一道离散证明可数题show that the set of all real numbers that are solutions of quadratic equations a^2x+bx+c=0,where a,b and c are integers,is countable.意思是证明当a,b,c都是整数时,a^2x+bx+c=0的所有的解释可数的,

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一道离散证明可数题showthatthesetofallrealnumbersthataresolutionsofquadraticequationsa^2x+bx+c=0,wherea,bandca

一道离散证明可数题show that the set of all real numbers that are solutions of quadratic equations a^2x+bx+c=0,where a,b and c are integers,is countable.意思是证明当a,b,c都是整数时,a^2x+bx+c=0的所有的解释可数的,
一道离散证明可数题
show that the set of all real numbers that are solutions of quadratic equations a^2x+bx+c=0,where a,b and c are integers,is countable.
意思是证明当a,b,c都是整数时,a^2x+bx+c=0的所有的解释可数的,

一道离散证明可数题show that the set of all real numbers that are solutions of quadratic equations a^2x+bx+c=0,where a,b and c are integers,is countable.意思是证明当a,b,c都是整数时,a^2x+bx+c=0的所有的解释可数的,
a,b,c固定时解有限(1或2,重根算1个)
a,b,c是可数的有限次方,可数,
按这个顺序排列可数个方程,其中的解未出现过的数依序列出即可(其实推广到有理数系数有限次多项式出来的代数数都是可数的)

因为三元组(a,b,c)是可数的(N*N*N=N,对角线法),而每个三元组(a,b,c)对应的方程ax^2+bx+c=0最多有两个实数根,那么所有这样的方程的根也是可数的

countable是有限数的意思啦,这道题其实就是讨论方程a^2x+bx+c=0的解的情况,很多网和参考书上都有哎

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