Suppose we start with a subset $A$ of $N$ i.e. all the things in $A$ are also in $N$. The induction axiom says if you know 1 is in $A$ and if you can prove that a number $n$ being in $A$ implies that $n+1$ is also in $A$ then $A$, is in fact, all of $N$.

When you are asked to solve a problem, the set $A$ is generally defined by some property $P$ of the numbers it contains. That is to say $A=\{n\in N:P(n)\text{is true}\}$. For example if $P(n)=`(n+1{)}^2=n^2+2n+1\text{'}$ then $A=\{n\in N:(n+1{)}^2=n^2+2n+1\}=N$ since $P(n)$ holds for every natural $n$.

The important thing, that you need to remember to solve problems, is:

If 1 has a particular property $P$ and the fact that any number $n$ has the property $P$ implies that $n+1$ has the property $P$ then every natural number has the property $P$. This is written in mathematical short-hand as:

If $P(1)$ and for all $n$ in $N$ $P(n)\Rightarrow P(n+1)$ then $P(m)$ holds for all $m$ in $N$.

Examples

Suppose you want to prove the statement $\sum _{i=1}^{n}i=(n^2+n)/2$ for every $n$ in $N$. We apply induction with $P(n)=`\sum _{i=1}^{n}i=(n^2+n)/2\text{'}$. There are two steps:

1. Is $P(1)$ true? - Yes because $\sum _{i=1}^{1}i=1=(1^2+1)/2$
2. Does $P(n)\Rightarrow P(n+1)$ for every $n$? - Yes because $\sum _{i=1}^{n+1}i=\sum _{i=1}^{n}i+n+1=(n^2+n)/2+n+1=((n^2+2n+1)+(n+1))/2=((n+1{)}^2+(n+1))/2$