设A={x|x=a2+b2,a,b∈Z},求证:(1)若s,t∈A,则st∈A.(2)若s,t∈A,t≠0,则st=p2+q2,其中p,q是有理数.-数学

题目简介

设A={x|x=a2+b2,a,b∈Z},求证:(1)若s,t∈A,则st∈A.(2)若s,t∈A,t≠0,则st=p2+q2,其中p,q是有理数.-数学

题目详情

设A={x|x=a2+b2,a,b∈Z},求证:
(1)若s,t∈A,则st∈A.
(2)若s,t∈A,t≠0,则
s
t
=p2+q2
,其中p,q是有理数.
题型:解答题难度:中档来源:不详

答案

(1)设s=a2+b2,t=c2+d2,则st=(a2+b2)(c2+d2)=a2c2+a2d2+b2c2+b2d2=(ac+bd)2+(ad-bc)2
所以st∈A.
(2)由(1)得st∈A,所以可设st=m2+n2,又t≠0,所以
class="stub"s
t
=class="stub"st
t2
=
m2+n2
t2
=(class="stub"m
t
)2+(class="stub"n
t
)2

p=class="stub"m
t
q=class="stub"n
t
,则class="stub"s
t
=p2+q2
,p,q为有理数.

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