设x=a+√2b (a∈Z,b∈Z),y=c+√2d (c∈Z,d∈Z),则x∈G,y∈G,∴x+y=(a+c)+√2(b+d),且(a+c)∈Z,(b+d)∈Z,故x+y∈G.设x=2+√2,显然x∈G,但1/x=1/(2+√2)=(2-√2)/2=1-(1/2)√2∵-1/2不属于Z,∴1/x不属于集合G,故1/x不...