已知A={a+√2 b},|a2-2b2｜=1，a∈Z，b∈Z，x∈A，y∈A，求證： (1)xy∈A (2)1/x∈A

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已知A={a+√2 b},|a^2-2b^2｜=1，a∈Z，b∈Z，x∈A，y∈A，求證： (1)xy∈A ,(2)1/x∈A設x=m+√2n ,y=s +√2t ,m∈Z,n∈Z,s∈Z,t∈Z所以xy = (m+√2n)( s +√2t) = (ms +2nt) + (sn+mt)*√2 因為ms+2nt∈Z,sn+mt∈Z ,所以 xy∈A因為 1/x = 1/(m+√2n) = (m-√2n)/(m^2-2n^2) = m-√2n所以1/x∈A