可量化的基数集项目 P&L: 0 (≃ 0 CNY)
将用途更普遍的具有可量化基数的集合引入数学中。
YAML
项目
这个项目是一个倡议,是根据一个特定的提议(我通过电子邮件收到了一个尚未公开的思想家发给的内容(对作者:请让我知道我是否可以公开披露源电子邮件)),定义和引入具有负基数的集合到数学和计算中。 它如下:
{1,2}+{3,4}={1,2,3,4} {1,2}+{2,3}={1,2,2,3}={1,2_2,3} 1,2+1,2=2*{1,2}={1,1,2,2}={1_2,2_2} {a_x}+{a_y}={a_(x+y)} {1,2,3}-{1}={2,3} {1,2}-{1,2}={}={1_0,2_0} {1,2}-{1,2,3,4}=-{3,4}={3_-1,4_-1} {1,2,3}-{3,4,5}={1,2}-{4,5}={1,2,4_-1,5_-1} {a_x}-{a_y}={a_(x-y)} 3*{1,2,3}={1,1,1,2,2,2,3,3,3}={1_3,2_3,3_3} -2*{1,2,3}={1_-2,2_-2,3_-2}=-{1_2,2_2,3_2} 0.5*{1,2,3}={1_0.5,2_0.5,3_0.5} 2*{1_0.5,2_0.5,3_0.5}={1,2,3} y*{a_x}={a_(x*y)} {a,b}*{c,d}={a+c,a+d,b+c,b+d} {a,b,c}*{d,e}={a+d,a+e,b+d,b+e,c+d,c+e} {a_x,b_y}*{c_z,d_t}={(a+c)_xz,(a+d)_xt,(b+c)_yz,(b+d)_yt} {{a},{b}}*{{c},{d}}={{a,c},{a,d},{b,c},{b,d}} {{a},{b}}^2={{a_2},{a,b}_2,{b_2}} P({a,b,c,d}),P({a,b}),P({c,d}): P({a,b})={0,{a},{b},{a,b}} P({c,d})={0,{c},{d},{c,d}} P({a,b,c,d})={0,{c},{d},{c,d},{a},{a,c},{a,d},{a,c,d},{b},{b,c},{b,d},{b,c,d},{a,b},{a,b,c},{a,b,d},{a,b,c,d}} P(A+B)=P(A)*P(B) a_{b}=a+b。{a}*{b}={a+b}={a_{b}} 0={} 1={0} 2=1+1={0}+{0}={0,0}={0_2} 3=2+1={0,0}+{0}={0,0,0}={0_3} n={0_n}。x={0_x}。 {2,4,6,...}/{1}={1,3,5,...} {1,2,3,...,}/{2,4,6,...}={0,-1} [0,∞)/[0,1)={0,1,2,3,...} x_{a}=x+a,x_{b}=x+b,x_{c}=x+c。A={m,n,p},{{a},{b},{c}}^A={{m+a},{m+b},{m+c}}*{{n+a},{n+b},{n+c}}*{{p+a},{p+b},{p+c}}。 {0,1}^A=P(A) {0,0}^A=2^A=2^|A| {1,1}^A={A_2^|A|} {0,1,2}^{a,b,c}={0,{a},{a_2}}*{0,{b},{b_2}}*{0,{c},{c_2}} {{c},{d}}^{a,b}={{a+c,b+c},{a+c,b+d},{a+d,b+c},{a+d,b+d}} {{c,d},{e,f}}^{a,b}={{a+c,a+d},{a+e,a+f}}*{{b+c,b+d},{b+e,b+f}} {{c},{d},{e},{f}}^{a,b}={{a+c},{a+d},{a+e},{a+f}}*{{b+c},{b+d},{b+e},{b+f}} {a_x,b_y}+{a_z,b_t}={a_(x+z),b_(y+t)} (a_x,b_y}*{c_z,d_t}={a+c_xz,a+d_xt,b+c_yz,b+d_yt} {{c},{d}}^{a,b}={{a+c},{a+d}}*{{b+c},{b+d}} P(A+B)=P(A)*P(B),A^(C*B)=(A^C)^B P(A)={0,1}^{a_x,b_y,c_z} P(A)={0,1}^{a_x}*{0,1}^{b_y}*{0,1}^{c_z} {0,1}^{a}={0,{a}},{0,1}^{a_x}=({0,1}^{a})^x,P(A)={0,{a}}^x*{0,{b}}^y*{0,{c}}^z {0,{a}}^x={0,{a}_x,{a_2}_x*(x-1)/2,...,{a_n}_x*(x-1)*...*(x-n+1)/n!,.....} P({a_x,b_y,c_z})={0,{a}_x,{a_2}_x*(x-1)/2,...,{a_n}_x*(x-1)*...*(x-n+1)/n!,.....}*{0,{b}_y,{b_2}_y*(y-1)/2,...,{b_n}_y*(y-1)*...*(y-n+1)/n!,.....}*{0,{c}_z,{c_2}_z*(z-1)/2,...,{c_n}_z*(z-1)*...*(z-n+1)/n!,.....} 1/{0,1}={0,1}^-1={0,1_-1,2,3_-1,4_1,.....}。 {0,1_-1,2,3_-1,4_1,.....}*{0,1}={0,1_-1,2,3_-1,4_1,.....,1,2_-1,3,4_-1,.......}={0}=1。1/{0,1,2}={0}+{1,2}*-1+{1,2}^2+{1,2}^3*-1+...,1-2+4-8+....=1/3。 1-n+n^2-n^3+....=1/(n+1)。 1/{0,1}^2={0}+{1}*-2+{2}*3+{3}*-4+....,1-2+3-4+....=1/4。 1/{0,1}^3={0}+{1}*-C(1,3)+{2}*C(2,4)+{3}*-C(3,5)+....,1-3+6-10+....=1/8。 C(0,n-1)-C(1,n)+C(2,n+1)-C(3,n+2)+.....=1/2^n。 C(n,n)*C(k,k+n-1)-C(n-1,n)*C(k+1,k+n)+,,,,+(-1)^i*C(n-i,n)*C(k+i,k+n-1+i)+......+(-1)^n*C(0,n)*C(k+n,k+2n-1)=0. 1/{-1,0_-1}=({-1}-1)^-1={1,2,3,4,.....} 1/{-2,0_-1}={2,4,6,8,.....} 1/{1,2,3,4,....}={-1,0_-1} {1,2,3,....}/{2,4,6,....}={-2,0_-1}/{-1,0_-1}={0,-1} {1,3,5,...}-{2,4,6,....}={1,2:-1,3,4:-1,5,6:-1,......}={1}/{1,0} a:1->b:1 a:10->b:10 a:3,b:7->c:4,d:6 a:-1->a:-1 a,b:0.5,c:-0.3->d:1.2 {a:-1}{} ={b,b:-1},b:-1->a:-1, aleph0+pi=aleph0。{a1,a2,a3,....}{a1,a2,a3,...,b:pi}。a1,a2,a3->b:3。a4->b:pi-3,a1:4-pi。a5->a1:pi-3,a2:4-pi。...a(n+4)->an:pi-3,a(n+1):4-pi。.... Aleph0*pi=aleph0。{a1:pi,a2:pi,a3:pi,....}{a1,a2,a3,...,} a(6i-5),a(6i-4),a(6i-3)->a(i):pi-3,a(2i-1):6-pi。a(6i-2),a(6i-1),a(6i)->a(i):pi-3,a(2i):6-pi。
如您所见,可量化的基数用下划线表示时,可进行集合操作的示例。据我了解,向我发送此提案是实现“负基数”这个想法的步骤之一,因此,让这个页面成为添加后续步骤的地方,以实现更广泛地验证和采用此概念。
记:以后用 %%LaTeX%% 重写。
[skihappy],您可以使用负基数对负质量进行建模,但是作为概念的负基数严格不等同于负质量,因此,它不是负质量。
对于会计师来说,负基数可能是负资产(负债),而其他专家可能是其他领域的概念。
[skihappy], you could model negative mass with negative cardinality, but negative cardinality as a concept is strictly is not equivalent to negative mass, so, it's not negative mass.
For an accountant, negative cardinality could be negative assets (liabilities), and other specialists it may be concepts in other domains.
// 负基数是负质量 ????
不。基数是集合中元素的数量(所谓的“集合大小”),因此,负基数将是元素少于 0 的集合的大小。
// negative cardinality is negative mass ????
No. Cardinality is the number of elements (so-called "size of the set") within a set, so, negative cardinality would be the size of the set that has less than 0 elements.
基数可以是质量的度量。那么负基数就是负质量。 ???这是什么意思?
Cardinality number can be a measure of mass. Then negative cardinality is negative mass. ???? What does it mean?
负基数是某种东西的赤字,一项尚未完成的任务。它与顺序、然后和时间有关。非常有意思。
Negative cardinality is deficit of something, a task yet to be done. It has to do with sequence, then, and time. Really interesting.