Mathematical Induction

Mathematical Induction is a method of proof typically used to establish that a given statement is true for all natural numbers. This is often the first type of proof statement seen by a high school student.

First Principle of Finite Induction
Let $S$ be a set of positive integers with the following properties:
(1) The integer 1 belongs to the set.
(2) Whenever the integer $k$ is in $S$, then the next integer $k+1$ must also be in $S$.
Then $S$ is the set of all positive integers.

This statement seems almost immediately obvious, and it can be proved by contradiction. Often, an analogy is drawn with a sequence of falling dominos; if the first domino falls, and each subsequent domino pushes it's neighbor, then all dominos in this sequence must eventually fall. For now, we will focus on how to prove statements by induction. These proofs often have a standard format that can be followed, as shown below:

Statement: Let $P_{n}$ be the proposition Induction Hypothesis, for $n$ in Domain.

Base Case: Consider Base Case.
LHS=LHS
RHS=RHS
Since LHS=RHS, thus Base Case is true.

Induction step: Assume $P_{k}$ is true for some $k$ in Domain. Consider $P_{k+1}$.
LHS $P_{k+1}$ =use Induction Hypothesis =RHS $P_{k+1}$.
Hence $P_{k}$ is true implies that $P_{k+1}$ is true.

Conclusion: By Mathematical Induction, since Base Case is true and $P_{k}$ is true implies $P_{k+1}$ is true, thus $P_{n}$ is true for all $n$ in Domain. $\square$

There are 4 main components in the proof, and you have to fill in the details of the various parts left in italics. Let us work through a specific example, to get a better sense of how this works.

Worked Examples

1. Show that for all positive integers $n$,

$\sum_{i=1} ^n i^2 = \frac {n(n+1)(2n+1)} {6}.$

Solution: This is a well known statement about the sum of the first $n$ squares, and we will show this via induction.

Statement: Let $P_{n}$ be the proposition that $\sum_{i=1} ^n i^2 = \frac {n(n+1)(2n+1)} {6}$, for all positive integers.

[We have stated what the Induction Hypothesis is, and specified the domain in which the statement is true.]

Base Case: Consider $n=1$.
LHS= $\sum_{i=1} ^ 1 1^2 = 1$.
RHS= $\frac{ 1 \times (1+1) \times (2 + 1) } {6} = 1$.
Since LHS=RHS, thus the Base Case is true.

[We have identified the Base Case, and shown that it is true.]

Induction Hypothesis: Assume $P_k$ is true for some positive integer $k$. Consider $P_{k+1}$.
$\begin{aligned} \mbox{LHS } P_{k+1} & = \sum_{i=1}^{k+1} i^2 \\ & = \left( \sum_{i=1}^k i^2 \right) + (k+1) ^2 \\ & = \frac{ k(k+1)(2k+1) } { 6} + (k+1) ^2 \\ & = (k+1) \left( \frac{ 2k^2 + k } { 6} + \frac{ 6 (k+1) } { 6} \right) \\ & = (k+1) \frac{ 2k^2 + 7k + 6 } { 6} \\ & = \frac { (k+1) (k+2) ( 2k + 3) }{6} \\ & = \mbox{RHS } P_{k+1} \\ \end{aligned}$ Hence $P_{k}$ is true implies that $P_{k+1}$ is true.

[We successfully used the induction hypothesis (in the 3rd equation) along with algebraic manipulation to show that the next statement is true.]

Conclusion: By Mathematical Induction, since $P_1$ is true and $P_{k}$ is true implies $P_{k+1}$ is true, thus $P_{n}$ is true for all positive integer $n$. $\square$

[We now sit back and drink a cup of coffee.]

2. Show that for all positive integers $n$, $3^{2n+1}+40n-3$ is a multiple of 64.

Solution: Statement: Let $P_n$ be the proposition that $3^{2n+1}+40n-3$ is a multiple of 64, for all positive integers $n$.

Base Case: Consider $n = 1$.
Since $3^{3} + 40 - 3 = 64$, it is a multiple of 64.
Hence, the Base case is true.

Induction Hypothesis: Assume $P_k$ is true for some positive integer $k$. Consider $P_{k+1}$.
$3^{2k+3} + 40(k+1) - 3 = 9 \times ( 3^{2k+1} + 40k - 3 ) -320k + 64$. Since $-320k+ 64 = 64 ( -5k + 1)$ is a multiple of 64, and the first term is a multiple of 64 by the Induction Hypothesis, hence the LHS is a multiple of $64$.
Hence $P_{k}$ is true implies that $P_{k+1}$ is true.

Conclusion: By Mathematical Induction, since $P(1)$ is true and $P_{k}$ is true implies $P_{k+1}$ is true, thus $P_{n}$ is true for all positive integer $n$. $\square$