可量化的基數集項目 P&L: 0 (≃ 0 TWD)
將用途更普遍的具有可量化基數的集合引入數學中。
YAML
項目
這個項目是一個倡議,是根據一個特定的提議(我通過電子郵件收到了一個尚未公開的思想家發給的內容(對作者:請讓我知道我是否可以公開披露源電子郵件)),定義和引入具有負基數的集合到數學和計算中。 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。
如您所見,可量化的基數用下劃線表示時,可進行集合操作的示例。據我瞭解,向我發送此提案是實現“負基數”這個想法的步驟之一,因此,讓這個頁面成爲添加後續步驟的地方,以實現更廣泛地驗證和採用此概念。
記:以後用 %%LaTeX%% 重寫。
[skihappy],您可以使用負基數對負質量進行建模,但是作爲概念的負基數嚴格不等同於負質量,因此,它不是負質量。
對於會計師來說,負基數可能是負資產(負債),而其他專家可能是其他領域的概念。
[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.
// 負基數是負質量 ????
不。基數是集合中元素的數量(所謂的“集合大小”),因此,負基數將是元素少於 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.
基數可以是質量的度量。那麼負基數就是負質量。 ???這是什麼意思?
Cardinality number can be a measure of mass. Then negative cardinality is negative mass. ???? What does it mean?
負基數是某種東西的赤字,一項尚未完成的任務。它與順序、然後和時間有關。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.