已知集合A={a+√2b Ia∈Z,b∈Z,且Ia^2-2b^2I=1},若X∈A,y∈A.求证:(1)XY∈A;(2)
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已知集合A={a+√2b Ia∈Z,b∈Z,且Ia^2-2b^2I=1},若X∈A,y∈A.求证:(1)XY∈A;(2)1/x∈A.
let x= a+√2b
y= a'+√2b'
xy= (a+√2b )(a'+√2b')
= (aa'+2bb')+ (ab'+a'b)√2
let a"= aa'+2bb' ∈Z
b"= ab'+a'b ∈Z
|a"^2-2b"^2|
=|(aa'+2bb')^2 - 2(ab'+a'b)^2|
=| (aa')²+4aa'bb'+4(bb')²- 2(ab')²-4aa'bb'-2(a'b)²|
=| a'²(a²-2b²)-2b'²(a²-2b²) |
=| a'²- 2b'²| ((a²-2b²)=1)
=1 ((a'²- 2b'²)=1)
∴ xy∈ A #
x∈A
x= c+√2d
1/x = 1/(c+√2d)
= (c-√2d)/(c²-2d²)
= c/(c²-2d²) + [ -d/(c²-2d²)]√2
= c-d√2 ( c²-2d² = 1 )
c' = c
d' = -d
|c'²-2d'²|
=|c²-2d²| =1
∴ 1/x∈ A #
y= a'+√2b'
xy= (a+√2b )(a'+√2b')
= (aa'+2bb')+ (ab'+a'b)√2
let a"= aa'+2bb' ∈Z
b"= ab'+a'b ∈Z
|a"^2-2b"^2|
=|(aa'+2bb')^2 - 2(ab'+a'b)^2|
=| (aa')²+4aa'bb'+4(bb')²- 2(ab')²-4aa'bb'-2(a'b)²|
=| a'²(a²-2b²)-2b'²(a²-2b²) |
=| a'²- 2b'²| ((a²-2b²)=1)
=1 ((a'²- 2b'²)=1)
∴ xy∈ A #
x∈A
x= c+√2d
1/x = 1/(c+√2d)
= (c-√2d)/(c²-2d²)
= c/(c²-2d²) + [ -d/(c²-2d²)]√2
= c-d√2 ( c²-2d² = 1 )
c' = c
d' = -d
|c'²-2d'²|
=|c²-2d²| =1
∴ 1/x∈ A #
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