commit e1a43ad11e539368c82ff8182efef2f7c101d555
parent 00fd6a40601cd5a859127c618f4888fad76467c4
Author: Sebastiano Tronto <sebastiano@tronto.net>
Date: Wed, 6 Sep 2023 23:26:29 +0200
fixed proof
Diffstat:
1 file changed, 7 insertions(+), 7 deletions(-)
diff --git a/src/speedcubing/slice-theory/slice-theory.md b/src/speedcubing/slice-theory/slice-theory.md
@@ -304,12 +304,12 @@ subsequence with sum 0, then k <= n.
*Clarification: non-trivial means that it contains at least one element
and it is not the whole sequence.*
-**Proof** (thanks to Chiara for the nice proof). the Theorem can be
-re-stated as follows: any sequence of n elements of Z/nZ has a
-subsequence whose sum is 0. To prove this equivalent statement,
-let, for l=1 to k, s_l = a_1 + ... + a_l. If s_i=0 for any i, we
-are done. Otherwise by the pigeonhole principle there must be s_i
-and s_j with s_i = s_j and, say, i < j. But then the subsequence
-a_(i+1), ..., a_j has sum s_j - s_i = 0. This proves the claim.
+**Proof** (thanks to Chiara for the nice proof). It is enough to
+prove that any sequence of n+1 elements of Z/nZ has a subsequence whose
+sum is 0. To prove this, let, for l=1 to n+1, s\_l = a\_1 + ... + a\_l.
+If s\_i=0 for any i, we are done. Otherwise by the pigeonhole principle
+there must be s\_i and s\_j with s\_i = s\_j and, say, i < j. But then
+the subsequence a\_(i+1), ..., a\_j has sum s\_j - s\_i = 0. This proves
+the claim.
More work needs to be done here.
