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Mathematical Induction

Induction is closely related to recursion. If a base case satisfies a property and a step carries the property from one element to the next, then the property holds for every element reachable from the base case by repeatedly applying the step.

Below is the principle of induction as presented by various authors.

We start with Charles Pinter 1.

Consider the following conditions:

  1. S1S_1 is true.
  2. For any k>0∈Zk>0 \in \mathbb{Z}, if SkS_k is true, then also Sk+1S_{k+1} is true.

If conditions 1 and 2 are satisfied, then SnS_n is true for every positive integer n.

Then, the definition by Halmos 2:

  1. 0∈ω0 \in \omega
  2. if n∈ωn\in \omega, then n+∈ωn^{+}\in \omega, where n+=n∪{n}n^+ = n \cup \{n\}
  3. if S⊂ωS\subset \omega, if 0⊂S0\subset S, and if n+∈Sn^+ \in S, whenever n∈Sn\in S, then S=ωS=\omega

Finally, Hamkins 3:

Least number principle: if there is a natural number with a property, then there is a smallest number with that property.

Common induction principle: Suppose that AA is a set of natural numbers with 0∈A0\in A. Whenever n∈An\in A, then also n+1∈An+1 \in A. Therefore, every natural number is in AA.

The base case (e.g., S1S_1 is true, or 0∈A0\in A), is referred to as the anchor, whereas the implication (e.g., Sk+1S_{k+1} is true or n+1∈An+1 \in A), is called the induction step.

Examples

The following is an inductive proof, written in Lean, of the fact that:

sum(n)=∑i=0n=n(n+1)2\text{sum}(n) = \sum_{i=0}^{n} = \frac{n(n+1)}{2}

This is also the method used to calculate the nth triangular number 3:

The following figure makes it easier to visualize the recursion used in the proof:

sum(n+1)=(n+1)+sum(n)\text{sum}(n+1) = (n + 1) + \text{sum}(n)

4th and 5th triangular numbers plotted using figural

Open in Lean-WebFollow the InfoView on the right. Hover over (or click) reserved words to see their descriptions

Finally, here is an example of an equality that I encountered in seventh grade while reading The Man Who Counted.

According to the legend narrated in this book, after King Iadava insisted, Sessa asked for one grain of wheat for the first cell, two for the second, four for the third, and so on as a reward for inventing the game of chess:

∑i=0632i\sum_{i=0}^{63} 2^i

The following Lean code proves that:

∑i=0n2i=2n+1−1\sum_{i=0}^{n} 2^i = 2^{n+1} - 1 Open in Lean-WebFollow the InfoView on the right. Hover over (or click) reserved words to see their descriptions

References

Footnotes

  1. Charles Pinter, A Book of Abstract Algegra, McGraw-Hill, 1982. ↩

  2. Paul R. Halmos, Naive Set Theory. Springer Verlag New York, 1974. ↩

  3. Joel David Hamkins, Proof and the Art of Mathematics. MIT Press, 2020. Chapter 4: Mathematical Induction. ↩ ↩2


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