1.
因m∈A,n∈A.
不妨设m=c+d√2,n=e+f√2,c、d、e、f都属于Q,
mn=(c+d√2)(e+f√2)
=(ce+2df)+(de+cf)√2
ce+2df∈Q,de+cf∈Q,
mn∈A,证毕;
2.
A○B=X○Y
={x+y|x∈X,y∈Y}
={x+y|x∈A,y∈B}
={x+a+b√3|x∈A,a、b∈Q}
A●B=X●Y
={xy|x∈X,y∈Y}
={xy|x∈A,y∈B}
={x(a+b√3)|x∈A,a、b∈Q}
={xa+xb√3|x∈A,a、b∈Q}
从形式上看A○B和A●B都是m+n√3,但X和A没有明确是什么,
所以不妨设A=B={0},则x=0,
A○B=X○Y={a+b√3|x∈A,a、b∈Q}
A●B=X●Y={0}
x=0∈A●B但属于A○B.
A●A=X●X
={xy|x∈X,y∈X}
={xy|x∈A,y∈A}
={0}
=A