# quiz3 - × 6 chessboard with three corners removed using 1...

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Name: Math 55 Quiz 3 July 1, 2009 GSI: Rob Bayer You have until 4:00 to complete this quiz. You must show your work. 1. (3 pts) Prove that the function f : R R given by f ( x ) = 3 x - 4 is a bijection. (injective) If f ( x ) = f ( y ) , then 3 x - 4 = 3 y - 4 so adding 4 gives 3 x = 3 y and thus x = y . Therefore, f ( x ) = f ( x ) x = y and f is injective. (surjective) Let y R . Then y +4 3 R and f ( y +4 3 ) = 3 y +4 3 - 4 = y , so y xf ( x ) = y and f is surjective. f is a bijection 2. (3 pts) Let f : D E be a function and let A,B D . Prove that f ( A B ) f ( A ) f ( B ) (Proof 1) Let y f ( A B ). So x A B such that f ( x ) = y . Since x A B , x A , so y = f ( x ) f ( A ). Similarly, x B so y f ( B ). Thus, y f ( A ) f ( B ). (Proof 2) A B A so f ( A B ) f ( A ). Similarly, f ( A B ) f ( B ). Thus, f ( A B ) f ( A ) f ( B ) 3. (4 pts) Show that it is not possible to tile a 6
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Unformatted text preview: × 6 chessboard with three corners removed using 1 × 3 rectangles We’ll color the board as follows: X R G B R X G B R G B R R G B R G B B R G B R G G B R G B R R G B R G X Note that anywhere you put a 1 × 3 triominoe on this board, it will cover exactly one red, one green, and one blue square. Since there are 33 squares to cover, any covering must use 11 triominoes. However, a quick count shows that there are 12 red squares, and thus they cannot all be covered by just 11 triominoes. Thus, there is no possible tiling of this board....
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