Mathematical Induction

A good informal analogy of mathematical induction is a line of up-ended dominos - if you can show that both of the following are true:

  1. If any one of the dominos fall, then the next domino will also fall
  2. The first domino must fall

then you can say with certainty that all the dominos will fall.

The "line of dominos" here is in fact some sequence of mathematical statements. For example, to assert the truth of the binomial theorem:

\begin{equation*} (x + y)^n = \sum_{i=0}^{n} {n \choose i} x^iy^{n-i}, n = 0, 1, 2, \dots \end{equation*}

is to assert the truth of each item of the sequence:

\begin{equation*} \begin{array} \\ (x + y)^0 &= \sum_{i=0}^{0} {0 \choose i} x^iy^{0-i} \\ (x + y)^1 &= \sum_{i=0}^{1} {1 \choose i} x^iy^{1-i} \\ (x + y)^2 &= \sum_{i=0}^{2} {2 \choose i} x^iy^{2-i} \\ \end{array} \end{equation*}

and so on. So if you can:

  • Show that IF the \(k\)th item in the sequence is true, then the \(k + 1\)th case must also be true
  • AND show that the first item in the sequence is true (when \(n = 0\))

then it can be said that the proposed theorem is true for all \(n \in \mathbb{N} \).

Strong Induction

The two parts of an Inductive proof are the base case and the inductive step. The base case is, as stated, the proof of the given statement for the first item in the sequence - typically when \(n = 0\) but, in general, when \(n\) is some lowest positive integer. And the inductive step is the assumption of the \(k\)th case and the subsequent proof of the \(k+1\)th case.

It may be the case that assumption of the \(k\)th case is insufficient in achieving the inductive step, and so invoking strong induction is called for. With strong induction you assume, not just the \(k\)th case, but each case up to and including the \(k\)th.

see also:Wikipedia, NRICH, The Method of Descent, proofwiki.org

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