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author committer David A. Madore 2016-05-26 18:24:25 +0200 David A. Madore 2016-05-26 18:24:25 +0200 37a912ae129a042a184f9e0a811a123122df51a5 (patch) d8cc2f4e155d825becda937eef4a2699bfc0913b 09af2d3256d2e2873b4533ef4e843a9674483b2d (diff) mitro206-37a912ae129a042a184f9e0a811a123122df51a5.tar.gzmitro206-37a912ae129a042a184f9e0a811a123122df51a5.tar.bz2mitro206-37a912ae129a042a184f9e0a811a123122df51a5.zip
Fix error in labeling of questions.
-rw-r--r--controle-20160421.tex4
1 files changed, 2 insertions, 2 deletions
 diff --git a/controle-20160421.tex b/controle-20160421.texindex 8720f49..4360c6b 100644--- a/controle-20160421.tex+++ b/controle-20160421.tex@@ -642,9 +642,9 @@ de l'exercice \ref{game-for-nim-product}. \smallbreak -(3) (b) Montrer que si $\alpha'\neq\alpha$ et $\beta'\neq\beta$, alors+(3) (a) Montrer que si $\alpha'\neq\alpha$ et $\beta'\neq\beta$, alors $(\alpha'\otimes\beta) \oplus (\alpha\otimes\beta') \oplus-(\alpha'\otimes\beta') \neq \alpha\otimes\beta$.\spaceout (a) En+(\alpha'\otimes\beta') \neq \alpha\otimes\beta$.\spaceout (b) En déduire que si$\alpha\otimes\beta = \alpha\otimes\beta'$alors$\alpha=0$ou bien$\beta=\beta'\$.