Section 2.1 Mathematical Induction
Suppose we wish to show that
\begin{equation*}
1 + 2 + \cdots + n = \frac{n(n + 1)}{2}
\end{equation*}
for any natural number \(n\text{.}\) This formula is easily verified for small numbers such as \(n = 1\text{,}\) \(2\text{,}\) \(3\text{,}\) or \(4\text{,}\) but it is impossible to verify for all natural numbers on a case-by-case basis. To prove the formula true in general, a more generic method is required.
Suppose we have verified the equation for the first \(n\) cases. We will attempt to show that we can generate the formula for the \((n + 1)\)th case from this knowledge. The formula is true for \(n = 1\) since
\begin{equation*}
1 = \frac{1(1 + 1)}{2}\text{.}
\end{equation*}
If we have verified the first \(n\) cases, then
\begin{align*}
1 + 2 + \cdots + n + (n + 1) & = \frac{n(n + 1)}{2} + n + 1\\
& = \frac{n^2 + 3n + 2}{2}\\
& = \frac{(n + 1)[(n + 1) + 1]}{2}\text{.}
\end{align*}
This is exactly the formula for the \((n + 1)\)th case.
This method of proof is known as mathematical induction. Instead of attempting to verify a statement about some subset \(S\) of the positive integers \({\mathbb N}\) on a case-by-case basis, an impossible task if \(S\) is an infinite set, we give a specific proof for the smallest integer being considered, followed by a generic argument showing that if the statement holds for a given case, then it must also hold for the next case in the sequence. We summarize mathematical induction in the following axiom.
Principle 2.1. First Principle of Mathematical Induction.
Let \(S(n)\) be a statement about integers for \(n \in {\mathbb N}\) and suppose \(S(n_0)\) is true for some integer \(n_0\text{.}\) If for all integers \(k\) with \(k \geq n_0\text{,}\) \(S(k)\) implies that \(S(k+1)\) is true, then \(S(n)\) is true for all integers \(n\) greater than or equal to \(n_0\text{.}\)
Example 2.2.
For all integers \(n \geq 3\text{,}\) \(2^n \gt n + 4\text{.}\) Since
\begin{equation*}
8 = 2^3 \gt 3 + 4 = 7\text{,}
\end{equation*}
the statement is true for \(n_0 = 3\text{.}\) Assume that \(2^k \gt k + 4\) for \(k \geq 3\text{.}\) Then \(2^{k + 1} = 2 \cdot 2^{k} \gt 2(k + 4)\text{.}\) But
\begin{equation*}
2(k + 4) = 2k + 8 \gt k + 5 = (k + 1) + 4
\end{equation*}
since \(k\) is positive. Hence, by induction, the statement holds for all integers \(n \geq 3\text{.}\)
Example 2.3.
Every integer \(10^{n + 1} + 3 \cdot 10^n + 5\) is divisible by \(9\) for \(n \in {\mathbb N}\text{.}\) For \(n = 1\text{,}\)
\begin{equation*}
10^{1 + 1} + 3 \cdot 10 + 5 = 135 = 9 \cdot 15
\end{equation*}
is divisible by \(9\text{.}\) Suppose that \(10^{k + 1} + 3 \cdot 10^k + 5\) is divisible by \(9\) for \(k \geq 1\text{.}\) Then
\begin{align*}
10^{(k + 1) + 1} + 3 \cdot 10^{k + 1} + 5& = 10^{k + 2} + 3 \cdot 10^{k + 1} + 50 - 45\\
& = 10 (10^{k + 1} + 3 \cdot 10^{k} + 5) - 45
\end{align*}
is divisible by \(9\text{.}\)
Example 2.4.
We will prove the binomial theorem using mathematical induction; that is,
\begin{equation*}
(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n - k}\text{,}
\end{equation*}
where \(a\) and \(b\) are real numbers, \(n \in \mathbb{N}\text{,}\) and
\begin{equation*}
\binom{n}{k} = \frac{n!}{k! (n - k)!}
\end{equation*}
is the binomial coefficient. We first show that
\begin{equation*}
\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}\text{.}
\end{equation*}
This result follows from
\begin{align*}
\binom{n}{k} + \binom{n}{k - 1} & = \frac{n!}{k!(n - k)!} +\frac{n!}{(k-1)!(n - k + 1)!}\\
& = \frac{(n + 1)!}{k!(n + 1 - k)!}\\
& =\binom{n + 1}{k}\text{.}
\end{align*}
If \(n = 1\text{,}\) the binomial theorem is easy to verify. Now assume that the result is true for \(n\) greater than or equal to \(1\text{.}\) Then
\begin{align*}
(a + b)^{n + 1} & = (a + b)(a + b)^n\\
& = (a + b) \left( \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n - k}\right)\\
& = \sum_{k = 0}^{n} \binom{n}{k} a^{k + 1} b^{n - k} + \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n + 1 - k}\\
& = a^{n + 1} + \sum_{k = 1}^{n} \binom{n}{k - 1} a^{k} b^{n + 1 - k} + \sum_{k = 1}^{n} \binom{n}{k} a^k b^{n + 1 - k} + b^{n + 1}\\
& = a^{n + 1} + \sum_{k = 1}^{n} \left[ \binom{n}{k - 1} + \binom{n}{k} \right]a^k b^{n + 1 - k} + b^{n + 1}\\
& = \sum_{k = 0}^{n + 1} \binom{n + 1}{k} a^k b^{n + 1- k}\text{.}
\end{align*}
We have an equivalent statement of the Principle of Mathematical Induction that is often very useful.
Principle 2.5. Second Principle of Mathematical Induction.
Let \(S(n)\) be a statement about integers for \(n \in {\mathbb N}\) and suppose \(S(n_0)\) is true for some integer \(n_0\text{.}\) If \(S(n_0), S(n_0 + 1), \ldots, S(k)\) imply that \(S(k + 1)\) for \(k \geq n_0\text{,}\) then the statement \(S(n)\) is true for all integers \(n \geq n_0\text{.}\)
A nonempty subset \(S\) of \({\mathbb Z}\) is well-ordered if \(S\) contains a least element. Notice that the set \({\mathbb Z}\) is not well-ordered since it does not contain a smallest element. However, the natural numbers are well-ordered.
Principle 2.6. Principle of Well-Ordering.
Every nonempty subset of the natural numbers is well-ordered.
The Principle of Well-Ordering is equivalent to the Principle of Mathematical Induction.
Lemma 2.7.
The Principle of Mathematical Induction implies that \(1\) is the least positive natural number.
Proof.
Let \(S = \{ n \in {\mathbb N} : n \geq 1 \}\text{.}\) Then \(1 \in S\text{.}\) Assume that \(n \in S\text{.}\) Since \(0 \lt 1\text{,}\) it must be the case that \(n = n + 0 \lt n + 1\text{.}\) Therefore, \(1 \leq n \lt n + 1\text{.}\) Consequently, if \(n \in S\text{,}\) then \(n + 1\) must also be in \(S\text{,}\) and by the Principle of Mathematical Induction, and we have \(S = \mathbb N\text{.}\)
Theorem 2.8.
The Principle of Mathematical Induction implies the Principle of Well-Ordering. That is, every nonempty subset of \(\mathbb N\) contains a least element.
Proof.
We must show that if
\(S\) is a nonempty subset of the natural numbers, then
\(S\) contains a least element. If
\(S\) contains 1, then the theorem is true by
Lemma 2.7. Assume that if
\(S\) contains an integer
\(k\) such that
\(1 \leq k \leq n\text{,}\) then
\(S\) contains a least element. We will show that if a set
\(S\) contains an integer less than or equal to
\(n + 1\text{,}\) then
\(S\) has a least element. If
\(S\) does not contain an integer less than
\(n+1\text{,}\) then
\(n+1\) is the smallest integer in
\(S\text{.}\) Otherwise, since
\(S\) is nonempty,
\(S\) must contain an integer less than or equal to
\(n\text{.}\) In this case, by induction,
\(S\) contains a least element.
Induction can also be very useful in formulating definitions. For instance, there are two ways to define \(n!\text{,}\) the factorial of a positive integer \(n\text{.}\)
Every good mathematician or computer scientist knows that looking at problems recursively, as opposed to explicitly, often results in better understanding of complex issues.
