12-240/Classnotes for Thursday October 18

DBN: Classes: 2012-13: 12-240 / Navigation Additions to this web site no longer count towards good deed points. # Week of... Notes and Links 1 Sep 10 About This Class, Tuesday, Thursday 2 Sep 17 HW1, Tuesday, Thursday, HW1 Solutions 3 Sep 24 HW2, Tuesday, Class Photo, Thursday 4 Oct 1 HW3, Tuesday, Thursday 5 Oct 8 HW4, Tuesday, Thursday 6 Oct 15 Tuesday, Thursday 7 Oct 22 HW5, Tuesday, Term Test was on Thursday. HW5 Solutions 8 Oct 29 Why LinAlg?, HW6, Tuesday, Thursday, Nov 4 is the last day to drop this class 9 Nov 5 Tuesday, Thursday 10 Nov 12 Monday-Tuesday is UofT November break, HW7, Thursday 11 Nov 19 HW8, Tuesday,Thursday 12 Nov 26 HW9, Tuesday , Thursday 13 Dec 3 Tuesday UofT Fall Semester ends Wednesday F Dec 10 The Final Exam (time, place, style, office hours times) Register of Good Deeds Add your name / see who's in!

Riddle Along

The game of 15 is played as follows. Two players alternate choosing cards numbered between 1 and 9, with repetitions forbidden, so that the game ends at most after 9 moves (or four and a half rounds). The first player to have within her/his cards a set of precisely 3 cards that add up to 15 wins.

Does this game have a winning strategy? What is it? Who wins, the first to move or the second? Why am I asking this question at this particular time?

See also a video and the transcript of that video.

Dror's notes above / Students' notes below

Theorems

1. If G generates, |G| $\geq \!\,$ n and G contains a basis, |G|=n then G is a basis

2. If L is linearly independent, |L| $\leq \!\,$ n and L can be extended to be a basis. |L|=n => L is a basis.

3.W $\subset \!\,$ V a subspace then W is finite dimensioned and dim W $\leq \!\,$ dim V

If dim W = dim V, then V = W If dim W < dim V, then any basis of W can be extended to be a basis of V

Proof of W is finite dimensioned:

Let L be a linearly independent subset of W which is of maximal size.

Fact about N

Every subset A of N, which is:

1. Non empty

2. Bounded : $\exists \!\,$ N $\in \!\,$ N, $\forall \!\,$ a $\in \!\,$ A, a $\leq \!\,$ N

has a maximal element: an element m $\in \!\,$ A, $\forall \!\,$ a $\in \!\,$ A, a $\leq \!\,$ m ( m + 1 $\notin \!\,$ A )

class note

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lecture note on oct 18, uploaded by starash

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