s09_t07

# s09_t07 - Inductive case : Prove p ( k ) p ( k + 1) for...

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UNIVERSITY OF WATERLOO School of Computer Science CS245 Logic and Computation Spring 2009 TUTORIAL 7 1. For each of the following p ( n ), prove n : N p ( n ) using mathematical induction. (a) Let p ( n ) be y 7 n - 2 n = 5 y , for n 0 Base case : Prove p (0) y 7 0 - 2 0 = 5 y WV y 1 - 1 = 5 y WV y 0 = 5 y WV true Inductive case : Prove p ( k ) p ( k + 1) for arbitrary k , k 0 I.e., prove ( y 7 k - 2 k = 5 y ) ( y 7 k +1 - 2 k +1 = 5 y ) 1. y 7 k - 2 k = 5 y assumption (induction hypothesis) 2. y u 7 k - 2 k = 5 y u assumption , using 1 3. 2 k = 7 k - 5 y u 2 , algebra 4. 7 k +1 - 2 k +1 = 7 k +1 - 2 k +1 = I 5. 7 k +1 - 2 k +1 = 7(7 k ) - 2(2 k ) 4 , algebra 6. 7 k +1 - 2 k +1 = 7(7 k ) - 2(7 k - 5 y u ) 3 , 5 , = E 7. 7 k +1 - 2 k +1 = 7(7 k ) - 2(7 k ) + 5(2 y u ) 6 , algebra 8. 7 k +1 - 2 k +1 = 5(7 k ) + 5(2 y u ) 7 , algebra 9. 7 k +1 - 2 k +1 = 5(7 k + 2 y u ) 8 , algebra 10. y 7 k +1 - 2 k +1 = 5 y 9 , I 11. y 7 k +1 - 2 k +1 = 5 y 2 - 10 , E 12. p ( k ) p ( k + 1) 1 - 11 , I 1

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(b) Let p ( n ) be y T ( n ) y · n , for n 2, where T( n ) = 3 if n = 1 T( n ) = T( n - 1) + 3 if n > 1 Base case : Prove p (2) y T (2) y · 2 WV y T (1) + 3 y · 2 WV y 3 + 3 y · 2 WV y 3 · 2 y · 2 WV true
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Unformatted text preview: Inductive case : Prove p ( k ) p ( k + 1) for arbitrary k , k 2 I.e., prove ( y T ( k ) y k ) ( y T ( k + 1) y ( k + 1)) 1. y T ( k ) y k assumption (induction hypothesis) 2. y u T ( k ) y u k assumption , using 1 3. T ( k + 1) = T ( k + 1) = I 4. T ( k + 1) = T ( k ) + 3 denition 5. T ( k + 1) y u k + 3 2 , 4 , = E 6. T ( k + 1) ( y u k + 3)( k + 1) 5 , algebra 7. y T ( k + 1) y ( k + 1) 6 , I 8. y T ( k + 1) y ( k + 1) 2-7 , E 9. p ( k ) p ( k + 1) 1-8 , I 2...
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## This note was uploaded on 01/19/2010 for the course CS 246 taught by Professor Wormer during the Spring '08 term at Waterloo.

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s09_t07 - Inductive case : Prove p ( k ) p ( k + 1) for...

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