ProofsPart3

# ProofsPart3 - MATH 135 Fall 2007 Proofs, Part III Some...

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MATH 135 Fall 2007 Proofs, Part III Some miscellaneous induction examples. .. Example Prove that n X r =1 f 2 r = f n f n +1 for all n P . We prove this by induction. Perhaps surprisingly, we do not need strong induction. Base Case If n = 1, 1 X r =1 f 2 r = f 2 1 = 1 and f 1 f 2 = 1, so the result is true. Induction Hypothesis Assume the result is true for n = k , for some k P , k 1. Induction Conclusion Consider n = k + 1 k +1 X r =1 f 2 r = " k X r =1 f 2 r # + f 2 k +1 = f k f k +1 + f 2 k +1 (by Induction Hypothesis) = f k +1 ( f k + f k +1 ) = f k +1 f k +2 so the result is true for n = k + 1. By POMI, the result is true for all n P . Example If n + 1 concrete blocks are placed in n knapsacks, then one knapsack must contain at least two concrete blocks. We prove this now by induction, and use a diﬀerent approach later. Base Case n = 1 : 1 knapsack, 2 concrete blocks Induction Hypothesis Assume the result holds for n = k , for some k P , k 1. Induction Conclusion

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## This note was uploaded on 02/05/2010 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.

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ProofsPart3 - MATH 135 Fall 2007 Proofs, Part III Some...

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