已知集合A={x/x=a+b根号2,a,b∈Z},设x1∈A,x2∈A,求证：x1x2∈A

设x1=a1+b1根号2,x2=a2+b2根号2
x1x2=(a1+b1根号2)( x2=a2+b2根号2)
=a1a2+2b1b2+(a1b2+a2b1)根号2
{x/x=a+b根号2,a,b∈Z}中
a=a1a2+2b1b2
b=a1b2+a2b1
所以x1x2∈A