已知集合A={x|x=a+b根號2,a,b∈Q},若X1∈A,X2∈A ,試問X1X2,X1/X2是否屬于A
設x1=a1+b1√2,x2=a2+b2√2(a1,a2,b1,b2∈Q)
則x1x2=a1a2+2b1b2+(a1b2+a2b1)√2
因為a1a2+2b1b2∈Q,a1b2+a2b1∈Q
所以x1x2∈A
x1/x2=(a1a2-2b1b2)/[a2^2-2(b2^2)]+{(a2b1-a1b2)/[a2^2-2(b2^2)]}*√2
因為(a1a2-2b1b2)/[a2^2-2(b2^2)]∈Q,(a2b1-a1b2)/[a2^2-2(b2^2)]∈Q
所以x1/x2∈A
請問：既然有理數(shù)除以有理數(shù)不一定為有理數(shù),為什么可以得出(a1a2-2b1b2)/[a2^2-2(b2^2)]∈Q,(a2b1-a1b2)/[a2^2-2(b2^2)]∈Q