The Well-Ordering Principle and Mathematical Induction
1.1. The Well-Ordering Principle and Mathematical Induction
🧭 Overview
🧠 One-sentence thesis
The well-ordering principle, pigeonhole principle, and mathematical induction are three fundamental proof tools for establishing properties of integers, with induction relying on well-ordering to show that properties holding for one natural number propagate to all subsequent numbers.
📌 Key points (3–5)
- Well-ordering principle: every non-empty set of natural numbers has a least element (smallest member).
- Two forms of induction: the first principle uses a single previous case (k implies k+1), while the second uses all previous cases (1 through k imply k+1).
- How induction works: prove a base case (usually n=1), then prove that if the statement holds for k, it must hold for k+1; this chain covers all natural numbers.
- Common confusion: the two induction principles differ in their inductive step—don't confuse "assuming P(k)" with "assuming P(1) through P(k)."
- Why these tools matter: they provide rigorous methods to prove statements about all integers without checking infinitely many cases individually.
🏗️ Foundational principles
🏗️ Well-ordering principle
Every non-empty set of natural numbers has a least element.
- This means: if you have any collection of natural numbers (1, 2, 3, ...) that isn't empty, there is always a smallest one.
- The excerpt notes this is "often taken as an axiom"—a starting assumption rather than something proved.
- Example: the set {5, 17, 3, 100} has least element 3; the set of all even numbers has least element 2.
🕳️ Pigeonhole principle
If s objects are placed in k boxes where s > k, then at least one box contains more than one object.
- Plain language: if you have more items than containers, at least one container must hold multiple items.
- The proof uses contradiction: if no box had more than one object, there would be at most k objects total, contradicting s > k.
- Example: if 10 students sit in 9 chairs, at least one chair must have 2 or more students.
🔁 First principle of mathematical induction
🔁 The structure
The first principle has two required steps:
| Step | What you prove | Why it matters |
|---|---|---|
| Base case | Property P(1) is true | Establishes a starting point |
| Inductive step | For any k, if P(k) is true then P(k+1) is true | Creates a chain: 1→2→3→... |
- If both steps succeed, the property holds for all natural numbers.
- The excerpt states it more formally: if a set S contains 1, and whenever S contains k it also contains k+1, then S equals all natural numbers.
🔗 How the proof works
- The proof uses well-ordering by contradiction.
- Assume some natural numbers are NOT in S; call this set T.
- By well-ordering, T has a least element ℓ.
- Since 1 is in S, we know ℓ > 1, so ℓ-1 is a natural number.
- Since ℓ is the least element not in S, ℓ-1 must be in S.
- But the inductive property says if ℓ-1 is in S, then (ℓ-1)+1 = ℓ must also be in S—contradiction!
- Therefore T is empty and S contains all natural numbers.
📐 Example: sum formula
The excerpt proves that the sum 1 + 2 + ... + n equals n(n+1)/2.
Base case (n=1):
- Left side: 1
- Right side: 1·2/2 = 1
- They match ✓
Inductive step:
- Assume the formula holds for some n (this is the "inductive hypothesis").
- Prove it for n+1: the sum 1 + 2 + ... + n + (n+1).
- Split this as [1 + 2 + ... + n] + (n+1).
- By the inductive hypothesis, the bracketed part equals n(n+1)/2.
- So the total is n(n+1)/2 + (n+1) = (n+1)(n+2)/2, which is the formula for n+1 ✓
📈 Example: factorial inequality
The excerpt proves n! ≤ n^n for all natural numbers n (where n! means n factorial and n^n means n to the nth power).
Base case (n=1):
- 1! = 1 and 1^1 = 1, so 1 ≤ 1 ✓
Inductive step:
- Assume k! ≤ k^k for some k.
- Prove (k+1)! ≤ (k+1)^(k+1).
- Note (k+1)! = (k+1)·k! ≤ (k+1)·k^k by the inductive hypothesis.
- Since k^k < (k+1)^k, we get (k+1)·k^k < (k+1)·(k+1)^k = (k+1)^(k+1) ✓
🔁🔁 Second principle of mathematical induction
🔁🔁 The stronger assumption
The second principle differs in the inductive step:
- First principle: assume P(k) is true, prove P(k+1).
- Second principle: assume P(1), P(2), ..., P(k) are ALL true, prove P(k+1).
The second form is "stronger" because you can use more information in your proof—not just the immediately previous case, but all previous cases.
🔗 Relationship to the first principle
- The excerpt proves the second principle using the first principle.
- Define a new property: "P(n) means that P(1), P(2), ..., P(n) are all true."
- Apply the first principle to this new property.
- Details are left to the reader, but the key idea is that the second principle is not fundamentally different—it's a consequence of the first.
🎯 When to use which
- Use the first principle when the step from n to n+1 depends only on the case n.
- Use the second principle when proving P(n+1) might require looking back at multiple earlier cases (e.g., P(n-1), P(n-2), etc.).
- Don't confuse: both principles are valid; the second just gives you more flexibility in the inductive step.