qjx0816
幼苗
共回答了16个问题采纳率:75% 举报
设x1=a+b√2 ∈M
x2=c+d√2 ∈M
a,b,c,d∈Q
则有:
x1+x2=(a+c)+(b+d)√2 ,因a+c,b+d都∈Q,所以x1+x2 ∈M
x1-x2=(a-c)+(b-d)√2 ,因a-c,b-d都∈Q,所以x1-x2 ∈M
x1x2=ac+2bd+(ad+bc)√2,因ac+2bd,ad+bc都∈Q,所以x1x2 ∈M
x1/x2=(a+b√2)(c-d√2)/(c²-2d²)
=(ac-2bd)/(c²-2d²)+(bc-ad)/(c²-2d²)√2
因为(ac-2bd)/(c²-2d²),(bc-ad)/(c²-2d²)都∈Q
所以x1/x2 ∈M
1年前
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