See my classical Chinese page文言 also my simple English page Simple English
「What I mean just technically speaking, it is incorrect to say Baidu blocks it. OK, I know how the Party and Baidu company evil, and I am f*cked by the holy Firewall almost everyday, but as a technical expert, I can still access Wikipedia, here, so that's all, and that's enough. So in an evil society, let us all be evil, that would be the best choice. I don't care about those non-technical things, as the ancient Chinese classics {左傳} says, 肉食者謀之，又何間焉。 Yao Ziyuan 21:47, 14 May 2006 (UTC)」
百度百科亦有排版不暢之疾，尤以數學公式。常以b/a代 b a {\displaystyle {\frac {b}{a}}} ，以(1011)2代 1011 ( 2 ) {\displaystyle 1011_{(2)}} ,以e^(-x^2)代 e − x 2 {\displaystyle e^{-x^{2}}} ,使人難閲之。
亦有排版不暢之疾，尤以數學公式。常代 b a {\displaystyle {\frac {b}{a}}} 之以b/a，代 1011 ( 2 ) {\displaystyle 1011_{(2)}} 之以(1011)2，代 e − x 2 {\displaystyle e^{-x^{2}}} 之以e^(-x^2)，使人難閲之。
孿生質數無窮性之證
由質數分佈定理可知，質數之距遞增。
設質數之距d,由定理可知
d ( p n , p n + m ) < d ( p n + k , p n + m + k ) {\displaystyle d_{(p_{n},p_{n+m})}<d_{(p_{n+k},p_{n+m+k})}}
暫令二者等距，且令 ( p n + k ) ∈ P , p n = r , {\displaystyle (p_{n}+k)\in \mathbb {P} \ ,p_{n}=r,} 則
Δ p n 2 = ( p n + k ) 2 − p n 2 {\displaystyle \Delta {p_{n}}^{2}=(p_{n}+k)^{2}-{p_{n}}^{2}}
= r 2 + 2 k r + k 2 − p n 2 {\displaystyle =r^{2}+2kr+k^{2}-{p_{n}}^{2}}
2 k r + k 2 {\displaystyle 2kr+k^{2}}
又質數平方內半質數（質數分佈缺陷）pp'有如下之分佈
p 3 2 {\displaystyle {p_{3}}^{2}} 52 ∅ {\displaystyle \varnothing }
p 4 2 {\displaystyle {p_{4}}^{2}} 72 5 ⋅ 7 {\displaystyle 5\cdot 7} C 2 2 {\displaystyle C_{2}^{2}}
p 5 2 {\displaystyle {p_{5}}^{2}} 112 5 ⋅ 7 5 ⋅ 11 7 ⋅ 11 {\displaystyle 5\cdot 7\quad 5\cdot 11\quad 7\cdot 11} C 3 2 {\displaystyle {\ce {C_3^2}}}
P 6 2 {\displaystyle {P_{6}}^{2}} 132 5 ⋅ 7 5 ⋅ 11 5 ⋅ 13 7 ⋅ 11 7 ⋅ 13 11 ⋅ 13 {\displaystyle 5\cdot 7\quad 5\cdot 11\quad 5\cdot 13\quad 7\cdot 11\quad 7\cdot 13\quad 11\cdot 13} C 4 2 {\displaystyle {\ce {C_4^2}}}
etc
P n 2 {\displaystyle {P_{n}}^{2}} C n − 2 2 {\displaystyle C_{n-2}^{2}}
C n − 2 2 = 1 2 ( n − 2 ) ( n − 3 ) = 1 2 [ n 2 − 5 n + 6 ] {\displaystyle C_{n-2}^{2}={\frac {1}{2}}(n-2)(n-3)={\frac {1}{2}}[n^{2}-5n+6]}
則 ( p n + k ) 2 {\displaystyle (p_{n}+k)^{2}} 滿足
n p p ′ < 1 2 ( r + k − 2 ) ( r + k − 3 ) {\displaystyle n_{pp'}<{\frac {1}{2}}(r+k-2)(r+k-3)}
= 1 2 [ r 2 + 2 k r + k 2 − 5 r − 5 k + 6 ] {\displaystyle ={\frac {1}{2}}[r^{2}+2kr+k^{2}-5r-5k+6]}
則
Δ n p p ′ < 1 2 [ r 2 + 2 k r + k 2 − 5 k − 5 r + 6 ] − 1 2 [ r 2 − 5 r + 6 ] {\displaystyle \Delta n_{pp'}<{\frac {1}{2}}[r^{2}+2kr+k^{2}-5k-5r+6]-{\frac {1}{2}}[r^{2}-5r+6]}
= 1 2 [ k 2 + 2 k r − 5 k ] {\displaystyle ={\frac {1}{2}}[k^{2}+2kr-5k]}
半質數之密度
Δ n p p ′ Δ p n 2 < 1 2 ⋅ [ k 2 + 2 k r − 5 k ] k 2 + 2 k r {\displaystyle {\frac {\Delta n_{pp'}}{\Delta {p_{n}}^{2}}}<{\frac {1}{2}}\cdot {\frac {[k^{2}+2kr-5k]}{k^{2}+2kr}}}
= 1 2 [ 1 − 5 k k 2 + 2 k r ] {\displaystyle ={\frac {1}{2}}[1-{\frac {5k}{k^{2}+2kr}}]}
lim k → ∞ lim r → ∞ Δ n p p ′ Δ p n 2 < 1 2 {\displaystyle \lim _{k\to \infty }\lim _{r\to \infty }{\frac {\Delta n_{pp'}}{\Delta {p_{n}}^{2}}}<{\frac {1}{2}}}
又質數之距遞增，故
lim r → ∞ Δ n p p ′ Δ p n 2 = 0 {\displaystyle \lim _{r\to \infty }{\frac {\Delta n_{pp'}}{\Delta {p_{n}}^{2}}}=0}
即半質數之比趋於零，故孿生質數{6n-1,6n+1}無窮
6n-1,6n+1
senary
duodecimal
則
52 
72 
112 
132 
![{\displaystyle C_{n-2}^{2}={\frac {1}{2}}(n-2)(n-3)={\frac {1}{2}}[n^{2}-5n+6]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/902b80f2ffaa09540ab45cdd51c649ac9edc2bf9)
滿足
![{\displaystyle ={\frac {1}{2}}[r^{2}+2kr+k^{2}-5r-5k+6]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a16db2b8777fe62c94dcd82031a90d5b627256f)
![{\displaystyle \Delta n_{pp'}<{\frac {1}{2}}[r^{2}+2kr+k^{2}-5k-5r+6]-{\frac {1}{2}}[r^{2}-5r+6]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b41231f08a477eb1d8c81b62a1e703090d879b4e)
![{\displaystyle ={\frac {1}{2}}[k^{2}+2kr-5k]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aadb8d8fbfd08d2ea86bf6ec372d2c4abfc9ced5)
![{\displaystyle {\frac {\Delta n_{pp'}}{\Delta {p_{n}}^{2}}}<{\frac {1}{2}}\cdot {\frac {[k^{2}+2kr-5k]}{k^{2}+2kr}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2259247c2c3872193469aca75b9297a6e194ca47)
![{\displaystyle ={\frac {1}{2}}[1-{\frac {5k}{k^{2}+2kr}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b247d50bd08b838e531682006e51645c9126a8d)
