Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true for all natural numbers (non-negative integers). It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.

The simplest and most common form of mathematical induction proves that a statement involving a natural number $n$ holds for all values of $n$. The proof consists of two steps:

1. The basis (base case): showing that the statement holds when $n$ is equal to the lowest value that $n$ is given in the question. Usually, $n=0$ or $n=1$.
2. The inductive step: showing that if the statement holds for some $n$, then the statement also holds when $n+1$ is substituted for $n$.

The assumption in the inductive step that the statement holds for some $n$, is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for $n+1$.

The choice between $n=0$ and $n=1$ in the base case is specific to the context of the proof: If $0$ is considered a natural number, as is common in the fields of combinatorics and mathematical logic, then $n=0$. If, on the other hand, $1$ is taken as the first natural number, then the base case is given by $n=1$.

This method works by first proving the statement is true for a starting value, and then proving that the process used to go from one value to the next is valid. If these are both proven, then any value can be obtained by performing the process repeatedly. 

Visualization

It may be helpful to think of the domino effect. If one is presented with a long row of dominoes standing on end, one can be sure that:

1. The first domino will fall
2. Whenever a domino falls, its next neighbor will also fall

So it is concluded that all of the dominoes will fall, and that this fact is inevitable.

Dominoes

Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes.

Example

Prove the following statement: For each positive integer $n$,

$\displaystyle{0 + 1 + 2 + \cdots + n = \frac{n(n+1)}{2}}$

Base Case: We first have to check that the statement holds for $n=0$. On the left-hand side of the equation, we have only $0$. On the right hand side of the equation we substitute $n=0$. Thus we have  $\displaystyle{0 = \frac{0(0+1)}{2}}$, which can be simplified to $0= 0$. We have proven that the statement holds true for the base case of $n=0$.

Inductive Step: Assume that the statement holds for $n=k$, and check if it holds for $n=k+1$ as well. In other words, we want to show that the statement holds true when we substitute $k+1$ for $n$:

$\displaystyle{0 + 1 + 2 + \cdots + k + (k+1) = \frac{\left(k+1\right)\left[\left(k+1\right)+1\right]}{2}}$

 Note that we can use the induction hypothesis that the statement holds for $n=k$ to rewrite the left-hand side of the equation:

$\displaystyle{0 + 1 + 2 + \cdots + k + (k+1) = \frac{k(k+1)}{2}} + (k+1)$

We can now rewrite this statement and show that it equals the right-hand side of the previous statement. In other words, if we can show that the following holds true, we will show that $\displaystyle{0 + 1 + 2 + \cdots + n = \frac{n(n+1)}{2}}$ holds true for $k+1$:

$\displaystyle{\frac{k(k+1)}{2} + (k+1) = \frac{\left(k+1\right)\left[\left(k+1\right)+1\right]}{2} }$

We can rewrite $\displaystyle{\frac{k(k+1)}{2} + (k+1)}$ algebraically as follows:

$\displaystyle{ \begin{aligned} \frac{k(k+1)}{2} + (k+1) &= \frac{k(k+1) + 2(k+1)}{2} \\ &=\frac{(k+1)(k+2)}{2} \\ &=\frac{\left(k+1\right)\left[\left(k+1\right)+1\right]}{2} \end{aligned} }$

This is exactly we wanted to prove. We have shown that $\displaystyle{0 + 1 + 2 + \cdots + n = \frac{n(n+1)}{2}}$ holds true for any $n = k + 1$ if it holds true for $n = k$. This completes the induction step. We conclude that the statement holds for every non-negative integer $n$.