离散数学搞不懂,就双射里面的单射来说,可以证明对于两个不同的点,(x1,y1)(x2,y2),x1≠x2,y1≠y2,映射为(x1/2+y1/2, x1/2-y1/2) ,(x2/2+y2/2, x2/2-y2/2)若这两个点重合,需满足x1+y1=x2+y2x1-y1=x2-y2两式相加,得到x1=x2与题设x1≠x2矛盾。若两个点为(x1,y1)(x2,y2),x1=x2,y1≠y2,(x1≠x2,y1=y2,与此类似)则映射为(x1/2+y1/2, x1/2-y1/2) ,(x2/2+y2/2, x2/2-y2/2)若重合需满足,x1+y1=x...
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