Discrete mathematics induction

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SUMMARY

The discussion centers on proving that for all integers \( a \geq 1 \), \( a^n - 1 \) is divisible by \( a - 1 \) for all \( n \geq 1 \). The proof utilizes mathematical induction, starting with the base case \( P(1) \) and assuming \( P(k) \) holds true. The inductive step demonstrates that if \( a^k - 1 \) is divisible by \( a - 1 \), then \( a^{k+1} - 1 \) is also divisible by \( a - 1 \). The conclusion confirms the statement holds for all integers \( a \geq 1 \).

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  • Understanding of mathematical induction
  • Familiarity with polynomial expressions
  • Basic knowledge of divisibility rules
  • Concept of integers and their properties
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  • Study advanced topics in mathematical induction
  • Explore polynomial long division techniques
  • Learn about the properties of divisibility in number theory
  • Investigate applications of induction in combinatorics
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Students of discrete mathematics, educators teaching mathematical proofs, and anyone interested in enhancing their understanding of mathematical induction and divisibility concepts.

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Homework Statement



Prove that for all integers a >= 1, a^n - 1 is divisible by a - 1 for all n >= 1.

Homework Equations



None.

The Attempt at a Solution



Proof - Let P(n): a^n - 1 is divisible by a - 1, then

P(1): a^1 - 1 is divisible by a - 1 is TRUE since a^1 - 1 = a - 1, and a - 1 = (1)(a - 1)

Suppose that P(k): a^k - 1 is divisible by a - 1 is TRUE. Then, there exists an integer q such that a^k - 1 = (a - 1)q or a^k = (a-1)q + 1

Now consider a^(k+1) - 1 = a(a^k) - 1
= a((a - 1)q + 1) - 1
= (a)(a - 1)q + a - 1
= (a - 1)aq + (a - 1)
= (a - 1)(aq + 1)

Thus, a^(k+1) - 1 is divisible by a - 1 whenever a^k - 1 is divisible by a - 1. Therefore, a^n - 1 is divisible by a -1 for all n >= 1.


The problem I'm having is I don't think I proved that for all a >= 1.
 
Physics news on Phys.org
No, there is no requirement that a- 1 be greater than 0 so there is no requirement that a be greater than 1.
 

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