Mathematical induction is a common method for proving
theorems about the positive integers, or just about any situation where
one case depends on previous cases. Here’s the basic idea, phrased
in terms of integers:
- You have a conjecture that you think is true for
every integer greater than 1. Show (by calculation or some other method)
that your conjecture definitely holds for 1.
- Show that whenever your conjecture holds for some
number, it must hold for the next number as well.
Why does this work? Well, you know your conjecture
holds for 1. You’ve shown that whenever it holds for some number,
it must hold for the next, so your conjecture also holds for 2. But
if it holds for 2, it must hold for the next number as well, so it holds
for 3. And so on, through all the integers greater than 1.
A few of things to notice:
The starting point (or base case) of 1 is common, but it’s
not the only choice. You might have a conjecture that holds for
every integer greater than 5, or for every integer greater than
1,001. The only thing that changes is the base case.
In practice, the way you do Step 2 is that you assume that for some number you call (n-1), your conjecture holds. (This is called the induction hypothesis.) Use that assumption to show that the conjecture
must hold for the number n as well. Some people complain that they are “assuming
what they want to prove.” That’s not the case, of course.
You are simply proving that if your claim is true for a particular case, then it is
true for the subsequent one. If no base case can be established,
then the link between cases is meaningless. The assumption is no
different than the one we make when we say “if a shape is
a parallelogram, then its adjacent angles are supplementary.”
We have not proven that any shapes are parallelograms. We are just
claiming that if a shape exists that matches our definition for
parallelogram, that it must also possess the stated property.
A slight variation on the induction hypothesis can be useful: assume
that for all integers k < n your conjecture holds. Then use that to prove
it holds for the number n as well. If your proof uses more than one previous step--for
example, it uses the fact that it holds at (n-1) and (n-2)--you’ll need to check more than one base case.
Mathematical induction and its variations are useful in proving
identities that are true for any value of integer, but they do not
help you see how someone figured out the identity at first place.
Mathematical induction is not only useful for proving algebraic
identities. It can be used any time you have a recursive relationship--one
where the current case depends on one or more of the previous cases.
See the second example below for a geometric application of induction. Here’s a more
general way to think about induction:
- You have a conjecture that you think is true for every case.
Show that your conjecture definitely holds for a base case.
- Show that whenever your conjecture holds for some case, it must
hold for the next case as well.
Here’s an example using integers. Someone discovered a formula that
seems to work for the sum of the integers:
You could check lots and lots of cases, but no matter
how long you worked, you could never check that the formula holds for
every integer greater than 1. With mathematical induction, you
can prove it does!
- Show that the conjecture holds for a base case.
Well, the sum on the left will just be 1. The formula on the right
gives = 1. So the formula holds
- Show that whenever your conjecture holds for some
number, it must hold for the next number as well. So, we’ll
start with the original formula and show that when it is true for
some n - 1, then the formula must also work for n:
Now, the equality will still hold if we add n to both sides of the equation:
And we do a little simplification on the right
side of the equation, trying to get it in the form of our original
which is exactly our original formula!
So the formula holds for 1. If it holds for 1, it
must hold for 2 (the next number). If it holds for 2, it must hold for
3 (the next number). And so on, and so on - by mathematical induction,
it holds for every integer greater than 1!
Here’s a geometric example: Someone noticed that
every polygon with n sides could be divided into n - 2 triangles. Again, you could never check every polygon,
but you don’t have to. Induction lets you prove it. Here’s
a sketch of the proof. Can you fill in the details?
- In this case, the base case will be quadrilaterals,
or polygons with four sides. Draw one diagonal, and you cut it into
2 triangles. (Make sure you believe you can do this for every quadrilateral--sometimes a diagonal will fall outside
of the figure. What then?)
- Now, assume that for every polygon with n - 1 sides, you can cut them into (n-1)-2 = n-2 triangles. Now think about a polygon with n sides.
- Find a diagonal that creates a triangle and
an (n - 1)-sided polygon. (Can you always do this?)
- You can divide the (n-1)-sided polygon into n-3 triangles. (Why?)
- So the total number of triangles is (n - 3) = 1 = n - 2.
You can try your hand at some mathematical induction
problems--some numeric and some not--at the Problems with a Point page on The
Principle of Mathematical Induction. You can find some harder problems
for which induction is useful at http://www.geocities.com/jespinos57/induction.htm.
More information and problems on Mathematical Induction
can be found at http://www.math.csusb.edu/notes/proofs/pfnot/node10.html
and in the articles "Teaching Mathematical Induction: An Alternative
Approach" and "When Memory Fails" in the September 2001
issue of Mathematics Teacher.