求下列集合的基数和幂集
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/21 10:20:20
(1)twenty-eighth(2)thirty-sixth
11的基数词eleven和13的基数词thirteen不明白的再问哟,请及时采纳,多谢!
参看http://zhidao.baidu.com/question/51321024.html
基数词序数词1(one)(first)2(two)(second)3(three)(third)4(four)(fourth)5(five)(fifth)6(six)(sixth)8(eight)(e
这个证明不难的,不过无限集的基数确实是个有意思的东西,比如偶数(2,4,6...2n)是正整数(1,2,3,...n)的一个真子集然而通过:2←→1,4←→2,6←→3,.,2n←→n,.它们之间建立
9nineninth22twenty-twotwenty-second14fourteenfourteenth79seventy-nineseventy-ninth345threehundredfor
one=【first】【1st】two=【second】【2nd】three=【third】【3rd】five=【fifth】【5th】nine=【ninthly】【9th】twelve=【Twelf
1.first2.second3.third4.fourth5.fifth6.sixth7.eighth8.twelfth9.twentieth10.fifty-first
1.WhenisTom'sbirthday?2.Yes,itis.3.DotheyhaveanEnglishpartyattheirschool?4.Whenisthebasketballgame?序
3,three,thethird(以后the省略)5,five,fifth8,eight,eighth9,nine,ninth11,eleven,eleventh12,twelve,twelfth13
first(one)fifth(five)eighth(eight)twelfth(twelve)second(two)ninth(nine)fourth(four)twentieth(twenty)
{∮,{∮},{a},{{a}},{∮,a},{∮,{a}},{a,{a}},{∮,a,{a}}}
nineninthtwelvethelfthfifteenfifteenthtuentytwentiethforty-oneforty-first
基数词:One,two,three,four,five,six,seven,eight,nine,ten,eleven,twelve,序数词:First,second,third,fourth,fif
基数词0nought;zero;O1one2two3three4four5five6six7seven8eight9nine10ten11eleven12twelve13thirteen14fourt
不知道你的集合是不是这两个1、{φ,a,{b}}2、{{1,{2,3}}}^21、集合三个元素φ,a,{b},也就是基数为3.它的幂集为:{空集,{φ},{a},{{b}},{φ,a},{φ,{b}}
非超越的无理数都是整系数方程的根,整系数方程集合可数,每个方程的解集可数,所以非超越无理数可数,而所有无理数的基数是N,所以超越数的基数是N
昨天回答了一类似的,他的第二个集合是{{1,{2,3}}}^2,你进参考资料的链接看看.1、集合三个元素φ,a,{b},也就是基数为3.它的幂集为:{空集,{φ},{a},{{b}},{φ,a},{φ
把自然数集的全体子集分成2类:一类是有限集,这类记成A,另一类是无限集,这类记成B,A显然是可数的;然后对于在B中的一个无限集M,用映射f(M)=∑(1/2)^k,这里求和号是对M中的全部k求和,这是
选A,AN*ρ({1,2,……,n})基数:阿列夫零.BN*(0,]少个数,如果填个正数,基数是C,C(2,4)Dρ(N)基数都是C再问:是的,B项是N*(0,1],能告诉我你是怎样判别的吗?谢谢再答