若A｛（X,Y)｜Y=x2-2｝,B｛（X,Y)｜X=Y2-2｝,求A∩B

求A∩B就是求Y=X²-2与X=Y²-2的公共解
两式相减得：Y-X=X²-Y²
故：(X-Y)(X+Y)+(X-Y)=0
故：(X-Y)(X+Y+1)=0
故：X=Y或X+Y+1=0
当X =Y时,代人Y=X²-2得：X=Y=2或X=Y=-1
当X+Y+1=0时,代人Y=X²-2得：X1=(-1+√5)/2,Y1=(-1-√5)/2或X2=(-1-√5)/2,Y2=(-1+√5)/2
故：A∩B={(2,2),(1,1),( (-1+√5)/2,(-1-√5)/2),( (-1-√5)/2,(-1+√5)/2)}