“Wait, induction? I thought math was deductive?” Well, yes, math is deductive and, in fact, mathematical induction is actually a deductive form of reasoning; if that doesn't make your brain hurt, it should. So, actually, mathematical induction seems like a misnomer, but really we give it that name because it reminds us of inductive reasoning in science.
I like to think of mathematical induction via an analogy. How can I convince you that I can climb a ladder? Well, first I show you that I can climb onto the first rung, which is obviously important. Then I convince you that for any rung, if I can get to that rung, then I can get to the next one.
Are you convinced? Well, let's see. I showed you I can get to the first rung, and then, by the second property, since I can get to the first rung, I can get to the second. Then, since I can get to the second rung, by the second property, I can get to the third, and so on and so forth. Thus I can get to any rung.
If \(S \subseteq \mathbb{N}\) with the properties:
then \(S = \mathbb{N}\text{.}\) A subset of \(\mathbb{N}\) is called inductive if it has the second property listed above. An inductive set is some tail of the set of natural numbers (i.e. it is the natural numbers less some initial segment)
If you are wondering about a proof of this theorem, then stop. We should actually call it a theorem, but instead, the definition of the natural numbers, but that's not important.
The cool thing about induction (we will henceforth drop the formality of “mathematical induction”) is that it allows us to prove infinitely many statements. How does it do this? Suppose we have a proposition \(P(n)\) for each natural number \(n \in \mathbb{N}\) and we want to prove that for all \(n \in \mathbb{N}\text{,}\) the statement \(P(n)\) is true. Well, we prove two things:
Then the set \(S = \{ n \in \mathbb{N} \mid P(n)\ \text{is true} \}\) satisfies the conditions of Theorem 2.4.1 and so \(S = \mathbb{N}\text{.}\)
Suppose we want to prove that for every \(n \in \mathbb{N}\text{,}\) \(3\) divides \(n^3 - n\text{.}\) Here, our proposition \(P(n)\) is that \(3\) divides \(n^3 - n\text{,}\) and we want to prove \((\forall n \in \mathbb{N})(P(n))\text{.}\)
We start by proving the base case. Note that \(1^3 - 1 = 0 = 3 \cdot 0\text{,}\) so the claim holds when \(n = 1\text{.}\)
We now prove the inductive step. Let \(n \in \mathbb{N}\) be an arbitrary integer and suppose that \(3\) divides \(n^3 - n\text{,}\) so that \(n^3 - n = 3k\) for some \(k \in \mathbb{Z}\text{.}\) Then Now notice that
\begin{align*} (n+1)^3 - (n+1) &= (n+1)\big((n+1)^2 - 1\big)\\ &= n(n+1)(n+2)\\ &= n^3 + 3n^2 + 2n\\ &= (n^3 - n) + 3n^2 + 3n\\ &= 3(k + n^2 + n) \end{align*}Thus \(3\) divides \((n+1)^3 - (n+1)\text{.}\)
By the Principle of Mathematical Induction (PMI), for every \(n \in \mathbb{N}\text{,}\) \(3\) divides \(n^3 - n\text{.}\)
Induction also allows us to define infinitely many things at the same time. For example, consider the function \(f(n) = n^2 - n + 2\text{.}\) We will define a sequence of numbers \(a_n\) by:
\begin{equation*} a_1 = 0, \qquad a_{n+1} = f(a_n). \end{equation*}For those of you familiar with computer science or programming, you may think of this as a recursively defined sequence. Induction and recursion are two sides of the same coin; we won't address the difference here.