Complete Induction: Proving using WOP/Induction

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SUMMARY

The discussion centers on proving the principle of complete induction using the well-ordering principle. The solution presented establishes that if a collection A of natural numbers contains 1 and is closed under the successor function, then the complement B of A cannot have a least element, leading to the conclusion that A must equal the set of natural numbers, ℕ. This proof effectively demonstrates the equivalence of complete induction and the well-ordering principle.

PREREQUISITES
  • Understanding of complete induction principles
  • Familiarity with the well-ordering principle
  • Knowledge of natural numbers and their properties
  • Basic proof techniques in mathematics
NEXT STEPS
  • Study the formal definitions of complete induction and the well-ordering principle
  • Explore examples of proofs using complete induction
  • Learn about other proof techniques such as contradiction and contrapositive
  • Investigate applications of induction in combinatorics and number theory
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Students of mathematics, particularly those studying mathematical proofs, educators teaching induction methods, and anyone interested in foundational concepts of number theory.

jgens
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Homework Statement



Prove the principle of complete induction using either the well-ordering principle or ordinary mathematical induction.

Homework Equations



N/A

The Attempt at a Solution



Suppose that [itex]A[/itex] is a collection of natural numbers such that [itex]1 \in A[/itex] and [itex]n+1 \in A[/itex] whenever [itex]1, \dots, n \in A[/itex]. Now suppose that [itex]B[/itex] is the collection of those natural numbers not in [itex]A[/itex]. If [itex]B \neq \emptyset[/itex], then by the well-ordering principle, [itex]B[/itex] must have some least element. Clearly, this least element cannot be 1 so suppose instead that [itex]k+1[/itex] is the least element in [itex]B[/itex], in which case [itex]1, \dots, k[/itex] are all in [itex]A[/itex]; however, this implies that [itex]k+1 \in A[/itex] which is a contradiction. Thus our assumption that [itex]B[/itex] was non-empty must have been incorrect and [itex]A = \mathbb{N}[/itex].

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