2
−
b
a
2 − 2b
2
√
2 ∈ Q[
√
2].