For simplicity, just think about the most basic form, as described before Theorem
Theorem 3.1.2.1. We proceed by contradiction.
Suppose the principle of mathematical induction is false. Then the set
\(S\) of positive integers for which
\(P(n)\) is false is nonempty.
By the well-ordering property,
\(S\) must have a least element. Let
\(m\) be the least element in
\(S\text{.}\) Then
\(P(m)\) is false.
Because
\(P(1)\) is true by the basis step, we have
\(m\ne 1\text{.}\) Because
\(m\) is positive and not
\(1\text{,}\) we must have that
\(m\) is at least
\(2\text{.}\) Then
\(m-1\) is a positive integer less than
\(m\text{,}\) and
\(P(m - 1)\) is true. By the inductive step, we know that
\(P(m-1)\) is true implies
\(P(m)\) is true. But
\(P(m)\) is true contradicts the proposition that
\(P(m)\) is true which we followed from the existence of the non-empty set
\(S\text{.}\) Thus such a set
\(S\) cannot exists which implies that induction is valid.
The proof the more general case as well as strong-induction is similar.