The simplest and most common form of mathematical induction proves that a statement holds for all natural numbers n and consists of two steps:
Table of contents |
2 Generalizations 3 Proof or reformulation of mathematical induction |
Now we have to show that if the statement holds when n = m, then it also holds when n = m + 1. This can be done as follows.
Assume the statement is true for n = m, i.e.,
This type of proof can be generalized in several ways. For instance, if we want to prove a statement not for all natural numbers but only for all numbers greater than or equal to a certain number b then the following steps are sufficient:
Another generalization allows that in the second step we not only assume that the statement holds for n = m but also for all n smaller than or equal to m. This leads to the following two steps:
The last two steps can be reformulated as one step:
Example
Suppose we wish to prove the statement:
for all natural numbers n.
This is a simple formula for the sum of the natural numbers up to the number n. The proof that the statement is true for all natural numbers n proceeds as follows.Proof
Check it is true for n = 0. Clearly, the sum of the first 0 natural numbers is 0, and 0.(0 + 1) / 2 = 0. So the statement is true for n = 0. We could define the statement as P(n), and thus we have that P(0) holds.
Adding m + 1 to both sides gives
By algebraic manipulation we have
Thus we have
This is that statement for n = m + 1. Note that it has not been proved as true: we made the assumption that P( m ) is true, and from that assumption we derived P( m + 1 ). Symbolically, we have shown that:
However, by induction, we may now conclude that the statement P(n) holds for all natural numbers n:Generalizations
This can be used, for example, to show that n2 > 2n for n ≥ 3. Note that this form of mathematical induction is actually a special case of the previous form because if the statement that we intend to prove is P(n) then proving it with these two rules is equivalent with proving P(n + b) for all natural numbers n with the first two steps.
This can be used, for example, to show that fib(n) = [Φn - (-1/Φ)n ] / 51/2 where fib(n) is the nth Fibonacci number and Φ = (1 + 51/2) / 2 (the so-called Golden mean). Since fib(m + 1) = fib(m) + fib(m - 1) it is straightforward to prove that the statement holds for m + 1 if we can assume that it already holds for both m and m - 1. Also for this generalization it holds that it is in fact just a special case of the first form; let P(n) be the statement that we intend to prove then proving it with these rules is equivalent with proving the statement ' P(m) for all m ≤ n ' for all natural numbers n with the first two steps.
This is in fact the most general form of mathematical induction and it can be shown that it is not only valid for statements about natural numbers, but for statements about elements of any well-founded set, that is, a set with a partial order that contains no infinite descending chains (where < is defined such that a < b iff a ≤ b and a ≠ b).
This form of induction, when applied to ordinals (which form a well-ordered and hence well-founded class), is called transfinite induction. It is an important proof technique in set theory, topology and other fields.
Proofs by transfinite induction typically need to distinguish three cases:
Note that the additional axiom is indeed an alternative formulation of principle of mathematical induction. That is, the two are equivalent. See proof of mathematical induction.