Base Case in Strong Induction Proofs

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SUMMARY

The discussion focuses on determining the base case in strong induction proofs related to forming postage values using 4-cent and 11-cent stamps. The base case is identified as 30 cents, as it is the smallest value from which all higher values can be formed using combinations of the two stamp denominations. The reasoning involves demonstrating that all consecutive postage values can be achieved starting from 30 cents, following the established combinations up to 28 cents. This approach aligns with the principles outlined in Rosen's "Discrete Mathematics."

PREREQUISITES
  • Understanding of strong induction proofs
  • Familiarity with the concept of postage problem in combinatorial mathematics
  • Knowledge of natural numbers and their properties
  • Basic principles of discrete mathematics as presented in Rosen's textbook
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  • Study the concept of strong induction in mathematical proofs
  • Explore the postage problem and its applications in combinatorial number theory
  • Learn how to establish base cases in induction proofs
  • Review examples of generating functions related to combinations of integers
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This discussion is beneficial for students of discrete mathematics, educators teaching induction proofs, and anyone interested in combinatorial problem-solving techniques.

velouria131
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I am dealing with sets of problems that go as such: "How many n-cent postages can be formed from x and y cent stamps"

For instance, I am doing a problem where x and y are 4 and 11 respectively. I don't understand how to determine a base case. I know that I must proof P(k + 1) for all P(i). The answer in my book is that the base case is a postage equal to 30 cents, but I don't follow. Could someone explain how I would go about finding a base case? The book seems to set this bound after showing 4 and 11 cent combinations up to 28. However, I do not see why 30 was chosen as a lower bound. Thanks.
 
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I'm sorry for the short reply, I'm running a slight fever, but i'll like to try to give you some advice since it's midday here and I've noticed no one else has tried to help you. I've encountered problems like this and if I do recall it was in Rosen's discrete mathematics. Anyway, the idea is what you need to find the number for which you can make every consecutive number appear. For example, the number cannot be 4, because you can't make 5, 6, 7 nor can it be 8 because you can't make 9 or 10. If you continue on this path, I think you'll find that 28 will be the least number you can have such that you can make 28 + k, where k is an natural number. To prove this, I also suspect you'll eventually need to show that 29, 30 also work, and proceed with the strong induction form. I hope that clears this up some.
 

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