试题

题目:
把下列各数分别填入相应的集合里.
-4,-|-
4
3
|,0,
22
7
π
2
,2013,-(+5),+1.88,0.010010001…,-2.33….
(1)正数集合:{                                          …};
(2)负数集合:{                                          …};
(3)整数集合:{                                          …};
(4)有理数集合:{                                         …}
(5)无理数集合:{                                         …}.
答案
解:-|-
4
3
|=-
4
3
,-(+5)=-5,
(1)正数集合:{
22
7
π
2
,2013,+1.88,0.010010001…,…};
(2)负数集合:{-4,-|-
4
3
|,-(+5),-2.33…,…};
(3)整数集合:{-4,0,2013,-(+5),…};
(4)有理数集合:{-4,-|-
4
3
|,0,
22
7
,2013,-(+5),+1.88,-2.33…,…}
(5)无理数集合:{
π
2
,0.010010001…,…}.
解:-|-
4
3
|=-
4
3
,-(+5)=-5,
(1)正数集合:{
22
7
π
2
,2013,+1.88,0.010010001…,…};
(2)负数集合:{-4,-|-
4
3
|,-(+5),-2.33…,…};
(3)整数集合:{-4,0,2013,-(+5),…};
(4)有理数集合:{-4,-|-
4
3
|,0,
22
7
,2013,-(+5),+1.88,-2.33…,…}
(5)无理数集合:{
π
2
,0.010010001…,…}.
考点梳理
实数.
由于实数包括有理数和无理数,有理数包括整数和分数,无限不循环小数是无理数;实数还可分为正实数、负实数和0.利用这些结论即可求解.
本题主要考查了实数的分类,特别注意:0.010010001…虽然有规律,但不循环,故为无理数.
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