Math 2.0

Mathematical logic and foundations Theorem

Induction from an arbitrary base case

Builds on Mathematical inductionOrder on the natural numbers

Statement

Let mm be a natural number and let P(n)P(n) be a property of natural numbers nn with n≥mn \ge m. Suppose that P(m)P(m) holds. Suppose also that for each natural number n≥mn \ge m, P(n)P(n) implies P(n+1)P(n + 1). Then P(n)P(n) holds for every natural number n≥mn \ge m.

Remarks

The proof applies ordinary induction to the property Q(k)Q(k): "P(m+k)P(m + k) holds". For example, induction from m=4m = 4 shows that 2n≥n22^n \ge n^2 for every natural number n≥4n \ge 4.

Reading path

To understand this page from the ground up, read these 8 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  1. Definition by recursion concept
  2. Mathematical induction concept
  3. Natural numbers object
  4. Peano axioms concept
  5. Successor function definition
  6. Addition of natural numbers definition
  7. Order on the natural numbers definition
  8. Induction from an arbitrary base case statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Mathematical_induction, Base case other than 0 or 1
About this page
Tag0GF6AJEBZ0
URIm2://0GF6AJEBZ0
Kindstatement, theorem
WritingW0 stub
MSC03F30
StatusThis copy of the site has no build data. Run m2 status 0GF6AJEBZ0.
UsesMathematical induction, Order on the natural numbers

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