Another common variant is known as strong induction. This allows you to assume all previous cases in your induction step, rather than just the last one.
It looks like the following:
Definition: [Strong induction] Let be a statement that depends on a natural number . Then if
is true, and
for all , if are all true, then is also true,
then is true for all .
Again, this can be viewed as a cleverly disguised version of ordinary induction: if we define the statement as follows: then proving by ordinary induction works out to be effectively the same thing as proving by strong induction.
Indeed, the base case is the same thing as the base case . The induction step looks like
In order to prove this, we assume are all true and have to prove that are all true. But then all of these except the last are assumptions: what is left is to prove assuming , and that’s exactly the induction step of a strong induction.
Here’s an example of strong induction in practice. First we’ll need a definition:
Definition: The Fibonacci numbers are defined by taking and , and then for by
Now for the result in question:
For all we have .
Let be the statement . We know that is true, since We also know that is true, since
Now we’ll show that for all we have that if are all true, then is true. So suppose that are all true.
We now have that which is exactly . This completes the induction step, and so finishes the proof.
That strong induction argument really had two base cases before the induction step.
So we proved , and we proved by proving , and then we proved by proving .
I like to think that the proof was arranged according to the shape of the definition of the Fibonacci numbers: that definition has two base cases and , and a step . This is not a rare coincidence.
Why induction works
In this section we make a few comments on why induction works. They may be helpful in thinking about when you can and when you can’t generalise induction to other settings.
Let’s introduce a definition:
Definition: A set (of numbers) is well-ordered if every nonempty subset has a least element.
For now, our main use of that is to say this:
Definition: The well-ordering principle for says that is well-ordered.
This is a very special property of . The integers , for example, are not well-ordered. Indeed, the subset of all integers does not have a least element: there is no least integer.
The main interest is this:
We’ll show first that we can derive the principle of strong induction from the well-ordering principle.
So suppose we had a statement for each , and we had a base case (that was true) and an induction step (that, for all if we have for all , we also have ). We need to show that is true for all .
We might argue as follows. Let be the set of natural numbers for which does not hold: So is the set of “counterexamples”.
If has any elements at all, it has a smallest element . But can’t be , because we have . But can’t be bigger than either: because is minimal, we have for all . Hence we have also, by the induction step. But that’s a contradiction: we assumed that .
Hence doesn’t have any elements, which is the same as saying that holds for all .
Now we’ll show the other half of the equivalence: that we can derive the well-ordering principle from the principle of strong induction.
Let be the statement, “any subset of which contains has a smallest element”. We’ll prove for all by strong induction.
Firstly, for a base case, we must prove (“any subset of which contains has a smallest element”). This is clearly true, as is the smallest natural number of all, so any such subset has as its smallest element.
Now we must prove the induction step: we assume for some that is true for all , and we prove that is true. Consider a subset which contains . If it contains some element , then by it has a least element. If, however, it contains no element , then is its least element: so in particular, it has a least element. This proves .
Hence we have for all by strong induction. So if is a nonempty subset of , it has at least one element: call it . But then by , the set has a least element: this proves the well-ordering principle.