问题 选择题

(x,y)称为数对,其中x,y都是任意实数,定义数对的加法、乘法运算如下:

(x1,y1)+(x2,y2)=(x1+x2,y1+y2

(x1,y1)•(x2,y2)=(x1x2-y1y2,x1y2+y1x2),则(  )不成立.

A.乘法交换律:(x1,y1)•(x2,y2)=(x2,y2)•(x1,y1

B.乘法结合律:(x1,y1)•(x2,y2)•(x3,y3)=(x1,y1)•[(x2,y2),(x3,y3)]

C.乘法对加法的分配律:(x,y)•[(x1,y1)+(x2,y2)]=[(x,y)•(x1,y1))+((x,y)•(x2,y2)]

D.加法对乘法的分配律:(x,y)+[(x1,y1)•(x2,y2)]=[(x,y)+(x1,y1)]•[(x,y)+(x2,y2)]

答案

A、由(x2,y2)•(x1,y1

=(x1x2-y1y2,x1y2+y1x2

=(x1,y1)•(x2,y2)可知,乘法交换律成立,A正确;

B、由[(x1,y1)•(x2,y2)]•(x3,y3

=(x1x2-y1y2,x1y2+y1x2)•(x3,y3

=(x1x2x3-y1y2x3-x1y2y3-y1x2y3,x1x2y3-y1y2x3+x1y2x3+y1x2x3

=(x1,y1)•(x2x3-y2y3,x2y3+y2x3)=(x1,y1)•[(x2,y2)•(x3y3)]可知,乘法结合律成立,B正确;

C、由(x,y)•[(x1,y1)+(x2,y2)]

=(x,y)•(x1+x2,y1+y2

=[x(x1+x2)-y(y1+y2),x(y1+y2)+y(x1+x2)]

=(xx1-yy1,xy1+yx1)+(xx2-yy2,xy2+yx2

=[(x,y)•(x1,y1)]+[(x,y)•(x2,y2)]可知,乘法对加法的分配律成立,C正确;

D、由(1,0)+[(1,0)•(1,0)]

=(1,0)+(1,0)

=(2,0)≠(2,0)•(2,0)

=[(1,0)+(1,0)•((1,0)+(1,0))]可知,加法对乘法的分配律不成立,D错误.

不成立的是D.

故选D.

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