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-rw-r--r--src/speedcubing/slice-theory/slice-theory.md14
1 files changed, 7 insertions, 7 deletions
diff --git a/src/speedcubing/slice-theory/slice-theory.md b/src/speedcubing/slice-theory/slice-theory.md
index 784f5d7..e537a0b 100644
--- 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.
304*Clarification: non-trivial means that it contains at least one element 304*Clarification: non-trivial means that it contains at least one element
305and it is not the whole sequence.* 305and it is not the whole sequence.*
306 306
307**Proof** (thanks to Chiara for the nice proof). the Theorem can be 307**Proof** (thanks to Chiara for the nice proof). It is enough to
308re-stated as follows: any sequence of n elements of Z/nZ has a 308prove that any sequence of n+1 elements of Z/nZ has a subsequence whose
309subsequence whose sum is 0. To prove this equivalent statement, 309sum is 0. To prove this, let, for l=1 to n+1, s\_l = a\_1 + ... + a\_l.
310let, for l=1 to k, s_l = a_1 + ... + a_l. If s_i=0 for any i, we 310If s\_i=0 for any i, we are done. Otherwise by the pigeonhole principle
311are done. Otherwise by the pigeonhole principle there must be s_i 311there must be s\_i and s\_j with s\_i = s\_j and, say, i < j. But then
312and s_j with s_i = s_j and, say, i < j. But then the subsequence 312the subsequence a\_(i+1), ..., a\_j has sum s\_j - s\_i = 0. This proves
313a_(i+1), ..., a_j has sum s_j - s_i = 0. This proves the claim. 313the claim.
314 314
315More work needs to be done here. 315More work needs to be done here.

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