1.Sperner Theorem 设A为n元集,A1,A2,...,Am为A的子集且两两互不包含,则m的最大值为(n[n/2]) proof: lemma: ∑mi=11(n|Ai|)≤1 proof of lemma: It is equivalent to ∑mi=1|Ai|!(n−|Ai|)!≤n! On the one hand, A中全排列有n!个 On the other hand, for eachAi,做A中全排列如下: x1x2...x|Ai|y1y2...yn−|Ai| 其中x1x2...x|Ai|是Ai中元素的全排列。 y1y2...yn−|Ai|是补集的全排列。 注意到,当i≠j时,对应的全排列不同。(否则两个子集有包含关系)
由lemma: m(n[n/2])≤∑mi=11(n|Ai|)≤1,得证。
2.Kummer Theorem n=(nknk−1...n0)p m=(mkmk−1...m0)p n−m=(dkdk−1...d0)p vp((nm)) equals to the aomunt of carry-bit: L when adding (n-m) and m. proof: vp(n!)=∑∞l=1[npl]=n1+n2(1+p)+...+nk(1+p2+...+pk−1)=n−(n0+n1+...+nk)p−1 Thus vp((nm))=vp(n!)−vp((n−m)!)−vp(m!)=∑ki=0(mi+di−ni)p−1=L